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Dual Nature of Radiation and Matter
Physics XII - Chapter 11: Complete NCERT Exercise Solutions
Master Dual Nature with NCERT solutions for photoelectric effect, matter waves, de Broglie wavelength, and particle-wave duality.
Question 11.1
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Answer & Explanation:
Step 1: Given data
Accelerating potential, \( V = 30 \text{ kV} = 30 \times 10^3 \text{ V} \)
Electron charge, \( e = 1.6 \times 10^{-19} \text{ C} \)
Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J s} \)
Speed of light, \( c = 3 \times 10^8 \text{ m/s} \)
Step 2: Maximum frequency
Energy of electron = \( eV \)
This energy is converted to photon energy: \( eV = h\nu_{\text{max}} \)
\( \nu_{\text{max}} = \frac{eV}{h} \)
\( = \frac{(1.6 \times 10^{-19}) \times (30 \times 10^3)}{6.626 \times 10^{-34}} \)
\( = \frac{4.8 \times 10^{-15}}{6.626 \times 10^{-34}} \)
\( = 7.24 \times 10^{18} \text{ Hz} \)
Step 3: Minimum wavelength
\( \lambda_{\text{min}} = \frac{hc}{eV} \)
\( = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{(1.6 \times 10^{-19}) \times (30 \times 10^3)} \)
\( = \frac{1.9878 \times 10^{-25}}{4.8 \times 10^{-15}} \)
\( = 4.14 \times 10^{-11} \text{ m} = 0.0414 \text{ nm} \)
Answers:
(a) Maximum frequency = \( 7.24 \times 10^{18} \text{ Hz} \)
(b) Minimum wavelength = \( 0.0414 \text{ nm} \)
Accelerating potential, \( V = 30 \text{ kV} = 30 \times 10^3 \text{ V} \)
Electron charge, \( e = 1.6 \times 10^{-19} \text{ C} \)
Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J s} \)
Speed of light, \( c = 3 \times 10^8 \text{ m/s} \)
Step 2: Maximum frequency
Energy of electron = \( eV \)
This energy is converted to photon energy: \( eV = h\nu_{\text{max}} \)
\( \nu_{\text{max}} = \frac{eV}{h} \)
\( = \frac{(1.6 \times 10^{-19}) \times (30 \times 10^3)}{6.626 \times 10^{-34}} \)
\( = \frac{4.8 \times 10^{-15}}{6.626 \times 10^{-34}} \)
\( = 7.24 \times 10^{18} \text{ Hz} \)
Step 3: Minimum wavelength
\( \lambda_{\text{min}} = \frac{hc}{eV} \)
\( = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{(1.6 \times 10^{-19}) \times (30 \times 10^3)} \)
\( = \frac{1.9878 \times 10^{-25}}{4.8 \times 10^{-15}} \)
\( = 4.14 \times 10^{-11} \text{ m} = 0.0414 \text{ nm} \)
Answers:
(a) Maximum frequency = \( 7.24 \times 10^{18} \text{ Hz} \)
(b) Minimum wavelength = \( 0.0414 \text{ nm} \)
Question 11.2
The work function of caesium metal is 2.14 eV. When light of frequency \(6 \times 10^{14}\) Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
Answer & Explanation:
Step 1: Given data
Work function, \( \phi = 2.14 \text{ eV} \)
Frequency of incident light, \( \nu = 6 \times 10^{14} \text{ Hz} \)
Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J s} = 4.14 \times 10^{-15} \text{ eV s} \)
Electron mass, \( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Electron charge, \( e = 1.6 \times 10^{-19} \text{ C} \)
Step 2: Energy of incident photon
\( E = h\nu = (4.14 \times 10^{-15}) \times (6 \times 10^{14}) \)
\( = 2.484 \text{ eV} \)
Step 3: Maximum kinetic energy
Using Einstein's photoelectric equation:
\( K_{\text{max}} = h\nu - \phi \)
\( = 2.484 - 2.14 = 0.344 \text{ eV} \)
Step 4: Stopping potential
\( K_{\text{max}} = eV_0 \)
\( V_0 = \frac{K_{\text{max}}}{e} = 0.344 \text{ V} \)
Step 5: Maximum speed
\( K_{\text{max}} = \frac{1}{2} m_e v_{\text{max}}^2 \)
Convert \( K_{\text{max}} \) to joules: \( 0.344 \text{ eV} = 0.344 \times 1.6 \times 10^{-19} = 5.504 \times 10^{-20} \text{ J} \)
\( v_{\text{max}} = \sqrt{\frac{2K_{\text{max}}}{m_e}} \)
\( = \sqrt{\frac{2 \times 5.504 \times 10^{-20}}{9.11 \times 10^{-31}}} \)
\( = \sqrt{1.208 \times 10^{11}} \)
\( = 3.476 \times 10^5 \text{ m/s} \)
Answers:
(a) Maximum kinetic energy = 0.344 eV
(b) Stopping potential = 0.344 V
(c) Maximum speed = \( 3.48 \times 10^5 \text{ m/s} \)
Work function, \( \phi = 2.14 \text{ eV} \)
Frequency of incident light, \( \nu = 6 \times 10^{14} \text{ Hz} \)
Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J s} = 4.14 \times 10^{-15} \text{ eV s} \)
Electron mass, \( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Electron charge, \( e = 1.6 \times 10^{-19} \text{ C} \)
Step 2: Energy of incident photon
\( E = h\nu = (4.14 \times 10^{-15}) \times (6 \times 10^{14}) \)
\( = 2.484 \text{ eV} \)
Step 3: Maximum kinetic energy
Using Einstein's photoelectric equation:
\( K_{\text{max}} = h\nu - \phi \)
\( = 2.484 - 2.14 = 0.344 \text{ eV} \)
Step 4: Stopping potential
\( K_{\text{max}} = eV_0 \)
\( V_0 = \frac{K_{\text{max}}}{e} = 0.344 \text{ V} \)
Step 5: Maximum speed
\( K_{\text{max}} = \frac{1}{2} m_e v_{\text{max}}^2 \)
Convert \( K_{\text{max}} \) to joules: \( 0.344 \text{ eV} = 0.344 \times 1.6 \times 10^{-19} = 5.504 \times 10^{-20} \text{ J} \)
\( v_{\text{max}} = \sqrt{\frac{2K_{\text{max}}}{m_e}} \)
\( = \sqrt{\frac{2 \times 5.504 \times 10^{-20}}{9.11 \times 10^{-31}}} \)
\( = \sqrt{1.208 \times 10^{11}} \)
\( = 3.476 \times 10^5 \text{ m/s} \)
Answers:
(a) Maximum kinetic energy = 0.344 eV
(b) Stopping potential = 0.344 V
(c) Maximum speed = \( 3.48 \times 10^5 \text{ m/s} \)
Question 11.3
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Answer & Explanation:
Step 1: Relationship between stopping potential and kinetic energy
The stopping potential \( V_0 \) is related to the maximum kinetic energy \( K_{\text{max}} \) by:
\( K_{\text{max}} = eV_0 \)
Step 2: Calculation
Given: \( V_0 = 1.5 \text{ V} \)
\( K_{\text{max}} = 1.5 \text{ eV} \) (since \( e \) in eV/V cancels)
Step 3: In joules
\( K_{\text{max}} = 1.5 \times 1.6 \times 10^{-19} = 2.4 \times 10^{-19} \text{ J} \)
Answer: Maximum kinetic energy = 1.5 eV or \( 2.4 \times 10^{-19} \text{ J} \)
The stopping potential \( V_0 \) is related to the maximum kinetic energy \( K_{\text{max}} \) by:
\( K_{\text{max}} = eV_0 \)
Step 2: Calculation
Given: \( V_0 = 1.5 \text{ V} \)
\( K_{\text{max}} = 1.5 \text{ eV} \) (since \( e \) in eV/V cancels)
Step 3: In joules
\( K_{\text{max}} = 1.5 \times 1.6 \times 10^{-19} = 2.4 \times 10^{-19} \text{ J} \)
Answer: Maximum kinetic energy = 1.5 eV or \( 2.4 \times 10^{-19} \text{ J} \)
Question 11.4
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Answer & Explanation:
Step 1: Given data
Wavelength, \( \lambda = 632.8 \text{ nm} = 6.328 \times 10^{-7} \text{ m} \)
Power, \( P = 9.42 \text{ mW} = 9.42 \times 10^{-3} \text{ W} \)
Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J s} \)
Speed of light, \( c = 3 \times 10^8 \text{ m/s} \)
Mass of hydrogen atom, \( m_H = 1.67 \times 10^{-27} \text{ kg} \)
Step 2: Energy of each photon
\( E = \frac{hc}{\lambda} \)
\( = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{6.328 \times 10^{-7}} \)
\( = \frac{1.9878 \times 10^{-25}}{6.328 \times 10^{-7}} \)
\( = 3.141 \times 10^{-19} \text{ J} \)
Step 3: Momentum of each photon
\( p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{6.328 \times 10^{-7}} \)
\( = 1.047 \times 10^{-27} \text{ kg m/s} \)
Step 4: Number of photons per second
Power = Energy per second = \( N \times E \)
\( N = \frac{P}{E} = \frac{9.42 \times 10^{-3}}{3.141 \times 10^{-19}} \)
\( = 3.0 \times 10^{16} \text{ photons/s} \)
Step 5: Speed of hydrogen atom with same momentum
\( p = m_H v \)
\( v = \frac{p}{m_H} = \frac{1.047 \times 10^{-27}}{1.67 \times 10^{-27}} \)
\( = 0.627 \text{ m/s} \)
Answers:
(a) Energy per photon = \( 3.14 \times 10^{-19} \text{ J} \), Momentum = \( 1.05 \times 10^{-27} \text{ kg m/s} \)
(b) Photons per second = \( 3.0 \times 10^{16} \)
(c) Speed of hydrogen atom = 0.627 m/s
Wavelength, \( \lambda = 632.8 \text{ nm} = 6.328 \times 10^{-7} \text{ m} \)
Power, \( P = 9.42 \text{ mW} = 9.42 \times 10^{-3} \text{ W} \)
Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J s} \)
Speed of light, \( c = 3 \times 10^8 \text{ m/s} \)
Mass of hydrogen atom, \( m_H = 1.67 \times 10^{-27} \text{ kg} \)
Step 2: Energy of each photon
\( E = \frac{hc}{\lambda} \)
\( = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{6.328 \times 10^{-7}} \)
\( = \frac{1.9878 \times 10^{-25}}{6.328 \times 10^{-7}} \)
\( = 3.141 \times 10^{-19} \text{ J} \)
Step 3: Momentum of each photon
\( p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{6.328 \times 10^{-7}} \)
\( = 1.047 \times 10^{-27} \text{ kg m/s} \)
Step 4: Number of photons per second
Power = Energy per second = \( N \times E \)
\( N = \frac{P}{E} = \frac{9.42 \times 10^{-3}}{3.141 \times 10^{-19}} \)
\( = 3.0 \times 10^{16} \text{ photons/s} \)
Step 5: Speed of hydrogen atom with same momentum
\( p = m_H v \)
\( v = \frac{p}{m_H} = \frac{1.047 \times 10^{-27}}{1.67 \times 10^{-27}} \)
\( = 0.627 \text{ m/s} \)
Answers:
(a) Energy per photon = \( 3.14 \times 10^{-19} \text{ J} \), Momentum = \( 1.05 \times 10^{-27} \text{ kg m/s} \)
(b) Photons per second = \( 3.0 \times 10^{16} \)
(c) Speed of hydrogen atom = 0.627 m/s
Question 11.5
The energy flux of sunlight reaching the surface of the earth is \(1.388 \times 10^3\) W/m\(^2\). How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer & Explanation:
Step 1: Given data
Energy flux, \( I = 1.388 \times 10^3 \text{ W/m}^2 \)
Average wavelength, \( \lambda = 550 \text{ nm} = 5.5 \times 10^{-7} \text{ m} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3 \times 10^8 \text{ m/s} \)
Step 2: Energy of one photon
\( E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{5.5 \times 10^{-7}} \)
\( = \frac{1.9878 \times 10^{-25}}{5.5 \times 10^{-7}} \)
\( = 3.614 \times 10^{-19} \text{ J} \)
Step 3: Number of photons per m² per second
\( N = \frac{I}{E} = \frac{1.388 \times 10^3}{3.614 \times 10^{-19}} \)
\( = 3.84 \times 10^{21} \text{ photons/(m²·s)} \)
Answer: Approximately \( 3.84 \times 10^{21} \) photons per square metre per second
Energy flux, \( I = 1.388 \times 10^3 \text{ W/m}^2 \)
Average wavelength, \( \lambda = 550 \text{ nm} = 5.5 \times 10^{-7} \text{ m} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3 \times 10^8 \text{ m/s} \)
Step 2: Energy of one photon
\( E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{5.5 \times 10^{-7}} \)
\( = \frac{1.9878 \times 10^{-25}}{5.5 \times 10^{-7}} \)
\( = 3.614 \times 10^{-19} \text{ J} \)
Step 3: Number of photons per m² per second
\( N = \frac{I}{E} = \frac{1.388 \times 10^3}{3.614 \times 10^{-19}} \)
\( = 3.84 \times 10^{21} \text{ photons/(m²·s)} \)
Answer: Approximately \( 3.84 \times 10^{21} \) photons per square metre per second
Question 11.6
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be \(4.12 \times 10^{-15}\) V s. Calculate the value of Planck's constant.
Answer & Explanation:
Step 1: Einstein's photoelectric equation
\( eV_0 = h\nu - \phi \)
This is of the form \( y = mx + c \)
\( V_0 = \left( \frac{h}{e} \right) \nu - \frac{\phi}{e} \)
Slope \( m = \frac{h}{e} \)
Step 2: Given slope
\( m = 4.12 \times 10^{-15} \text{ V s} \)
\( e = 1.6 \times 10^{-19} \text{ C} \)
Step 3: Calculate Planck's constant
\( h = m \times e \)
\( = (4.12 \times 10^{-15}) \times (1.6 \times 10^{-19}) \)
\( = 6.592 \times 10^{-34} \text{ J s} \)
Answer: Planck's constant \( h \approx \) \( 6.59 \times 10^{-34} \text{ J s} \)
\( eV_0 = h\nu - \phi \)
This is of the form \( y = mx + c \)
\( V_0 = \left( \frac{h}{e} \right) \nu - \frac{\phi}{e} \)
Slope \( m = \frac{h}{e} \)
Step 2: Given slope
\( m = 4.12 \times 10^{-15} \text{ V s} \)
\( e = 1.6 \times 10^{-19} \text{ C} \)
Step 3: Calculate Planck's constant
\( h = m \times e \)
\( = (4.12 \times 10^{-15}) \times (1.6 \times 10^{-19}) \)
\( = 6.592 \times 10^{-34} \text{ J s} \)
Answer: Planck's constant \( h \approx \) \( 6.59 \times 10^{-34} \text{ J s} \)
Question 11.7
A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm.
(a) What is the energy per photon associated with the sodium light?
(b) At what rate are the photons delivered to the sphere?
(a) What is the energy per photon associated with the sodium light?
(b) At what rate are the photons delivered to the sphere?
Answer & Explanation:
Step 1: Given data
Power of lamp, \( P = 100 \text{ W} \)
Wavelength, \( \lambda = 589 \text{ nm} = 5.89 \times 10^{-7} \text{ m} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3 \times 10^8 \text{ m/s} \)
Step 2: Energy per photon
\( E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{5.89 \times 10^{-7}} \)
\( = \frac{1.9878 \times 10^{-25}}{5.89 \times 10^{-7}} \)
\( = 3.375 \times 10^{-19} \text{ J} \)
Step 3: Rate of photon delivery
Since the lamp radiates uniformly in all directions, all photons reach the sphere.
Number of photons per second, \( N = \frac{P}{E} \)
\( = \frac{100}{3.375 \times 10^{-19}} \)
\( = 2.96 \times 10^{20} \text{ photons/s} \)
Answers:
(a) Energy per photon = \( 3.375 \times 10^{-19} \text{ J} \)
(b) Photon delivery rate = \( 2.96 \times 10^{20} \) photons per second
Power of lamp, \( P = 100 \text{ W} \)
Wavelength, \( \lambda = 589 \text{ nm} = 5.89 \times 10^{-7} \text{ m} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3 \times 10^8 \text{ m/s} \)
Step 2: Energy per photon
\( E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{5.89 \times 10^{-7}} \)
\( = \frac{1.9878 \times 10^{-25}}{5.89 \times 10^{-7}} \)
\( = 3.375 \times 10^{-19} \text{ J} \)
Step 3: Rate of photon delivery
Since the lamp radiates uniformly in all directions, all photons reach the sphere.
Number of photons per second, \( N = \frac{P}{E} \)
\( = \frac{100}{3.375 \times 10^{-19}} \)
\( = 2.96 \times 10^{20} \text{ photons/s} \)
Answers:
(a) Energy per photon = \( 3.375 \times 10^{-19} \text{ J} \)
(b) Photon delivery rate = \( 2.96 \times 10^{20} \) photons per second
Question 11.8
The threshold frequency for a certain metal is \(3.3 \times 10^{14}\) Hz. If light of frequency \(8.2 \times 10^{14}\) Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer & Explanation:
Step 1: Given data
Threshold frequency, \( \nu_0 = 3.3 \times 10^{14} \text{ Hz} \)
Incident frequency, \( \nu = 8.2 \times 10^{14} \text{ Hz} \)
\( h = 4.14 \times 10^{-15} \text{ eV s} \) (using eV for convenience)
Step 2: Maximum kinetic energy
\( K_{\text{max}} = h(\nu - \nu_0) \)
\( = (4.14 \times 10^{-15}) \times (8.2 \times 10^{14} - 3.3 \times 10^{14}) \)
\( = (4.14 \times 10^{-15}) \times (4.9 \times 10^{14}) \)
\( = 2.0286 \text{ eV} \)
Step 3: Cut-off voltage
\( K_{\text{max}} = eV_0 \)
\( V_0 = \frac{K_{\text{max}}}{e} = 2.0286 \text{ V} \)
Answer: Cut-off voltage ≈ 2.03 V
Threshold frequency, \( \nu_0 = 3.3 \times 10^{14} \text{ Hz} \)
Incident frequency, \( \nu = 8.2 \times 10^{14} \text{ Hz} \)
\( h = 4.14 \times 10^{-15} \text{ eV s} \) (using eV for convenience)
Step 2: Maximum kinetic energy
\( K_{\text{max}} = h(\nu - \nu_0) \)
\( = (4.14 \times 10^{-15}) \times (8.2 \times 10^{14} - 3.3 \times 10^{14}) \)
\( = (4.14 \times 10^{-15}) \times (4.9 \times 10^{14}) \)
\( = 2.0286 \text{ eV} \)
Step 3: Cut-off voltage
\( K_{\text{max}} = eV_0 \)
\( V_0 = \frac{K_{\text{max}}}{e} = 2.0286 \text{ V} \)
Answer: Cut-off voltage ≈ 2.03 V
Question 11.9
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Answer & Explanation:
Step 1: Given data
Work function, \( \phi = 4.2 \text{ eV} \)
Wavelength, \( \lambda = 330 \text{ nm} = 3.30 \times 10^{-7} \text{ m} \)
\( h = 4.14 \times 10^{-15} \text{ eV s} \), \( c = 3 \times 10^8 \text{ m/s} \)
Step 2: Energy of incident photon
\( E = \frac{hc}{\lambda} = \frac{(4.14 \times 10^{-15}) \times (3 \times 10^8)}{3.30 \times 10^{-7}} \)
\( = \frac{1.242 \times 10^{-6}}{3.30 \times 10^{-7}} \)
\( = 3.764 \text{ eV} \)
Step 3: Check condition for photoelectric emission
For emission: \( E > \phi \)
Here: \( 3.764 \text{ eV} < 4.2 \text{ eV} \)
Photon energy is less than the work function.
Answer: No, photoelectric emission will not occur because the photon energy (3.76 eV) is less than the work function (4.2 eV).
Work function, \( \phi = 4.2 \text{ eV} \)
Wavelength, \( \lambda = 330 \text{ nm} = 3.30 \times 10^{-7} \text{ m} \)
\( h = 4.14 \times 10^{-15} \text{ eV s} \), \( c = 3 \times 10^8 \text{ m/s} \)
Step 2: Energy of incident photon
\( E = \frac{hc}{\lambda} = \frac{(4.14 \times 10^{-15}) \times (3 \times 10^8)}{3.30 \times 10^{-7}} \)
\( = \frac{1.242 \times 10^{-6}}{3.30 \times 10^{-7}} \)
\( = 3.764 \text{ eV} \)
Step 3: Check condition for photoelectric emission
For emission: \( E > \phi \)
Here: \( 3.764 \text{ eV} < 4.2 \text{ eV} \)
Photon energy is less than the work function.
Answer: No, photoelectric emission will not occur because the photon energy (3.76 eV) is less than the work function (4.2 eV).
Question 11.10
Light of frequency \(7.21 \times 10^{14}\) Hz is incident on a metal surface. Electrons with a maximum speed of \(6.0 \times 10^5\) m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer & Explanation:
Step 1: Given data
Incident frequency, \( \nu = 7.21 \times 10^{14} \text{ Hz} \)
Maximum speed of electrons, \( v_{\text{max}} = 6.0 \times 10^5 \text{ m/s} \)
Electron mass, \( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J s} \)
Step 2: Maximum kinetic energy in joules
\( K_{\text{max}} = \frac{1}{2} m_e v_{\text{max}}^2 \)
\( = \frac{1}{2} \times (9.11 \times 10^{-31}) \times (6.0 \times 10^5)^2 \)
\( = 0.5 \times 9.11 \times 10^{-31} \times 3.6 \times 10^{11} \)
\( = 1.6398 \times 10^{-19} \text{ J} \)
Step 3: Convert \( K_{\text{max}} \) to eV
\( K_{\text{max}} (\text{eV}) = \frac{1.6398 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.0249 \text{ eV} \)
Step 4: Using Einstein's equation
\( h\nu = \phi + K_{\text{max}} \)
\( h\nu_0 = \phi \) (threshold)
So: \( h\nu_0 = h\nu - K_{\text{max}} \)
Step 5: Calculate threshold frequency
\( \nu_0 = \nu - \frac{K_{\text{max}}}{h} \)
First find \( \frac{K_{\text{max}}}{h} \):
\( = \frac{1.6398 \times 10^{-19}}{6.626 \times 10^{-34}} = 2.475 \times 10^{14} \text{ Hz} \)
\( \nu_0 = 7.21 \times 10^{14} - 2.475 \times 10^{14} \)
\( = 4.735 \times 10^{14} \text{ Hz} \)
Answer: Threshold frequency = \( 4.74 \times 10^{14} \text{ Hz} \)
Incident frequency, \( \nu = 7.21 \times 10^{14} \text{ Hz} \)
Maximum speed of electrons, \( v_{\text{max}} = 6.0 \times 10^5 \text{ m/s} \)
Electron mass, \( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J s} \)
Step 2: Maximum kinetic energy in joules
\( K_{\text{max}} = \frac{1}{2} m_e v_{\text{max}}^2 \)
\( = \frac{1}{2} \times (9.11 \times 10^{-31}) \times (6.0 \times 10^5)^2 \)
\( = 0.5 \times 9.11 \times 10^{-31} \times 3.6 \times 10^{11} \)
\( = 1.6398 \times 10^{-19} \text{ J} \)
Step 3: Convert \( K_{\text{max}} \) to eV
\( K_{\text{max}} (\text{eV}) = \frac{1.6398 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.0249 \text{ eV} \)
Step 4: Using Einstein's equation
\( h\nu = \phi + K_{\text{max}} \)
\( h\nu_0 = \phi \) (threshold)
So: \( h\nu_0 = h\nu - K_{\text{max}} \)
Step 5: Calculate threshold frequency
\( \nu_0 = \nu - \frac{K_{\text{max}}}{h} \)
First find \( \frac{K_{\text{max}}}{h} \):
\( = \frac{1.6398 \times 10^{-19}}{6.626 \times 10^{-34}} = 2.475 \times 10^{14} \text{ Hz} \)
\( \nu_0 = 7.21 \times 10^{14} - 2.475 \times 10^{14} \)
\( = 4.735 \times 10^{14} \text{ Hz} \)
Answer: Threshold frequency = \( 4.74 \times 10^{14} \text{ Hz} \)
Question 11.11
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Answer & Explanation:
Step 1: Given data
Wavelength, \( \lambda = 488 \text{ nm} = 4.88 \times 10^{-7} \text{ m} \)
Stopping potential, \( V_0 = 0.38 \text{ V} \)
\( h = 6.626 \times 10^{-34} \text{ J s} = 4.14 \times 10^{-15} \text{ eV s} \)
\( c = 3 \times 10^8 \text{ m/s} \), \( e = 1.6 \times 10^{-19} \text{ C} \)
Step 2: Energy of incident photon
\( E = \frac{hc}{\lambda} = \frac{(4.14 \times 10^{-15}) \times (3 \times 10^8)}{4.88 \times 10^{-7}} \)
\( = \frac{1.242 \times 10^{-6}}{4.88 \times 10^{-7}} \)
\( = 2.545 \text{ eV} \)
Step 3: Maximum kinetic energy
\( K_{\text{max}} = eV_0 = 0.38 \text{ eV} \)
Step 4: Work function
Using Einstein's equation: \( h\nu = \phi + K_{\text{max}} \)
\( \phi = h\nu - K_{\text{max}} = 2.545 - 0.38 = 2.165 \text{ eV} \)
Answer: Work function = 2.165 eV
Wavelength, \( \lambda = 488 \text{ nm} = 4.88 \times 10^{-7} \text{ m} \)
Stopping potential, \( V_0 = 0.38 \text{ V} \)
\( h = 6.626 \times 10^{-34} \text{ J s} = 4.14 \times 10^{-15} \text{ eV s} \)
\( c = 3 \times 10^8 \text{ m/s} \), \( e = 1.6 \times 10^{-19} \text{ C} \)
Step 2: Energy of incident photon
\( E = \frac{hc}{\lambda} = \frac{(4.14 \times 10^{-15}) \times (3 \times 10^8)}{4.88 \times 10^{-7}} \)
\( = \frac{1.242 \times 10^{-6}}{4.88 \times 10^{-7}} \)
\( = 2.545 \text{ eV} \)
Step 3: Maximum kinetic energy
\( K_{\text{max}} = eV_0 = 0.38 \text{ eV} \)
Step 4: Work function
Using Einstein's equation: \( h\nu = \phi + K_{\text{max}} \)
\( \phi = h\nu - K_{\text{max}} = 2.545 - 0.38 = 2.165 \text{ eV} \)
Answer: Work function = 2.165 eV
Question 11.12
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer & Explanation:
Step 1: Given data
Potential difference, \( V = 56 \text{ V} \)
Electron charge, \( e = 1.6 \times 10^{-19} \text{ C} \)
Electron mass, \( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J s} \)
Step 2: Kinetic energy
\( K = eV = 56 \text{ eV} = 56 \times 1.6 \times 10^{-19} = 8.96 \times 10^{-18} \text{ J} \)
Step 3: Momentum
\( p = \sqrt{2m_e K} \)
\( = \sqrt{2 \times (9.11 \times 10^{-31}) \times (8.96 \times 10^{-18})} \)
\( = \sqrt{1.632 \times 10^{-47}} \)
\( = 4.04 \times 10^{-24} \text{ kg m/s} \)
Step 4: de Broglie wavelength
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{4.04 \times 10^{-24}} \)
\( = 1.64 \times 10^{-10} \text{ m} = 0.164 \text{ nm} \)
Answers:
(a) Momentum = \( 4.04 \times 10^{-24} \text{ kg m/s} \)
(b) de Broglie wavelength = 0.164 nm
Potential difference, \( V = 56 \text{ V} \)
Electron charge, \( e = 1.6 \times 10^{-19} \text{ C} \)
Electron mass, \( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J s} \)
Step 2: Kinetic energy
\( K = eV = 56 \text{ eV} = 56 \times 1.6 \times 10^{-19} = 8.96 \times 10^{-18} \text{ J} \)
Step 3: Momentum
\( p = \sqrt{2m_e K} \)
\( = \sqrt{2 \times (9.11 \times 10^{-31}) \times (8.96 \times 10^{-18})} \)
\( = \sqrt{1.632 \times 10^{-47}} \)
\( = 4.04 \times 10^{-24} \text{ kg m/s} \)
Step 4: de Broglie wavelength
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{4.04 \times 10^{-24}} \)
\( = 1.64 \times 10^{-10} \text{ m} = 0.164 \text{ nm} \)
Answers:
(a) Momentum = \( 4.04 \times 10^{-24} \text{ kg m/s} \)
(b) de Broglie wavelength = 0.164 nm
Question 11.13
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Answer & Explanation:
Step 1: Given data
Kinetic energy, \( K = 120 \text{ eV} = 120 \times 1.6 \times 10^{-19} = 1.92 \times 10^{-17} \text{ J} \)
\( m_e = 9.11 \times 10^{-31} \text{ kg} \), \( h = 6.626 \times 10^{-34} \text{ J s} \)
Step 2: Momentum
\( p = \sqrt{2m_e K} \)
\( = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.92 \times 10^{-17}} \)
\( = \sqrt{3.498 \times 10^{-47}} \)
\( = 5.915 \times 10^{-24} \text{ kg m/s} \)
Step 3: Speed
\( v = \frac{p}{m_e} = \frac{5.915 \times 10^{-24}}{9.11 \times 10^{-31}} \)
\( = 6.49 \times 10^6 \text{ m/s} \)
Step 4: de Broglie wavelength
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{5.915 \times 10^{-24}} \)
\( = 1.12 \times 10^{-10} \text{ m} = 0.112 \text{ nm} \)
Answers:
(a) Momentum = \( 5.92 \times 10^{-24} \text{ kg m/s} \)
(b) Speed = \( 6.49 \times 10^6 \text{ m/s} \)
(c) de Broglie wavelength = 0.112 nm
Kinetic energy, \( K = 120 \text{ eV} = 120 \times 1.6 \times 10^{-19} = 1.92 \times 10^{-17} \text{ J} \)
\( m_e = 9.11 \times 10^{-31} \text{ kg} \), \( h = 6.626 \times 10^{-34} \text{ J s} \)
Step 2: Momentum
\( p = \sqrt{2m_e K} \)
\( = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.92 \times 10^{-17}} \)
\( = \sqrt{3.498 \times 10^{-47}} \)
\( = 5.915 \times 10^{-24} \text{ kg m/s} \)
Step 3: Speed
\( v = \frac{p}{m_e} = \frac{5.915 \times 10^{-24}}{9.11 \times 10^{-31}} \)
\( = 6.49 \times 10^6 \text{ m/s} \)
Step 4: de Broglie wavelength
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{5.915 \times 10^{-24}} \)
\( = 1.12 \times 10^{-10} \text{ m} = 0.112 \text{ nm} \)
Answers:
(a) Momentum = \( 5.92 \times 10^{-24} \text{ kg m/s} \)
(b) Speed = \( 6.49 \times 10^6 \text{ m/s} \)
(c) de Broglie wavelength = 0.112 nm
Question 11.14
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.
Answer & Explanation:
Step 1: Given data
Wavelength, \( \lambda = 589 \text{ nm} = 5.89 \times 10^{-7} \text{ m} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \)
Mass of electron, \( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Mass of neutron, \( m_n = 1.675 \times 10^{-27} \text{ kg} \)
Step 2: General formula
de Broglie wavelength: \( \lambda = \frac{h}{p} \)
Momentum: \( p = \frac{h}{\lambda} \)
Kinetic energy: \( K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2} \)
Step 3: For electron
\( K_e = \frac{h^2}{2m_e \lambda^2} \)
\( = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (5.89 \times 10^{-7})^2} \)
\( = \frac{4.390 \times 10^{-67}}{2 \times 9.11 \times 10^{-31} \times 3.469 \times 10^{-13}} \)
\( = \frac{4.390 \times 10^{-67}}{6.317 \times 10^{-43}} \)
\( = 6.95 \times 10^{-25} \text{ J} \)
In eV: \( K_e = \frac{6.95 \times 10^{-25}}{1.6 \times 10^{-19}} = 4.34 \times 10^{-6} \text{ eV} \)
Step 4: For neutron
\( K_n = \frac{h^2}{2m_n \lambda^2} \)
Since \( m_n \approx 1839 m_e \), \( K_n = \frac{K_e}{1839} \)
\( = \frac{6.95 \times 10^{-25}}{1839} = 3.78 \times 10^{-28} \text{ J} \)
In eV: \( K_n = 2.36 \times 10^{-9} \text{ eV} \)
Answers:
(a) Electron kinetic energy = \( 4.34 \times 10^{-6} \text{ eV} \)
(b) Neutron kinetic energy = \( 2.36 \times 10^{-9} \text{ eV} \)
Wavelength, \( \lambda = 589 \text{ nm} = 5.89 \times 10^{-7} \text{ m} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \)
Mass of electron, \( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Mass of neutron, \( m_n = 1.675 \times 10^{-27} \text{ kg} \)
Step 2: General formula
de Broglie wavelength: \( \lambda = \frac{h}{p} \)
Momentum: \( p = \frac{h}{\lambda} \)
Kinetic energy: \( K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2} \)
Step 3: For electron
\( K_e = \frac{h^2}{2m_e \lambda^2} \)
\( = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (5.89 \times 10^{-7})^2} \)
\( = \frac{4.390 \times 10^{-67}}{2 \times 9.11 \times 10^{-31} \times 3.469 \times 10^{-13}} \)
\( = \frac{4.390 \times 10^{-67}}{6.317 \times 10^{-43}} \)
\( = 6.95 \times 10^{-25} \text{ J} \)
In eV: \( K_e = \frac{6.95 \times 10^{-25}}{1.6 \times 10^{-19}} = 4.34 \times 10^{-6} \text{ eV} \)
Step 4: For neutron
\( K_n = \frac{h^2}{2m_n \lambda^2} \)
Since \( m_n \approx 1839 m_e \), \( K_n = \frac{K_e}{1839} \)
\( = \frac{6.95 \times 10^{-25}}{1839} = 3.78 \times 10^{-28} \text{ J} \)
In eV: \( K_n = 2.36 \times 10^{-9} \text{ eV} \)
Answers:
(a) Electron kinetic energy = \( 4.34 \times 10^{-6} \text{ eV} \)
(b) Neutron kinetic energy = \( 2.36 \times 10^{-9} \text{ eV} \)
Question 11.15
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass \(1.0 \times 10^{-9}\) kg drifting with a speed of 2.2 m/s?
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(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass \(1.0 \times 10^{-9}\) kg drifting with a speed of 2.2 m/s?
Answer & Explanation:
Step 1: Formula
de Broglie wavelength: \( \lambda = \frac{h}{p} = \frac{h}{mv} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \)
Step 2: (a) Bullet
\( m = 0.040 \text{ kg} \), \( v = 1.0 \text{ km/s} = 1000 \text{ m/s} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{0.040 \times 1000} \)
\( = \frac{6.626 \times 10^{-34}}{40} = 1.6565 \times 10^{-35} \text{ m} \)
Step 3: (b) Ball
\( m = 0.060 \text{ kg} \), \( v = 1.0 \text{ m/s} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{0.060 \times 1.0} \)
\( = 1.104 \times 10^{-32} \text{ m} \)
Step 4: (c) Dust particle
\( m = 1.0 \times 10^{-9} \text{ kg} \), \( v = 2.2 \text{ m/s} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{(1.0 \times 10^{-9}) \times 2.2} \)
\( = \frac{6.626 \times 10^{-34}}{2.2 \times 10^{-9}} \)
\( = 3.012 \times 10^{-25} \text{ m} \)
Answers:
(a) Bullet: \( \lambda = \) \( 1.66 \times 10^{-35} \text{ m} \)
(b) Ball: \( \lambda = \) \( 1.10 \times 10^{-32} \text{ m} \)
(c) Dust particle: \( \lambda = \) \( 3.01 \times 10^{-25} \text{ m} \)
de Broglie wavelength: \( \lambda = \frac{h}{p} = \frac{h}{mv} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \)
Step 2: (a) Bullet
\( m = 0.040 \text{ kg} \), \( v = 1.0 \text{ km/s} = 1000 \text{ m/s} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{0.040 \times 1000} \)
\( = \frac{6.626 \times 10^{-34}}{40} = 1.6565 \times 10^{-35} \text{ m} \)
Step 3: (b) Ball
\( m = 0.060 \text{ kg} \), \( v = 1.0 \text{ m/s} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{0.060 \times 1.0} \)
\( = 1.104 \times 10^{-32} \text{ m} \)
Step 4: (c) Dust particle
\( m = 1.0 \times 10^{-9} \text{ kg} \), \( v = 2.2 \text{ m/s} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{(1.0 \times 10^{-9}) \times 2.2} \)
\( = \frac{6.626 \times 10^{-34}}{2.2 \times 10^{-9}} \)
\( = 3.012 \times 10^{-25} \text{ m} \)
Answers:
(a) Bullet: \( \lambda = \) \( 1.66 \times 10^{-35} \text{ m} \)
(b) Ball: \( \lambda = \) \( 1.10 \times 10^{-32} \text{ m} \)
(c) Dust particle: \( \lambda = \) \( 3.01 \times 10^{-25} \text{ m} \)
Question 11.16
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Answer & Explanation:
Step 1: Given data
Wavelength, \( \lambda = 1.00 \text{ nm} = 1.00 \times 10^{-9} \text{ m} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3 \times 10^8 \text{ m/s} \)
Electron mass, \( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Step 2: (a) Momenta
For both electron and photon: \( p = \frac{h}{\lambda} \)
\( p = \frac{6.626 \times 10^{-34}}{1.00 \times 10^{-9}} = 6.626 \times 10^{-25} \text{ kg m/s} \)
Both have same momentum.
Step 3: (b) Energy of photon
\( E_{\text{photon}} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{1.00 \times 10^{-9}} \)
\( = 1.9878 \times 10^{-16} \text{ J} \)
In eV: \( E = \frac{1.9878 \times 10^{-16}}{1.6 \times 10^{-19}} = 1242 \text{ eV} = 1.242 \text{ keV} \)
Step 4: (c) Kinetic energy of electron
\( K_e = \frac{p^2}{2m_e} = \frac{(6.626 \times 10^{-25})^2}{2 \times 9.11 \times 10^{-31}} \)
\( = \frac{4.390 \times 10^{-49}}{1.822 \times 10^{-30}} \)
\( = 2.409 \times 10^{-19} \text{ J} \)
In eV: \( K_e = \frac{2.409 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.506 \text{ eV} \)
Answers:
(a) Momentum = \( 6.63 \times 10^{-25} \text{ kg m/s} \) (same for both)
(b) Photon energy = 1.242 keV
(c) Electron kinetic energy = 1.506 eV
Wavelength, \( \lambda = 1.00 \text{ nm} = 1.00 \times 10^{-9} \text{ m} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3 \times 10^8 \text{ m/s} \)
Electron mass, \( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Step 2: (a) Momenta
For both electron and photon: \( p = \frac{h}{\lambda} \)
\( p = \frac{6.626 \times 10^{-34}}{1.00 \times 10^{-9}} = 6.626 \times 10^{-25} \text{ kg m/s} \)
Both have same momentum.
Step 3: (b) Energy of photon
\( E_{\text{photon}} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{1.00 \times 10^{-9}} \)
\( = 1.9878 \times 10^{-16} \text{ J} \)
In eV: \( E = \frac{1.9878 \times 10^{-16}}{1.6 \times 10^{-19}} = 1242 \text{ eV} = 1.242 \text{ keV} \)
Step 4: (c) Kinetic energy of electron
\( K_e = \frac{p^2}{2m_e} = \frac{(6.626 \times 10^{-25})^2}{2 \times 9.11 \times 10^{-31}} \)
\( = \frac{4.390 \times 10^{-49}}{1.822 \times 10^{-30}} \)
\( = 2.409 \times 10^{-19} \text{ J} \)
In eV: \( K_e = \frac{2.409 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.506 \text{ eV} \)
Answers:
(a) Momentum = \( 6.63 \times 10^{-25} \text{ kg m/s} \) (same for both)
(b) Photon energy = 1.242 keV
(c) Electron kinetic energy = 1.506 eV
Question 11.17
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be \(1.40 \times 10^{-10}\) m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of \((3/2) kT\) at 300 K.
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of \((3/2) kT\) at 300 K.
Answer & Explanation:
Step 1: Given data
\( \lambda = 1.40 \times 10^{-10} \text{ m} \)
Neutron mass, \( m_n = 1.675 \times 10^{-27} \text{ kg} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \)
Boltzmann constant, \( k = 1.38 \times 10^{-23} \text{ J/K} \)
Temperature, \( T = 300 \text{ K} \)
Step 2: (a) Kinetic energy from de Broglie wavelength
\( \lambda = \frac{h}{p} \Rightarrow p = \frac{h}{\lambda} \)
\( K = \frac{p^2}{2m_n} = \frac{h^2}{2m_n \lambda^2} \)
\( = \frac{(6.626 \times 10^{-34})^2}{2 \times 1.675 \times 10^{-27} \times (1.40 \times 10^{-10})^2} \)
\( = \frac{4.390 \times 10^{-67}}{2 \times 1.675 \times 10^{-27} \times 1.96 \times 10^{-20}} \)
\( = \frac{4.390 \times 10^{-67}}{6.566 \times 10^{-47}} \)
\( = 6.686 \times 10^{-21} \text{ J} \)
In eV: \( K = \frac{6.686 \times 10^{-21}}{1.6 \times 10^{-19}} = 0.0418 \text{ eV} \)
Step 3: (b) Thermal neutron wavelength
Average \( K = \frac{3}{2} kT = 1.5 \times 1.38 \times 10^{-23} \times 300 \)
\( = 6.21 \times 10^{-21} \text{ J} \)
Momentum: \( p = \sqrt{2m_n K} \)
\( = \sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}} \)
\( = \sqrt{2.079 \times 10^{-47}} = 4.56 \times 10^{-24} \text{ kg m/s} \)
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{4.56 \times 10^{-24}} \)
\( = 1.453 \times 10^{-10} \text{ m} = 0.1453 \text{ nm} \)
Answers:
(a) Kinetic energy = 0.0418 eV
(b) de Broglie wavelength = 0.145 nm
\( \lambda = 1.40 \times 10^{-10} \text{ m} \)
Neutron mass, \( m_n = 1.675 \times 10^{-27} \text{ kg} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \)
Boltzmann constant, \( k = 1.38 \times 10^{-23} \text{ J/K} \)
Temperature, \( T = 300 \text{ K} \)
Step 2: (a) Kinetic energy from de Broglie wavelength
\( \lambda = \frac{h}{p} \Rightarrow p = \frac{h}{\lambda} \)
\( K = \frac{p^2}{2m_n} = \frac{h^2}{2m_n \lambda^2} \)
\( = \frac{(6.626 \times 10^{-34})^2}{2 \times 1.675 \times 10^{-27} \times (1.40 \times 10^{-10})^2} \)
\( = \frac{4.390 \times 10^{-67}}{2 \times 1.675 \times 10^{-27} \times 1.96 \times 10^{-20}} \)
\( = \frac{4.390 \times 10^{-67}}{6.566 \times 10^{-47}} \)
\( = 6.686 \times 10^{-21} \text{ J} \)
In eV: \( K = \frac{6.686 \times 10^{-21}}{1.6 \times 10^{-19}} = 0.0418 \text{ eV} \)
Step 3: (b) Thermal neutron wavelength
Average \( K = \frac{3}{2} kT = 1.5 \times 1.38 \times 10^{-23} \times 300 \)
\( = 6.21 \times 10^{-21} \text{ J} \)
Momentum: \( p = \sqrt{2m_n K} \)
\( = \sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}} \)
\( = \sqrt{2.079 \times 10^{-47}} = 4.56 \times 10^{-24} \text{ kg m/s} \)
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{4.56 \times 10^{-24}} \)
\( = 1.453 \times 10^{-10} \text{ m} = 0.1453 \text{ nm} \)
Answers:
(a) Kinetic energy = 0.0418 eV
(b) de Broglie wavelength = 0.145 nm
Question 11.18
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Answer & Explanation:
Step 1: For electromagnetic radiation
For a photon: Energy \( E = h\nu \) ...(1)
Momentum \( p = \frac{E}{c} = \frac{h\nu}{c} \) ...(2)
Step 2: Wavelength of EM radiation
Standard relation: \( c = \nu\lambda_{\text{EM}} \)
\( \lambda_{\text{EM}} = \frac{c}{\nu} \) ...(3)
Step 3: de Broglie wavelength of photon
de Broglie wavelength: \( \lambda_{\text{dB}} = \frac{h}{p} \)
Substitute \( p \) from (2): \( \lambda_{\text{dB}} = \frac{h}{(h\nu/c)} = \frac{c}{\nu} \) ...(4)
Step 4: Comparison
From (3) and (4): \( \lambda_{\text{EM}} = \lambda_{\text{dB}} = \frac{c}{\nu} \)
Conclusion: The wavelength of electromagnetic radiation is indeed equal to the de Broglie wavelength of its photon quantum.
For a photon: Energy \( E = h\nu \) ...(1)
Momentum \( p = \frac{E}{c} = \frac{h\nu}{c} \) ...(2)
Step 2: Wavelength of EM radiation
Standard relation: \( c = \nu\lambda_{\text{EM}} \)
\( \lambda_{\text{EM}} = \frac{c}{\nu} \) ...(3)
Step 3: de Broglie wavelength of photon
de Broglie wavelength: \( \lambda_{\text{dB}} = \frac{h}{p} \)
Substitute \( p \) from (2): \( \lambda_{\text{dB}} = \frac{h}{(h\nu/c)} = \frac{c}{\nu} \) ...(4)
Step 4: Comparison
From (3) and (4): \( \lambda_{\text{EM}} = \lambda_{\text{dB}} = \frac{c}{\nu} \)
Conclusion: The wavelength of electromagnetic radiation is indeed equal to the de Broglie wavelength of its photon quantum.
Question 11.19
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Answer & Explanation:
Step 1: Given data
Temperature, \( T = 300 \text{ K} \)
Atomic mass of N = 14.0076 u
1 u = \( 1.66 \times 10^{-27} \text{ kg} \)
\( k = 1.38 \times 10^{-23} \text{ J/K} \), \( h = 6.626 \times 10^{-34} \text{ J s} \)
Step 2: Mass of N₂ molecule
\( m = 2 \times 14.0076 \times 1.66 \times 10^{-27} \)
\( = 46.5 \times 10^{-27} \text{ kg} = 4.65 \times 10^{-26} \text{ kg} \)
Step 3: RMS speed
\( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \)
\( = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{4.65 \times 10^{-26}}} \)
\( = \sqrt{\frac{1.242 \times 10^{-20}}{4.65 \times 10^{-26}}} \)
\( = \sqrt{2.67 \times 10^5} = 517 \text{ m/s} \)
Step 4: de Broglie wavelength
\( \lambda = \frac{h}{mv_{\text{rms}}} = \frac{6.626 \times 10^{-34}}{4.65 \times 10^{-26} \times 517} \)
\( = \frac{6.626 \times 10^{-34}}{2.404 \times 10^{-23}} \)
\( = 2.756 \times 10^{-11} \text{ m} = 0.0276 \text{ nm} \)
Answer: de Broglie wavelength = 0.0276 nm
Temperature, \( T = 300 \text{ K} \)
Atomic mass of N = 14.0076 u
1 u = \( 1.66 \times 10^{-27} \text{ kg} \)
\( k = 1.38 \times 10^{-23} \text{ J/K} \), \( h = 6.626 \times 10^{-34} \text{ J s} \)
Step 2: Mass of N₂ molecule
\( m = 2 \times 14.0076 \times 1.66 \times 10^{-27} \)
\( = 46.5 \times 10^{-27} \text{ kg} = 4.65 \times 10^{-26} \text{ kg} \)
Step 3: RMS speed
\( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \)
\( = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{4.65 \times 10^{-26}}} \)
\( = \sqrt{\frac{1.242 \times 10^{-20}}{4.65 \times 10^{-26}}} \)
\( = \sqrt{2.67 \times 10^5} = 517 \text{ m/s} \)
Step 4: de Broglie wavelength
\( \lambda = \frac{h}{mv_{\text{rms}}} = \frac{6.626 \times 10^{-34}}{4.65 \times 10^{-26} \times 517} \)
\( = \frac{6.626 \times 10^{-34}}{2.404 \times 10^{-23}} \)
\( = 2.756 \times 10^{-11} \text{ m} = 0.0276 \text{ nm} \)
Answer: de Broglie wavelength = 0.0276 nm
Question 11.20
(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its \( e/m \) is given to be \( 1.76 \times 10^{11} \, \text{C kg}^{-1} \).
(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
Answer & Explanation:
Step 1: (a) For 500 V potential
Given: \( V = 500 \text{ V} \), \( \frac{e}{m} = 1.76 \times 10^{11} \text{ C/kg} \)
Classical formula: \( \frac{1}{2}mv^2 = eV \)
\( v = \sqrt{\frac{2eV}{m}} = \sqrt{2\left(\frac{e}{m}\right)V} \)
\( = \sqrt{2 \times 1.76 \times 10^{11} \times 500} \)
\( = \sqrt{1.76 \times 10^{14}} \)
\( = 1.327 \times 10^7 \text{ m/s} \)
Step 2: (b) For 10 MV potential
\( V = 10 \text{ MV} = 10^7 \text{ V} \)
Using same formula: \( v = \sqrt{2 \times 1.76 \times 10^{11} \times 10^7} \)
\( = \sqrt{3.52 \times 10^{18}} = 1.876 \times 10^9 \text{ m/s} \)
Step 3: Problem identified
The calculated speed \( 1.876 \times 10^9 \text{ m/s} \) exceeds the speed of light \( c = 3 \times 10^8 \text{ m/s} \).
This is impossible according to special relativity.
Step 4: Modification needed
For high speeds (comparable to \( c \)), we must use relativistic mechanics:
\( K = (\gamma - 1)m_0c^2 \) where \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \)
For \( eV \ll m_0c^2 \) (511 keV), classical formula works.
For \( eV \approx \) or \( > m_0c^2 \), relativistic formula must be used.
Answers:
(a) Speed = \( 1.33 \times 10^7 \text{ m/s} \)
(b) Classical formula gives impossible result (> speed of light). Must use relativistic formula for high energies.
Given: \( V = 500 \text{ V} \), \( \frac{e}{m} = 1.76 \times 10^{11} \text{ C/kg} \)
Classical formula: \( \frac{1}{2}mv^2 = eV \)
\( v = \sqrt{\frac{2eV}{m}} = \sqrt{2\left(\frac{e}{m}\right)V} \)
\( = \sqrt{2 \times 1.76 \times 10^{11} \times 500} \)
\( = \sqrt{1.76 \times 10^{14}} \)
\( = 1.327 \times 10^7 \text{ m/s} \)
Step 2: (b) For 10 MV potential
\( V = 10 \text{ MV} = 10^7 \text{ V} \)
Using same formula: \( v = \sqrt{2 \times 1.76 \times 10^{11} \times 10^7} \)
\( = \sqrt{3.52 \times 10^{18}} = 1.876 \times 10^9 \text{ m/s} \)
Step 3: Problem identified
The calculated speed \( 1.876 \times 10^9 \text{ m/s} \) exceeds the speed of light \( c = 3 \times 10^8 \text{ m/s} \).
This is impossible according to special relativity.
Step 4: Modification needed
For high speeds (comparable to \( c \)), we must use relativistic mechanics:
\( K = (\gamma - 1)m_0c^2 \) where \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \)
For \( eV \ll m_0c^2 \) (511 keV), classical formula works.
For \( eV \approx \) or \( > m_0c^2 \), relativistic formula must be used.
Answers:
(a) Speed = \( 1.33 \times 10^7 \text{ m/s} \)
(b) Classical formula gives impossible result (> speed of light). Must use relativistic formula for high energies.
Question 11.21
(a) A monoenergetic electron beam with electron speed of \( 5.20 \times 10^6 \, \text{m s}^{-1} \) is subject to a magnetic field of \( 1.30 \times 10^{-4} \, \text{T} \) normal to the beam velocity. What is the radius of the circle traced by the beam, given \( e/m \) for electron equals \( 1.76 \times 10^{11} \, \text{C kg}^{-1} \).
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
Answer & Explanation:
Step 1: (a) Classical calculation
Given: \( v = 5.20 \times 10^6 \text{ m/s} \), \( B = 1.30 \times 10^{-4} \text{ T} \)
\( \frac{e}{m} = 1.76 \times 10^{11} \text{ C/kg} \)
Magnetic force provides centripetal force: \( Bev = \frac{mv^2}{r} \)
\( r = \frac{mv}{eB} = \frac{v}{(e/m)B} \)
\( = \frac{5.20 \times 10^6}{(1.76 \times 10^{11}) \times (1.30 \times 10^{-4})} \)
\( = \frac{5.20 \times 10^6}{2.288 \times 10^7} = 0.227 \text{ m} = 22.7 \text{ cm} \)
Step 2: (b) For 20 MeV electron beam
Rest mass energy of electron = 0.511 MeV
20 MeV ≫ 0.511 MeV, so electron is relativistic.
Step 3: Modification needed
Classical formula \( r = \frac{mv}{eB} \) uses rest mass \( m_0 \).
For relativistic electrons: \( m = \gamma m_0 \) where \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \)
Relativistic momentum: \( p = \gamma m_0 v \)
Correct formula: \( r = \frac{p}{eB} = \frac{\gamma m_0 v}{eB} \)
Answers:
(a) Radius = 22.7 cm
(b) Formula is not valid for 20 MeV electrons. Must use relativistic momentum in radius formula.
Given: \( v = 5.20 \times 10^6 \text{ m/s} \), \( B = 1.30 \times 10^{-4} \text{ T} \)
\( \frac{e}{m} = 1.76 \times 10^{11} \text{ C/kg} \)
Magnetic force provides centripetal force: \( Bev = \frac{mv^2}{r} \)
\( r = \frac{mv}{eB} = \frac{v}{(e/m)B} \)
\( = \frac{5.20 \times 10^6}{(1.76 \times 10^{11}) \times (1.30 \times 10^{-4})} \)
\( = \frac{5.20 \times 10^6}{2.288 \times 10^7} = 0.227 \text{ m} = 22.7 \text{ cm} \)
Step 2: (b) For 20 MeV electron beam
Rest mass energy of electron = 0.511 MeV
20 MeV ≫ 0.511 MeV, so electron is relativistic.
Step 3: Modification needed
Classical formula \( r = \frac{mv}{eB} \) uses rest mass \( m_0 \).
For relativistic electrons: \( m = \gamma m_0 \) where \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \)
Relativistic momentum: \( p = \gamma m_0 v \)
Correct formula: \( r = \frac{p}{eB} = \frac{\gamma m_0 v}{eB} \)
Answers:
(a) Radius = 22.7 cm
(b) Formula is not valid for 20 MeV electrons. Must use relativistic momentum in radius formula.
Question 11.22
An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (\( -10^{-2} \, \text{mm} \) of Hg). A magnetic field of \( 2.83 \times 10^{-4} \, \text{T} \) curves the path of the electrons in a circular orbit of radius 12.0 cm. Determine \( e/m \) from the data.
Answer & Explanation:
Step 1: Given data
Accelerating voltage, \( V = 100 \text{ V} \)
Magnetic field, \( B = 2.83 \times 10^{-4} \text{ T} \)
Radius, \( r = 12.0 \text{ cm} = 0.12 \text{ m} \)
Step 2: Relationship between parameters
Kinetic energy: \( \frac{1}{2}mv^2 = eV \) ...(1)
Magnetic force: \( Bev = \frac{mv^2}{r} \) ...(2)
Step 3: Solve for \( e/m \)
From (2): \( v = \frac{eBr}{m} \)
Substitute into (1): \( \frac{1}{2}m\left(\frac{eBr}{m}\right)^2 = eV \)
\( \frac{1}{2} \frac{e^2 B^2 r^2}{m} = eV \)
\( \frac{e}{m} = \frac{2V}{B^2 r^2} \)
Step 4: Calculation
\( \frac{e}{m} = \frac{2 \times 100}{(2.83 \times 10^{-4})^2 \times (0.12)^2} \)
\( = \frac{200}{(8.009 \times 10^{-8}) \times 0.0144} \)
\( = \frac{200}{1.153 \times 10^{-9}} \)
\( = 1.735 \times 10^{11} \text{ C/kg} \)
Answer: \( e/m = \) \( 1.74 \times 10^{11} \text{ C/kg} \)
Accelerating voltage, \( V = 100 \text{ V} \)
Magnetic field, \( B = 2.83 \times 10^{-4} \text{ T} \)
Radius, \( r = 12.0 \text{ cm} = 0.12 \text{ m} \)
Step 2: Relationship between parameters
Kinetic energy: \( \frac{1}{2}mv^2 = eV \) ...(1)
Magnetic force: \( Bev = \frac{mv^2}{r} \) ...(2)
Step 3: Solve for \( e/m \)
From (2): \( v = \frac{eBr}{m} \)
Substitute into (1): \( \frac{1}{2}m\left(\frac{eBr}{m}\right)^2 = eV \)
\( \frac{1}{2} \frac{e^2 B^2 r^2}{m} = eV \)
\( \frac{e}{m} = \frac{2V}{B^2 r^2} \)
Step 4: Calculation
\( \frac{e}{m} = \frac{2 \times 100}{(2.83 \times 10^{-4})^2 \times (0.12)^2} \)
\( = \frac{200}{(8.009 \times 10^{-8}) \times 0.0144} \)
\( = \frac{200}{1.153 \times 10^{-9}} \)
\( = 1.735 \times 10^{11} \text{ C/kg} \)
Answer: \( e/m = \) \( 1.74 \times 10^{11} \text{ C/kg} \)
Question 11.23
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Answer & Explanation:
Step 1: (a) Maximum photon energy
Short wavelength end, \( \lambda_{\text{min}} = 0.45 \text{ Å} = 0.45 \times 10^{-10} \text{ m} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3 \times 10^8 \text{ m/s} \)
\( E_{\text{max}} = \frac{hc}{\lambda_{\text{min}}} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{0.45 \times 10^{-10}} \)
\( = \frac{1.9878 \times 10^{-25}}{4.5 \times 10^{-11}} = 4.417 \times 10^{-15} \text{ J} \)
In eV: \( E = \frac{4.417 \times 10^{-15}}{1.6 \times 10^{-19}} = 27,606 \text{ eV} \approx 27.6 \text{ keV} \)
Step 2: (b) Accelerating voltage
Maximum photon energy comes from electrons losing all their kinetic energy in one collision.
So, \( eV = E_{\text{max}} \)
\( V = \frac{E_{\text{max}}}{e} = 27.6 \text{ kV} \)
Answers:
(a) Maximum photon energy = 27.6 keV
(b) Accelerating voltage should be of the order of 30 kV
Short wavelength end, \( \lambda_{\text{min}} = 0.45 \text{ Å} = 0.45 \times 10^{-10} \text{ m} \)
\( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3 \times 10^8 \text{ m/s} \)
\( E_{\text{max}} = \frac{hc}{\lambda_{\text{min}}} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{0.45 \times 10^{-10}} \)
\( = \frac{1.9878 \times 10^{-25}}{4.5 \times 10^{-11}} = 4.417 \times 10^{-15} \text{ J} \)
In eV: \( E = \frac{4.417 \times 10^{-15}}{1.6 \times 10^{-19}} = 27,606 \text{ eV} \approx 27.6 \text{ keV} \)
Step 2: (b) Accelerating voltage
Maximum photon energy comes from electrons losing all their kinetic energy in one collision.
So, \( eV = E_{\text{max}} \)
\( V = \frac{E_{\text{max}}}{e} = 27.6 \text{ kV} \)
Answers:
(a) Maximum photon energy = 27.6 keV
(b) Accelerating voltage should be of the order of 30 kV
Question 11.24
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1 BeV = \( 10^9 \, \text{eV} \))
Answer & Explanation:
Step 1: Given data
Total energy = 10.2 BeV = 10.2 × 10⁹ eV = 1.02 × 10¹⁰ eV
Two γ-rays of equal energy are produced.
\( h = 4.14 \times 10^{-15} \text{ eV s} \), \( c = 3 \times 10^8 \text{ m/s} \)
Step 2: Energy per γ-ray
\( E_\gamma = \frac{\text{Total energy}}{2} = \frac{1.02 \times 10^{10}}{2} = 5.1 \times 10^9 \text{ eV} \)
Step 3: Wavelength of each γ-ray
\( \lambda = \frac{hc}{E_\gamma} = \frac{(4.14 \times 10^{-15}) \times (3 \times 10^8)}{5.1 \times 10^9} \)
\( = \frac{1.242 \times 10^{-6}}{5.1 \times 10^9} = 2.435 \times 10^{-16} \text{ m} \)
Answer: Wavelength of each γ-ray = \( 2.44 \times 10^{-16} \text{ m} \)
Total energy = 10.2 BeV = 10.2 × 10⁹ eV = 1.02 × 10¹⁰ eV
Two γ-rays of equal energy are produced.
\( h = 4.14 \times 10^{-15} \text{ eV s} \), \( c = 3 \times 10^8 \text{ m/s} \)
Step 2: Energy per γ-ray
\( E_\gamma = \frac{\text{Total energy}}{2} = \frac{1.02 \times 10^{10}}{2} = 5.1 \times 10^9 \text{ eV} \)
Step 3: Wavelength of each γ-ray
\( \lambda = \frac{hc}{E_\gamma} = \frac{(4.14 \times 10^{-15}) \times (3 \times 10^8)}{5.1 \times 10^9} \)
\( = \frac{1.242 \times 10^{-6}}{5.1 \times 10^9} = 2.435 \times 10^{-16} \text{ m} \)
Answer: Wavelength of each γ-ray = \( 2.44 \times 10^{-16} \text{ m} \)
Question 11.25
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never 'count photons', even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (\( 10^{-10} \, \text{W m}^{-2} \)). Take the area of the pupil to be about 0.4 cm², and the average frequency of white light to be about \( 6 \times 10^{14} \, \text{Hz} \).
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (\( 10^{-10} \, \text{W m}^{-2} \)). Take the area of the pupil to be about 0.4 cm², and the average frequency of white light to be about \( 6 \times 10^{14} \, \text{Hz} \).
Answer & Explanation:
Step 1: (a) Radio transmitter photons
Power, \( P = 10 \text{ kW} = 10^4 \text{ W} \)
Wavelength, \( \lambda = 500 \text{ m} \)
Energy per photon: \( E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{500} \)
\( = \frac{1.9878 \times 10^{-25}}{500} = 3.9756 \times 10^{-28} \text{ J} \)
Number of photons per second: \( N = \frac{P}{E} = \frac{10^4}{3.9756 \times 10^{-28}} \)
\( = 2.515 \times 10^{31} \text{ photons/s} \)
This enormous number means radio waves behave classically, not as individual photons.
Step 2: (b) Eye sensitivity
Minimum intensity, \( I = 10^{-10} \text{ W/m}^2 \)
Pupil area, \( A = 0.4 \text{ cm}^2 = 4 \times 10^{-5} \text{ m}^2 \)
Power entering eye: \( P_{\text{eye}} = I \times A = 10^{-10} \times 4 \times 10^{-5} = 4 \times 10^{-15} \text{ W} \)
Average frequency, \( \nu = 6 \times 10^{14} \text{ Hz} \)
Energy per photon: \( E = h\nu = 6.626 \times 10^{-34} \times 6 \times 10^{14} = 3.9756 \times 10^{-19} \text{ J} \)
Photons per second: \( N = \frac{P_{\text{eye}}}{E} = \frac{4 \times 10^{-15}}{3.9756 \times 10^{-19}} \)
\( = 1.006 \times 10^4 \approx 10,000 \text{ photons/s} \)
Since the eye needs ~100 photons to trigger a signal, and they arrive over many receptors, we cannot "count" individual photons.
Answers:
(a) \( \approx 2.5 \times 10^{31} \) photons/s – too many to worry about individually
(b) \( \approx 10,000 \) photons/s – eye integrates over time, cannot count individually
Power, \( P = 10 \text{ kW} = 10^4 \text{ W} \)
Wavelength, \( \lambda = 500 \text{ m} \)
Energy per photon: \( E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{500} \)
\( = \frac{1.9878 \times 10^{-25}}{500} = 3.9756 \times 10^{-28} \text{ J} \)
Number of photons per second: \( N = \frac{P}{E} = \frac{10^4}{3.9756 \times 10^{-28}} \)
\( = 2.515 \times 10^{31} \text{ photons/s} \)
This enormous number means radio waves behave classically, not as individual photons.
Step 2: (b) Eye sensitivity
Minimum intensity, \( I = 10^{-10} \text{ W/m}^2 \)
Pupil area, \( A = 0.4 \text{ cm}^2 = 4 \times 10^{-5} \text{ m}^2 \)
Power entering eye: \( P_{\text{eye}} = I \times A = 10^{-10} \times 4 \times 10^{-5} = 4 \times 10^{-15} \text{ W} \)
Average frequency, \( \nu = 6 \times 10^{14} \text{ Hz} \)
Energy per photon: \( E = h\nu = 6.626 \times 10^{-34} \times 6 \times 10^{14} = 3.9756 \times 10^{-19} \text{ J} \)
Photons per second: \( N = \frac{P_{\text{eye}}}{E} = \frac{4 \times 10^{-15}}{3.9756 \times 10^{-19}} \)
\( = 1.006 \times 10^4 \approx 10,000 \text{ photons/s} \)
Since the eye needs ~100 photons to trigger a signal, and they arrive over many receptors, we cannot "count" individual photons.
Answers:
(a) \( \approx 2.5 \times 10^{31} \) photons/s – too many to worry about individually
(b) \( \approx 10,000 \) photons/s – eye integrates over time, cannot count individually
Question 11.26
Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (\( 10^5 \, \text{W m}^{-2} \)) red light of wavelength 6328 Å produced by a He-Ne laser?
Answer & Explanation:
Step 1: Work function from UV light data
\( \lambda = 2271 \text{ Å} = 2.271 \times 10^{-7} \text{ m} \)
Stopping potential, \( V_0 = 1.3 \text{ V} \) (magnitude)
\( hc = 1240 \text{ eV·nm} \) (convenient conversion)
First convert: 2271 Å = 227.1 nm
Photon energy: \( E = \frac{1240}{227.1} = 5.46 \text{ eV} \)
\( K_{\text{max}} = eV_0 = 1.3 \text{ eV} \)
Work function: \( \phi = E - K_{\text{max}} = 5.46 - 1.3 = 4.16 \text{ eV} \)
Step 2: Red light check
\( \lambda_{\text{red}} = 6328 \text{ Å} = 632.8 \text{ nm} \)
Photon energy: \( E_{\text{red}} = \frac{1240}{632.8} = 1.96 \text{ eV} \)
Since \( 1.96 \text{ eV} < 4.16 \text{ eV} \) (work function)
Red light cannot cause photoelectric emission regardless of intensity.
Answers:
Work function ≈ 4.16 eV
Photo-cell will not respond to red light even at high intensity because photon energy is below threshold.
\( \lambda = 2271 \text{ Å} = 2.271 \times 10^{-7} \text{ m} \)
Stopping potential, \( V_0 = 1.3 \text{ V} \) (magnitude)
\( hc = 1240 \text{ eV·nm} \) (convenient conversion)
First convert: 2271 Å = 227.1 nm
Photon energy: \( E = \frac{1240}{227.1} = 5.46 \text{ eV} \)
\( K_{\text{max}} = eV_0 = 1.3 \text{ eV} \)
Work function: \( \phi = E - K_{\text{max}} = 5.46 - 1.3 = 4.16 \text{ eV} \)
Step 2: Red light check
\( \lambda_{\text{red}} = 6328 \text{ Å} = 632.8 \text{ nm} \)
Photon energy: \( E_{\text{red}} = \frac{1240}{632.8} = 1.96 \text{ eV} \)
Since \( 1.96 \text{ eV} < 4.16 \text{ eV} \) (work function)
Red light cannot cause photoelectric emission regardless of intensity.
Answers:
Work function ≈ 4.16 eV
Photo-cell will not respond to red light even at high intensity because photon energy is below threshold.
Question 11.27
Monochromatic radiation of wavelength 640.2 nm (1nm = \( 10^{-9} \) m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Answer & Explanation:
Step 1: Find work function from first measurement
\( \lambda_1 = 640.2 \text{ nm} \)
\( V_{01} = 0.54 \text{ V} \)
Using \( hc = 1240 \text{ eV·nm} \)
\( E_1 = \frac{1240}{640.2} = 1.937 \text{ eV} \)
\( K_{\text{max1}} = eV_{01} = 0.54 \text{ eV} \)
\( \phi = E_1 - K_{\text{max1}} = 1.937 - 0.54 = 1.397 \text{ eV} \)
Step 2: New stopping voltage for iron source
\( \lambda_2 = 427.2 \text{ nm} \)
\( E_2 = \frac{1240}{427.2} = 2.902 \text{ eV} \)
\( K_{\text{max2}} = E_2 - \phi = 2.902 - 1.397 = 1.505 \text{ eV} \)
\( V_{02} = \frac{K_{\text{max2}}}{e} = 1.505 \text{ V} \)
Answer: New stopping voltage = 1.51 V
\( \lambda_1 = 640.2 \text{ nm} \)
\( V_{01} = 0.54 \text{ V} \)
Using \( hc = 1240 \text{ eV·nm} \)
\( E_1 = \frac{1240}{640.2} = 1.937 \text{ eV} \)
\( K_{\text{max1}} = eV_{01} = 0.54 \text{ eV} \)
\( \phi = E_1 - K_{\text{max1}} = 1.937 - 0.54 = 1.397 \text{ eV} \)
Step 2: New stopping voltage for iron source
\( \lambda_2 = 427.2 \text{ nm} \)
\( E_2 = \frac{1240}{427.2} = 2.902 \text{ eV} \)
\( K_{\text{max2}} = E_2 - \phi = 2.902 - 1.397 = 1.505 \text{ eV} \)
\( V_{02} = \frac{K_{\text{max2}}}{e} = 1.505 \text{ V} \)
Answer: New stopping voltage = 1.51 V
Question 11.28
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
\( \lambda_1 = 3650 \, \text{Å}, \, \lambda_2 = 4047 \, \text{Å}, \, \lambda_3 = 4358 \, \text{Å}, \, \lambda_4 = 5461 \, \text{Å}, \, \lambda_5 = 6907 \, \text{Å} \)
The stopping voltages, respectively, were measured to be:
\( V_{01} = 1.28 \, \text{V}, \, V_{02} = 0.95 \, \text{V}, \, V_{03} = 0.74 \, \text{V}, \, V_{04} = 0.16 \, \text{V}, \, V_{05} = 0 \, \text{V} \)
Determine the value of Planck's constant \( h \), the threshold frequency and work function for the material.
\( \lambda_1 = 3650 \, \text{Å}, \, \lambda_2 = 4047 \, \text{Å}, \, \lambda_3 = 4358 \, \text{Å}, \, \lambda_4 = 5461 \, \text{Å}, \, \lambda_5 = 6907 \, \text{Å} \)
The stopping voltages, respectively, were measured to be:
\( V_{01} = 1.28 \, \text{V}, \, V_{02} = 0.95 \, \text{V}, \, V_{03} = 0.74 \, \text{V}, \, V_{04} = 0.16 \, \text{V}, \, V_{05} = 0 \, \text{V} \)
Determine the value of Planck's constant \( h \), the threshold frequency and work function for the material.
Answer & Explanation:
Step 1: Convert wavelengths to frequencies
\( c = 3 \times 10^8 \text{ m/s} \)
\( \nu = \frac{c}{\lambda} \)
\( \nu_1 = \frac{3 \times 10^8}{3650 \times 10^{-10}} = 8.219 \times 10^{14} \text{ Hz} \)
\( \nu_2 = \frac{3 \times 10^8}{4047 \times 10^{-10}} = 7.413 \times 10^{14} \text{ Hz} \)
\( \nu_3 = \frac{3 \times 10^8}{4358 \times 10^{-10}} = 6.884 \times 10^{14} \text{ Hz} \)
\( \nu_4 = \frac{3 \times 10^8}{5461 \times 10^{-10}} = 5.494 \times 10^{14} \text{ Hz} \)
\( \nu_5 = \frac{3 \times 10^8}{6907 \times 10^{-10}} = 4.343 \times 10^{14} \text{ Hz} \)
Step 2: Use Einstein's equation
\( eV_0 = h\nu - \phi \)
Plot \( V_0 \) vs \( \nu \), slope = \( h/e \), intercept = \( \phi/e \)
Using two points (for accuracy):
Point 1: \( \nu_1 = 8.219 \times 10^{14} \), \( V_{01} = 1.28 \)
Point 4: \( \nu_4 = 5.494 \times 10^{14} \), \( V_{04} = 0.16 \)
Slope = \( \frac{1.28 - 0.16}{(8.219 - 5.494) \times 10^{14}} = \frac{1.12}{2.725 \times 10^{14}} \)
\( = 4.11 \times 10^{-15} \text{ V·s} \)
\( h = e \times \text{slope} = 1.6 \times 10^{-19} \times 4.11 \times 10^{-15} \)
\( = 6.576 \times 10^{-34} \text{ J·s} \)
Step 3: Work function and threshold frequency
From \( V_0 = 0 \), threshold \( \nu_0 = 4.343 \times 10^{14} \text{ Hz} \) (from \( \lambda_5 \))
\( \phi = h\nu_0 = (6.576 \times 10^{-34}) \times (4.343 \times 10^{14}) \)
\( = 2.856 \times 10^{-19} \text{ J} = 1.785 \text{ eV} \)
Answers:
Planck's constant \( h \approx \) \( 6.58 \times 10^{-34} \text{ J·s} \)
Threshold frequency \( \nu_0 \approx \) \( 4.34 \times 10^{14} \text{ Hz} \)
Work function \( \phi \approx \) 1.79 eV
\( c = 3 \times 10^8 \text{ m/s} \)
\( \nu = \frac{c}{\lambda} \)
\( \nu_1 = \frac{3 \times 10^8}{3650 \times 10^{-10}} = 8.219 \times 10^{14} \text{ Hz} \)
\( \nu_2 = \frac{3 \times 10^8}{4047 \times 10^{-10}} = 7.413 \times 10^{14} \text{ Hz} \)
\( \nu_3 = \frac{3 \times 10^8}{4358 \times 10^{-10}} = 6.884 \times 10^{14} \text{ Hz} \)
\( \nu_4 = \frac{3 \times 10^8}{5461 \times 10^{-10}} = 5.494 \times 10^{14} \text{ Hz} \)
\( \nu_5 = \frac{3 \times 10^8}{6907 \times 10^{-10}} = 4.343 \times 10^{14} \text{ Hz} \)
Step 2: Use Einstein's equation
\( eV_0 = h\nu - \phi \)
Plot \( V_0 \) vs \( \nu \), slope = \( h/e \), intercept = \( \phi/e \)
Using two points (for accuracy):
Point 1: \( \nu_1 = 8.219 \times 10^{14} \), \( V_{01} = 1.28 \)
Point 4: \( \nu_4 = 5.494 \times 10^{14} \), \( V_{04} = 0.16 \)
Slope = \( \frac{1.28 - 0.16}{(8.219 - 5.494) \times 10^{14}} = \frac{1.12}{2.725 \times 10^{14}} \)
\( = 4.11 \times 10^{-15} \text{ V·s} \)
\( h = e \times \text{slope} = 1.6 \times 10^{-19} \times 4.11 \times 10^{-15} \)
\( = 6.576 \times 10^{-34} \text{ J·s} \)
Step 3: Work function and threshold frequency
From \( V_0 = 0 \), threshold \( \nu_0 = 4.343 \times 10^{14} \text{ Hz} \) (from \( \lambda_5 \))
\( \phi = h\nu_0 = (6.576 \times 10^{-34}) \times (4.343 \times 10^{14}) \)
\( = 2.856 \times 10^{-19} \text{ J} = 1.785 \text{ eV} \)
Answers:
Planck's constant \( h \approx \) \( 6.58 \times 10^{-34} \text{ J·s} \)
Threshold frequency \( \nu_0 \approx \) \( 4.34 \times 10^{14} \text{ Hz} \)
Work function \( \phi \approx \) 1.79 eV
Question 11.29
The work function for the following metals is given:
Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV.
Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV.
Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
Answer & Explanation:
Step 1: Energy of incident radiation
\( \lambda = 3300 \text{ Å} = 330 \text{ nm} \)
Using \( hc = 1240 \text{ eV·nm} \)
\( E = \frac{1240}{330} = 3.758 \text{ eV} \)
Step 2: Compare with work functions
Condition for emission: \( E > \phi \)
For Na: \( 3.758 > 2.75 \) → emission occurs
For K: \( 3.758 > 2.30 \) → emission occurs
For Mo: \( 3.758 > 4.17 \) → NO emission (3.758 < 4.17)
For Ni: \( 3.758 > 5.15 \) → NO emission (3.758 < 5.15)
Step 3: Effect of bringing laser closer
Photoelectric effect depends only on photon energy (frequency), not intensity or distance.
Bringing laser closer increases intensity but not photon energy.
So Mo and Ni will still not emit even if laser is brought closer.
Answers:
Mo and Ni will not give photoelectric emission.
Bringing laser closer does not change this result (intensity doesn't affect threshold).
\( \lambda = 3300 \text{ Å} = 330 \text{ nm} \)
Using \( hc = 1240 \text{ eV·nm} \)
\( E = \frac{1240}{330} = 3.758 \text{ eV} \)
Step 2: Compare with work functions
Condition for emission: \( E > \phi \)
For Na: \( 3.758 > 2.75 \) → emission occurs
For K: \( 3.758 > 2.30 \) → emission occurs
For Mo: \( 3.758 > 4.17 \) → NO emission (3.758 < 4.17)
For Ni: \( 3.758 > 5.15 \) → NO emission (3.758 < 5.15)
Step 3: Effect of bringing laser closer
Photoelectric effect depends only on photon energy (frequency), not intensity or distance.
Bringing laser closer increases intensity but not photon energy.
So Mo and Ni will still not emit even if laser is brought closer.
Answers:
Mo and Ni will not give photoelectric emission.
Bringing laser closer does not change this result (intensity doesn't affect threshold).
Question 11.30
Light of intensity \( 10^{-5} \, \text{W m}^{-2} \) falls on a sodium photo-cell of surface area 2 cm². Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Answer & Explanation:
Step 1: Calculate power absorbed
Intensity, \( I = 10^{-5} \text{ W/m}^2 \)
Area, \( A = 2 \text{ cm}^2 = 2 \times 10^{-4} \text{ m}^2 \)
Total power incident: \( P_{\text{inc}} = I \times A = 10^{-5} \times 2 \times 10^{-4} = 2 \times 10^{-9} \text{ W} \)
Assuming 5 layers absorb energy, but wave picture assumes energy distributes uniformly.
For estimation, assume all incident energy absorbed by surface electrons.
Step 2: Energy needed per electron
Work function, \( \phi = 2 \text{ eV} = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \text{ J} \)
Step 3: Time to accumulate energy for one electron
Power per electron? Need number of surface electrons.
Approximate: Atomic spacing ~ 2 Å = 2 × 10⁻¹⁰ m
Atoms per m² ≈ \( (1/(2 \times 10^{-10}))^2 = 2.5 \times 10^{19} \)
For 2 cm² = 2 × 10⁻⁴ m²: atoms ≈ \( 2.5 \times 10^{19} \times 2 \times 10^{-4} = 5 \times 10^{15} \)
If each atom contributes 1 electron: 5 × 10¹⁵ electrons
Power per electron = \( \frac{2 \times 10^{-9}}{5 \times 10^{15}} = 4 \times 10^{-25} \text{ W} \)
Time to accumulate 3.2 × 10⁻¹⁹ J:
\( t = \frac{3.2 \times 10^{-19}}{4 \times 10^{-25}} = 8 \times 10^5 \text{ s} \approx 9.3 \text{ days} \)
Step 4: Implication
Actual photoelectric emission is instantaneous (~10⁻⁹ s).
Wave picture predicts delay of days, contradicting experiment.
This shows wave theory fails; photon picture (energy quantized) is needed.
Answer:
Estimated time ≈ 9 days
Implication: Wave theory fails to explain instantaneous emission; supports photon concept.
Intensity, \( I = 10^{-5} \text{ W/m}^2 \)
Area, \( A = 2 \text{ cm}^2 = 2 \times 10^{-4} \text{ m}^2 \)
Total power incident: \( P_{\text{inc}} = I \times A = 10^{-5} \times 2 \times 10^{-4} = 2 \times 10^{-9} \text{ W} \)
Assuming 5 layers absorb energy, but wave picture assumes energy distributes uniformly.
For estimation, assume all incident energy absorbed by surface electrons.
Step 2: Energy needed per electron
Work function, \( \phi = 2 \text{ eV} = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \text{ J} \)
Step 3: Time to accumulate energy for one electron
Power per electron? Need number of surface electrons.
Approximate: Atomic spacing ~ 2 Å = 2 × 10⁻¹⁰ m
Atoms per m² ≈ \( (1/(2 \times 10^{-10}))^2 = 2.5 \times 10^{19} \)
For 2 cm² = 2 × 10⁻⁴ m²: atoms ≈ \( 2.5 \times 10^{19} \times 2 \times 10^{-4} = 5 \times 10^{15} \)
If each atom contributes 1 electron: 5 × 10¹⁵ electrons
Power per electron = \( \frac{2 \times 10^{-9}}{5 \times 10^{15}} = 4 \times 10^{-25} \text{ W} \)
Time to accumulate 3.2 × 10⁻¹⁹ J:
\( t = \frac{3.2 \times 10^{-19}}{4 \times 10^{-25}} = 8 \times 10^5 \text{ s} \approx 9.3 \text{ days} \)
Step 4: Implication
Actual photoelectric emission is instantaneous (~10⁻⁹ s).
Wave picture predicts delay of days, contradicting experiment.
This shows wave theory fails; photon picture (energy quantized) is needed.
Answer:
Estimated time ≈ 9 days
Implication: Wave theory fails to explain instantaneous emission; supports photon concept.
Question 11.31
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (\( m_e = 9.11 \times 10^{-31} \text{ kg} \)).
Answer & Explanation:
Step 1: X-ray photon energy
\( \lambda = 1 \text{ Å} = 10^{-10} \text{ m} \)
\( E_{\text{X-ray}} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{10^{-10}} \)
\( = 1.9878 \times 10^{-15} \text{ J} \)
In eV: \( E_{\text{X-ray}} = \frac{1.9878 \times 10^{-15}}{1.6 \times 10^{-19}} = 12,424 \text{ eV} \approx 12.4 \text{ keV} \)
Step 2: Electron energy for same wavelength
de Broglie wavelength: \( \lambda = \frac{h}{p} \)
\( p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{10^{-10}} = 6.626 \times 10^{-24} \text{ kg·m/s} \)
Kinetic energy: \( K = \frac{p^2}{2m_e} = \frac{(6.626 \times 10^{-24})^2}{2 \times 9.11 \times 10^{-31}} \)
\( = \frac{4.390 \times 10^{-47}}{1.822 \times 10^{-30}} = 2.409 \times 10^{-17} \text{ J} \)
In eV: \( K = \frac{2.409 \times 10^{-17}}{1.6 \times 10^{-19}} = 150.6 \text{ eV} \)
Step 3: Comparison
X-ray energy = 12.4 keV
Electron energy = 150.6 eV
X-ray has much greater energy (~82 times more).
Answer: X-rays have greater energy (12.4 keV vs 151 eV for electrons at same wavelength).
\( \lambda = 1 \text{ Å} = 10^{-10} \text{ m} \)
\( E_{\text{X-ray}} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{10^{-10}} \)
\( = 1.9878 \times 10^{-15} \text{ J} \)
In eV: \( E_{\text{X-ray}} = \frac{1.9878 \times 10^{-15}}{1.6 \times 10^{-19}} = 12,424 \text{ eV} \approx 12.4 \text{ keV} \)
Step 2: Electron energy for same wavelength
de Broglie wavelength: \( \lambda = \frac{h}{p} \)
\( p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{10^{-10}} = 6.626 \times 10^{-24} \text{ kg·m/s} \)
Kinetic energy: \( K = \frac{p^2}{2m_e} = \frac{(6.626 \times 10^{-24})^2}{2 \times 9.11 \times 10^{-31}} \)
\( = \frac{4.390 \times 10^{-47}}{1.822 \times 10^{-30}} = 2.409 \times 10^{-17} \text{ J} \)
In eV: \( K = \frac{2.409 \times 10^{-17}}{1.6 \times 10^{-19}} = 150.6 \text{ eV} \)
Step 3: Comparison
X-ray energy = 12.4 keV
Electron energy = 150.6 eV
X-ray has much greater energy (~82 times more).
Answer: X-rays have greater energy (12.4 keV vs 151 eV for electrons at same wavelength).
Question 11.32
(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (\( m_n = 1.675 \times 10^{-27} \text{ kg} \))
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Answer & Explanation:
Step 1: (a) Neutron with 150 eV energy
\( K = 150 \text{ eV} = 150 \times 1.6 \times 10^{-19} = 2.4 \times 10^{-17} \text{ J} \)
\( m_n = 1.675 \times 10^{-27} \text{ kg} \)
\( p = \sqrt{2m_n K} = \sqrt{2 \times 1.675 \times 10^{-27} \times 2.4 \times 10^{-17}} \)
\( = \sqrt{8.04 \times 10^{-44}} = 2.835 \times 10^{-22} \text{ kg·m/s} \)
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{2.835 \times 10^{-22}} = 2.337 \times 10^{-12} \text{ m} = 0.00234 \text{ nm} \)
This is much smaller than typical lattice spacing (~0.1–0.3 nm).
Step 2: Suitability for diffraction
For diffraction: \( \lambda \) should be comparable to lattice spacing (~0.1 nm).
Electron at 150 eV: \( \lambda \approx 0.1 \text{ nm} \) (from Q11.13)
Neutron at 150 eV: \( \lambda \approx 0.0023 \text{ nm} \) → too small.
Not suitable – wavelength too short for crystal diffraction.
Step 3: (b) Thermal neutrons
\( T = 27^\circ \text{C} = 300 \text{ K} \)
Average \( K = \frac{3}{2} kT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \text{ J} \)
\( p = \sqrt{2m_n K} = \sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}} \)
\( = \sqrt{2.079 \times 10^{-47}} = 4.56 \times 10^{-24} \text{ kg·m/s} \)
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{4.56 \times 10^{-24}} = 1.453 \times 10^{-10} \text{ m} = 0.145 \text{ nm} \)
This matches lattice spacing well.
Step 4: Need for thermalisation
Fast neutrons have high energy → short λ → poor diffraction.
Thermal neutrons (~0.025 eV) have λ ~ 0.1–0.2 nm, ideal for diffraction.
So fast neutrons must be slowed (thermalised) to useful wavelengths.
Answers:
(a) Neutron λ = 0.0023 nm – not suitable (too short compared to lattice)
(b) Thermal neutron λ = 0.145 nm – ideal; fast neutrons need thermalisation to get right λ.
\( K = 150 \text{ eV} = 150 \times 1.6 \times 10^{-19} = 2.4 \times 10^{-17} \text{ J} \)
\( m_n = 1.675 \times 10^{-27} \text{ kg} \)
\( p = \sqrt{2m_n K} = \sqrt{2 \times 1.675 \times 10^{-27} \times 2.4 \times 10^{-17}} \)
\( = \sqrt{8.04 \times 10^{-44}} = 2.835 \times 10^{-22} \text{ kg·m/s} \)
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{2.835 \times 10^{-22}} = 2.337 \times 10^{-12} \text{ m} = 0.00234 \text{ nm} \)
This is much smaller than typical lattice spacing (~0.1–0.3 nm).
Step 2: Suitability for diffraction
For diffraction: \( \lambda \) should be comparable to lattice spacing (~0.1 nm).
Electron at 150 eV: \( \lambda \approx 0.1 \text{ nm} \) (from Q11.13)
Neutron at 150 eV: \( \lambda \approx 0.0023 \text{ nm} \) → too small.
Not suitable – wavelength too short for crystal diffraction.
Step 3: (b) Thermal neutrons
\( T = 27^\circ \text{C} = 300 \text{ K} \)
Average \( K = \frac{3}{2} kT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \text{ J} \)
\( p = \sqrt{2m_n K} = \sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}} \)
\( = \sqrt{2.079 \times 10^{-47}} = 4.56 \times 10^{-24} \text{ kg·m/s} \)
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{4.56 \times 10^{-24}} = 1.453 \times 10^{-10} \text{ m} = 0.145 \text{ nm} \)
This matches lattice spacing well.
Step 4: Need for thermalisation
Fast neutrons have high energy → short λ → poor diffraction.
Thermal neutrons (~0.025 eV) have λ ~ 0.1–0.2 nm, ideal for diffraction.
So fast neutrons must be slowed (thermalised) to useful wavelengths.
Answers:
(a) Neutron λ = 0.0023 nm – not suitable (too short compared to lattice)
(b) Thermal neutron λ = 0.145 nm – ideal; fast neutrons need thermalisation to get right λ.
Question 11.33
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Answer & Explanation:
Step 1: Electron wavelength (relativistic correction needed)
\( V = 50 \text{ kV} = 5 \times 10^4 \text{ V} \)
Rest mass energy of electron = 511 keV = 0.511 MeV
Since 50 keV ≈ 0.05 MeV < 0.511 MeV, non-relativistic approximation OK.
\( \lambda = \frac{h}{\sqrt{2m_e eV}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 5 \times 10^4}} \)
Denominator: \( \sqrt{2 \times 9.11 \times 10^{-31} \times 8 \times 10^{-15}} = \sqrt{1.4576 \times 10^{-44}} \)
\( = 1.207 \times 10^{-22} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{1.207 \times 10^{-22}} = 5.49 \times 10^{-12} \text{ m} = 0.00549 \text{ nm} \)
Step 2: Optical microscope wavelength
Yellow light: \( \lambda_{\text{light}} \approx 550 \text{ nm} = 5.5 \times 10^{-7} \text{ m} \)
Step 3: Resolving power comparison
Resolving power ∝ \( \frac{1}{\lambda} \) (smaller λ gives better resolution)
Ratio: \( \frac{\lambda_{\text{light}}}{\lambda_{\text{electron}}} = \frac{5.5 \times 10^{-7}}{5.49 \times 10^{-12}} \approx 10^5 \)
Electron microscope has ~100,000 times better resolution.
Answers:
Electron λ ≈ 0.0055 nm
Electron microscope resolution is ~10⁵ times better than optical microscope.
\( V = 50 \text{ kV} = 5 \times 10^4 \text{ V} \)
Rest mass energy of electron = 511 keV = 0.511 MeV
Since 50 keV ≈ 0.05 MeV < 0.511 MeV, non-relativistic approximation OK.
\( \lambda = \frac{h}{\sqrt{2m_e eV}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 5 \times 10^4}} \)
Denominator: \( \sqrt{2 \times 9.11 \times 10^{-31} \times 8 \times 10^{-15}} = \sqrt{1.4576 \times 10^{-44}} \)
\( = 1.207 \times 10^{-22} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{1.207 \times 10^{-22}} = 5.49 \times 10^{-12} \text{ m} = 0.00549 \text{ nm} \)
Step 2: Optical microscope wavelength
Yellow light: \( \lambda_{\text{light}} \approx 550 \text{ nm} = 5.5 \times 10^{-7} \text{ m} \)
Step 3: Resolving power comparison
Resolving power ∝ \( \frac{1}{\lambda} \) (smaller λ gives better resolution)
Ratio: \( \frac{\lambda_{\text{light}}}{\lambda_{\text{electron}}} = \frac{5.5 \times 10^{-7}}{5.49 \times 10^{-12}} \approx 10^5 \)
Electron microscope has ~100,000 times better resolution.
Answers:
Electron λ ≈ 0.0055 nm
Electron microscope resolution is ~10⁵ times better than optical microscope.
Question 11.34
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of \( 10^{-15} \) m or less. This structure was first probed in early 1970's using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)
Answer & Explanation:
Step 1: Required wavelength
To probe structure of size \( d \approx 10^{-15} \text{ m} \), need \( \lambda \lesssim d \).
Take \( \lambda \approx 10^{-15} \text{ m} \).
Step 2: Relativistic electrons
For such short wavelengths, electrons are highly relativistic.
de Broglie: \( \lambda = \frac{h}{p} \)
For relativistic particles: \( E \approx pc \) (when \( E \gg m_0c^2 \))
So \( E \approx \frac{hc}{\lambda} \)
Step 3: Calculate energy
\( E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{10^{-15}} \)
\( = 1.9878 \times 10^{-10} \text{ J} \)
In eV: \( E = \frac{1.9878 \times 10^{-10}}{1.6 \times 10^{-19}} = 1.242 \times 10^9 \text{ eV} = 1.242 \text{ GeV} \)
Step 4: Check against rest mass
Rest mass energy = 0.511 MeV = 0.000511 GeV
1.242 GeV ≫ 0.000511 GeV, so ultra-relativistic approximation valid.
Answer: Electron beam energy was of the order of 1 GeV (giga-electronvolt).
To probe structure of size \( d \approx 10^{-15} \text{ m} \), need \( \lambda \lesssim d \).
Take \( \lambda \approx 10^{-15} \text{ m} \).
Step 2: Relativistic electrons
For such short wavelengths, electrons are highly relativistic.
de Broglie: \( \lambda = \frac{h}{p} \)
For relativistic particles: \( E \approx pc \) (when \( E \gg m_0c^2 \))
So \( E \approx \frac{hc}{\lambda} \)
Step 3: Calculate energy
\( E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{10^{-15}} \)
\( = 1.9878 \times 10^{-10} \text{ J} \)
In eV: \( E = \frac{1.9878 \times 10^{-10}}{1.6 \times 10^{-19}} = 1.242 \times 10^9 \text{ eV} = 1.242 \text{ GeV} \)
Step 4: Check against rest mass
Rest mass energy = 0.511 MeV = 0.000511 GeV
1.242 GeV ≫ 0.000511 GeV, so ultra-relativistic approximation valid.
Answer: Electron beam energy was of the order of 1 GeV (giga-electronvolt).
Question 11.35
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.
Answer & Explanation:
Step 1: de Broglie wavelength of He atom
\( T = 27^\circ \text{C} = 300 \text{ K} \)
Mass of He atom: \( m = 4 \text{ u} = 4 \times 1.66 \times 10^{-27} = 6.64 \times 10^{-27} \text{ kg} \)
Average kinetic energy: \( \overline{K} = \frac{3}{2} kT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \text{ J} \)
RMS speed: \( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{6.64 \times 10^{-27}}} \)
\( = \sqrt{\frac{1.242 \times 10^{-20}}{6.64 \times 10^{-27}}} = \sqrt{1.87 \times 10^6} = 1368 \text{ m/s} \)
Momentum: \( p = mv_{\text{rms}} = 6.64 \times 10^{-27} \times 1368 = 9.08 \times 10^{-24} \text{ kg·m/s} \)
\( \lambda_{\text{dB}} = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{9.08 \times 10^{-24}} = 7.30 \times 10^{-11} \text{ m} = 0.073 \text{ nm} \)
Step 2: Mean separation between atoms
Ideal gas at STP: 1 mole = 22.4 L = 0.0224 m³ contains \( 6.022 \times 10^{23} \) atoms.
Volume per atom: \( v = \frac{0.0224}{6.022 \times 10^{23}} = 3.72 \times 10^{-26} \text{ m}^3 \)
Mean separation: \( d = v^{1/3} = (3.72 \times 10^{-26})^{1/3} \)
\( = (37.2 \times 10^{-27})^{1/3} = 3.34 \times 10^{-9} \text{ m} = 3.34 \text{ nm} \)
Step 3: Comparison
\( \lambda_{\text{dB}} \approx 0.073 \text{ nm} \), \( d \approx 3.34 \text{ nm} \)
\( \lambda_{\text{dB}} \ll d \) (by factor ~45)
Wave packets don't overlap significantly.
Answers:
de Broglie λ ≈ 0.073 nm
Mean separation ≈ 3.34 nm
λ ≪ separation → wave packets are non-overlapping.
\( T = 27^\circ \text{C} = 300 \text{ K} \)
Mass of He atom: \( m = 4 \text{ u} = 4 \times 1.66 \times 10^{-27} = 6.64 \times 10^{-27} \text{ kg} \)
Average kinetic energy: \( \overline{K} = \frac{3}{2} kT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \text{ J} \)
RMS speed: \( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{6.64 \times 10^{-27}}} \)
\( = \sqrt{\frac{1.242 \times 10^{-20}}{6.64 \times 10^{-27}}} = \sqrt{1.87 \times 10^6} = 1368 \text{ m/s} \)
Momentum: \( p = mv_{\text{rms}} = 6.64 \times 10^{-27} \times 1368 = 9.08 \times 10^{-24} \text{ kg·m/s} \)
\( \lambda_{\text{dB}} = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{9.08 \times 10^{-24}} = 7.30 \times 10^{-11} \text{ m} = 0.073 \text{ nm} \)
Step 2: Mean separation between atoms
Ideal gas at STP: 1 mole = 22.4 L = 0.0224 m³ contains \( 6.022 \times 10^{23} \) atoms.
Volume per atom: \( v = \frac{0.0224}{6.022 \times 10^{23}} = 3.72 \times 10^{-26} \text{ m}^3 \)
Mean separation: \( d = v^{1/3} = (3.72 \times 10^{-26})^{1/3} \)
\( = (37.2 \times 10^{-27})^{1/3} = 3.34 \times 10^{-9} \text{ m} = 3.34 \text{ nm} \)
Step 3: Comparison
\( \lambda_{\text{dB}} \approx 0.073 \text{ nm} \), \( d \approx 3.34 \text{ nm} \)
\( \lambda_{\text{dB}} \ll d \) (by factor ~45)
Wave packets don't overlap significantly.
Answers:
de Broglie λ ≈ 0.073 nm
Mean separation ≈ 3.34 nm
λ ≪ separation → wave packets are non-overlapping.
Question 11.36
Compute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about \( 2 \times 10^{-10} \) m.
Answer & Explanation:
Step 1: de Broglie wavelength of electron in metal
\( T = 300 \text{ K} \)
\( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Thermal kinetic energy: \( K = \frac{3}{2} kT = 6.21 \times 10^{-21} \text{ J} \) (same as before)
But electrons in metal follow Fermi-Dirac statistics, not Maxwell-Boltzmann.
Fermi energy \( E_F \) is more relevant (~few eV).
For estimation, use thermal energy:
\( p = \sqrt{2m_e K} = \sqrt{2 \times 9.11 \times 10^{-31} \times 6.21 \times 10^{-21}} \)
\( = \sqrt{1.131 \times 10^{-50}} = 1.063 \times 10^{-25} \text{ kg·m/s} \)
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{1.063 \times 10^{-25}} = 6.23 \times 10^{-9} \text{ m} = 6.23 \text{ nm} \)
Step 2: More accurate using Fermi energy
Typical \( E_F \approx 5 \text{ eV} = 8 \times 10^{-19} \text{ J} \)
\( p = \sqrt{2m_e E_F} = \sqrt{2 \times 9.11 \times 10^{-31} \times 8 \times 10^{-19}} \)
\( = \sqrt{1.4576 \times 10^{-48}} = 1.207 \times 10^{-24} \text{ kg·m/s} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{1.207 \times 10^{-24}} = 5.49 \times 10^{-10} \text{ m} = 0.549 \text{ nm} \)
Step 3: Compare with mean electron separation
Given: mean separation \( d \approx 2 \times 10^{-10} \text{ m} = 0.2 \text{ nm} \)
Using λ ≈ 0.55 nm (Fermi energy estimate):
λ > d (by factor ~2.75)
Wave packets overlap significantly.
Answers:
de Broglie λ ≈ 0.55 nm (using Fermi energy)
Mean separation ≈ 0.2 nm
λ > d → wave packets strongly overlap → electrons are indistinguishable.
\( T = 300 \text{ K} \)
\( m_e = 9.11 \times 10^{-31} \text{ kg} \)
Thermal kinetic energy: \( K = \frac{3}{2} kT = 6.21 \times 10^{-21} \text{ J} \) (same as before)
But electrons in metal follow Fermi-Dirac statistics, not Maxwell-Boltzmann.
Fermi energy \( E_F \) is more relevant (~few eV).
For estimation, use thermal energy:
\( p = \sqrt{2m_e K} = \sqrt{2 \times 9.11 \times 10^{-31} \times 6.21 \times 10^{-21}} \)
\( = \sqrt{1.131 \times 10^{-50}} = 1.063 \times 10^{-25} \text{ kg·m/s} \)
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{1.063 \times 10^{-25}} = 6.23 \times 10^{-9} \text{ m} = 6.23 \text{ nm} \)
Step 2: More accurate using Fermi energy
Typical \( E_F \approx 5 \text{ eV} = 8 \times 10^{-19} \text{ J} \)
\( p = \sqrt{2m_e E_F} = \sqrt{2 \times 9.11 \times 10^{-31} \times 8 \times 10^{-19}} \)
\( = \sqrt{1.4576 \times 10^{-48}} = 1.207 \times 10^{-24} \text{ kg·m/s} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{1.207 \times 10^{-24}} = 5.49 \times 10^{-10} \text{ m} = 0.549 \text{ nm} \)
Step 3: Compare with mean electron separation
Given: mean separation \( d \approx 2 \times 10^{-10} \text{ m} = 0.2 \text{ nm} \)
Using λ ≈ 0.55 nm (Fermi energy estimate):
λ > d (by factor ~2.75)
Wave packets overlap significantly.
Answers:
de Broglie λ ≈ 0.55 nm (using Fermi energy)
Mean separation ≈ 0.2 nm
λ > d → wave packets strongly overlap → electrons are indistinguishable.
Question 11.37
Answer the following questions:
(a) Quarks inside protons and neutrons are thought to carry fractional charges [\( (+2/3)e : (-1/3)e \)]. Why do they not show up in Millikan's oil-drop experiment?
(b) What is so special about the combination \( e/m \)? Why do we not simply talk of \( e \) and \( m \) separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
\( E = h \nu, p = \frac{h}{\lambda} \)
But while the value of \( \lambda \) is physically significant, the value of \( \nu \) (and therefore, the value of the phase speed \( \nu \lambda \)) has no physical significance. Why?
(a) Quarks inside protons and neutrons are thought to carry fractional charges [\( (+2/3)e : (-1/3)e \)]. Why do they not show up in Millikan's oil-drop experiment?
(b) What is so special about the combination \( e/m \)? Why do we not simply talk of \( e \) and \( m \) separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
\( E = h \nu, p = \frac{h}{\lambda} \)
But while the value of \( \lambda \) is physically significant, the value of \( \nu \) (and therefore, the value of the phase speed \( \nu \lambda \)) has no physical significance. Why?
Answer & Explanation:
(a) Quarks and Millikan's experiment:
Quarks are confined inside hadrons (protons/neutrons) by the strong nuclear force.
They cannot exist as free particles at low energies.
Millikan's experiment measures charge on free particles (oil drops, ions, electrons).
Since quarks are never free under ordinary conditions, their fractional charges don't appear.
(b) Special nature of \( e/m \):
• \( e/m \) appears in fundamental equations: Lorentz force (\( F = e(v × B) \)), cyclotron motion (\( r = mv/eB \)).
• It can be measured precisely without knowing \( e \) and \( m \) individually.
• Many phenomena (cathode rays, Thomson's experiment) directly give \( e/m \).
• Historically, \( e/m \) was measured before individual values of \( e \) and \( m \).
(c) Gas conduction at low pressure:
At ordinary pressures: frequent collisions between gas molecules and ions.
Ions cannot gain enough energy between collisions to cause ionization.
At very low pressures: mean free path increases dramatically.
Ions accelerated by electric field gain sufficient energy to ionize molecules upon collision.
This starts avalanche multiplication → gas breakdown/conduction.
(d) Energy distribution of photoelectrons:
• Photons give energy \( h\nu \) to electrons.
• Electrons in metal have different initial energies (Fermi distribution).
• Electrons deeper inside need extra energy to reach surface.
• Surface conditions, scattering, and emission directions vary.
• Result: range of kinetic energies from 0 to \( K_{\text{max}} = h\nu - \phi \).
(e) Significance of λ vs ν for matter waves:
• λ determines interference/diffraction patterns (observable).
• ν is related to total energy \( E = h\nu \), but for matter waves \( E = \frac{p^2}{2m} + V \).
• Phase velocity \( v_p = \nu\lambda = \frac{E}{p} = \frac{p}{2m} + \frac{V}{p} \) depends on V and is not constant.
• Group velocity \( v_g = \frac{d\omega}{dk} = \frac{dE}{dp} = \frac{p}{m} \) equals particle velocity (physically meaningful).
• Only group velocity is observable; phase velocity is not directly measurable.
Quarks are confined inside hadrons (protons/neutrons) by the strong nuclear force.
They cannot exist as free particles at low energies.
Millikan's experiment measures charge on free particles (oil drops, ions, electrons).
Since quarks are never free under ordinary conditions, their fractional charges don't appear.
(b) Special nature of \( e/m \):
• \( e/m \) appears in fundamental equations: Lorentz force (\( F = e(v × B) \)), cyclotron motion (\( r = mv/eB \)).
• It can be measured precisely without knowing \( e \) and \( m \) individually.
• Many phenomena (cathode rays, Thomson's experiment) directly give \( e/m \).
• Historically, \( e/m \) was measured before individual values of \( e \) and \( m \).
(c) Gas conduction at low pressure:
At ordinary pressures: frequent collisions between gas molecules and ions.
Ions cannot gain enough energy between collisions to cause ionization.
At very low pressures: mean free path increases dramatically.
Ions accelerated by electric field gain sufficient energy to ionize molecules upon collision.
This starts avalanche multiplication → gas breakdown/conduction.
(d) Energy distribution of photoelectrons:
• Photons give energy \( h\nu \) to electrons.
• Electrons in metal have different initial energies (Fermi distribution).
• Electrons deeper inside need extra energy to reach surface.
• Surface conditions, scattering, and emission directions vary.
• Result: range of kinetic energies from 0 to \( K_{\text{max}} = h\nu - \phi \).
(e) Significance of λ vs ν for matter waves:
• λ determines interference/diffraction patterns (observable).
• ν is related to total energy \( E = h\nu \), but for matter waves \( E = \frac{p^2}{2m} + V \).
• Phase velocity \( v_p = \nu\lambda = \frac{E}{p} = \frac{p}{2m} + \frac{V}{p} \) depends on V and is not constant.
• Group velocity \( v_g = \frac{d\omega}{dk} = \frac{dE}{dp} = \frac{p}{m} \) equals particle velocity (physically meaningful).
• Only group velocity is observable; phase velocity is not directly measurable.
📘 Exam Preparation Tip:
These exercise questions will help you understand quantum phenomena and wave-particle duality. You'll learn to solve photoelectric effect problems using Einstein's equation, calculate de Broglie wavelength for particles, understand Davisson-Germer experiment, analyze graphs of photoelectric current versus voltage, and appreciate the dual nature of radiation and matter. Fundamental for quantum mechanics introduction.