Constants used: h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C, c = 3 × 10⁸ m/s, mₑ = 9.1 × 10⁻³¹ kg, 1 eV = 1.6 × 10⁻¹⁹ J.
Exercises
(11.1) Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.
Answer
Given V = 30 kV = 3 × 10⁴ V. An electron accelerated through V gains energy eV; the most energetic X-ray photon carries all of it.
(a) hν_max = eV
⇒ ν_max = eV/h = (1.6 × 10⁻¹⁹ × 3 × 10⁴)/(6.63 × 10⁻³⁴)
⇒ ν_max ≈ 7.24 × 10¹⁸ Hz
(b) λ_min = c/ν_max = (3 × 10⁸)/(7.24 × 10¹⁸)
⇒ λ_min ≈ 4.14 × 10⁻¹¹ m = 0.0414 nm
(11.2) The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10¹⁴ 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?
Answer
Given φ₀ = 2.14 eV, ν = 6 × 10¹⁴ Hz.
Photon energy hν = 6.63 × 10⁻³⁴ × 6 × 10¹⁴ = 3.98 × 10⁻¹⁹ J
⇒ hν = 3.98 × 10⁻¹⁹/1.6 × 10⁻¹⁹ ≈ 2.49 eV
(a) K_max = hν − φ₀ = 2.49 − 2.14
⇒ K_max ≈ 0.35 eV (≈ 5.5 × 10⁻²⁰ J)
(b) eV₀ = K_max → V₀ ≈ 0.35 V
(c) ½mv²_max = K_max
⇒ v_max = √(2K_max/m) = √(2 × 0.346 × 1.6 × 10⁻¹⁹ / 9.1 × 10⁻³¹)
⇒ v_max ≈ 3.49 × 10⁵ m/s (about 349 km/s)
(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
K_max = eV₀ = 1.6 × 10⁻¹⁹ × 1.5
⇒ K_max = 2.4 × 10⁻¹⁹ J = 1.5 eV
(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?
Answer
Given λ = 632.8 × 10⁻⁹ m, P = 9.42 × 10⁻³ W.
(a) E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(632.8 × 10⁻⁹)
⇒ E ≈ 3.14 × 10⁻¹⁹ J (≈ 1.96 eV)
p = h/λ = 6.63 × 10⁻³⁴/632.8 × 10⁻⁹
⇒ p ≈ 1.05 × 10⁻²⁷ kg m/s
(b) Number of photons per second N = P/E = 9.42 × 10⁻³/3.14 × 10⁻¹⁹
⇒ N ≈ 3.0 × 10¹⁶ photons/s. Since the whole beam falls on the target, this many photons reach it each second.
(c) Mass of hydrogen atom m_H ≈ 1.66 × 10⁻²⁷ kg. v = p/m_H = 1.05 × 10⁻²⁷/1.66 × 10⁻²⁷
⇒ v ≈ 0.63 m/s
(11.5) In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10⁻¹⁵ V s. Calculate the value of Planck's constant.
Answer
From Einstein's equation, eV₀ = hν − φ₀
⇒ V₀ = (h/e)ν − φ₀/e, so slope = h/e.
h = e × slope = 1.6 × 10⁻¹⁹ × 4.12 × 10⁻¹⁵
⇒ h ≈ 6.59 × 10⁻³⁴ J s
(11.6) The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer
Given ν₀ = 3.3 × 10¹⁴ Hz, ν = 8.2 × 10¹⁴ Hz.
eV₀ = h(ν − ν₀)
⇒ V₀ = h(ν − ν₀)/e = 6.63 × 10⁻³⁴ × 4.9 × 10¹⁴/1.6 × 10⁻¹⁹
⇒ V₀ ≈ 2.03 V
(11.7) 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
Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(330 × 10⁻⁹) = 6.03 × 10⁻¹⁹ J
⇒ E = 6.03 × 10⁻¹⁹/1.6 × 10⁻¹⁹ ≈ 3.77 eV
Since 3.77 eV is less than the work function 4.2 eV, no photoelectric emission takes place, however intense the light.
(11.8) Light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer
Given ν = 7.21 × 10¹⁴ Hz, v_max = 6.0 × 10⁵ m/s.
½mv²_max = h(ν − ν₀)
⇒ ν₀ = ν − mv²_max/2h
½mv²_max = 0.5 × 9.1 × 10⁻³¹ × (6.0 × 10⁵)² = 1.64 × 10⁻¹⁹ J
⇒ 1.64 × 10⁻¹⁹/6.63 × 10⁻³⁴ = 2.47 × 10¹⁴ Hz
ν₀ = 7.21 × 10¹⁴ − 2.47 × 10¹⁴
⇒ ν₀ ≈ 4.74 × 10¹⁴ Hz
(11.9) 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
Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(488 × 10⁻⁹) = 4.08 × 10⁻¹⁹ J ≈ 2.55 eV
φ₀ = hν − eV₀ = 2.55 eV − 0.38 eV
⇒ φ₀ ≈ 2.17 eV (about 2.16 eV with more precise constants)
(11.10) 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 × 10⁻⁹ kg drifting with a speed of 2.2 m/s?
Answer
λ = h/p = h/mv
(a) λ = 6.63 × 10⁻³⁴/(0.040 × 1000)
⇒ λ ≈ 1.66 × 10⁻³⁵ m
(b) λ = 6.63 × 10⁻³⁴/(0.060 × 1.0)
⇒ λ ≈ 1.10 × 10⁻³² m
(c) λ = 6.63 × 10⁻³⁴/(1.0 × 10⁻⁹ × 2.2)
⇒ λ ≈ 3.0 × 10⁻²⁵ m
These wavelengths are far too small to be observed, so wave behaviour of everyday objects is not noticeable.
(11.11) Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Answer
A photon of radiation of frequency ν and wavelength λ has energy E = hν and travels at speed c.
Its momentum is p = E/c = hν/c
⇒ since c = νλ, p = h/λ
De Broglie wavelength of the photon λ_dB = h/p = h/(h/λ)
⇒ λ_dB = λ. Hence the two wavelengths are equal.
Important Extra Questions
Define work function and threshold frequency.
Answer
Work function φ₀ is the minimum energy needed to free an electron from a metal surface. Threshold frequency ν₀ is the minimum frequency of incident light that can cause photoemission; φ₀ = hν₀.
How does the stopping potential depend on (a) intensity and (b) frequency of incident light?
Answer
(a) It is independent of intensity (intensity changes only the photocurrent). (b) It increases linearly with frequency above ν₀, since eV₀ = hν − φ₀.
Why can the wave picture of light not explain the photoelectric effect?
Answer
Wave theory predicts that any frequency should eject electrons if intense enough, that K_max should grow with intensity and that there should be a time lag. Experiments show a threshold frequency, K_max independent of intensity and instantaneous emission, which only the photon picture explains.
An electron and a proton have the same kinetic energy. Which has the longer de Broglie wavelength?
Answer
λ = h/√(2mK). For equal K, λ ∝ 1/√m, so the lighter electron has the longer wavelength.
Find the de Broglie wavelength of an electron accelerated through 100 V.
Answer
λ = 1.227/√V nm = 1.227/√100 nm
⇒ λ ≈ 0.123 nm