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Quantum Physics: Question 2

Syllabus 22.2

Structured A2 10 marks

A photocell contains a caesium metal surface with work function ϕ=2.14 eV\phi = 2.14\text{ eV}, housed inside an evacuated glass tube.

(a) State what is meant by the work function of a metal. [1]

(b) Show that the threshold frequency for photoelectric emission from this caesium surface is f0=5.16×1014 Hzf_0 = 5.16\times10^{14}\text{ Hz}. [2]

(c) The caesium surface is now illuminated with violet light of wavelength 380 nm380\text{ nm}. Calculate the maximum kinetic energy of the photoelectrons emitted, giving your answer in both joules and electron-volts. [3]

(d) Calculate the maximum speed of these photoelectrons. (mass of electron =9.11×10−31 kg= 9.11\times10^{-31}\text{ kg}) [2]

(e) The intensity of the violet light is now increased, while its frequency is kept the same. State and explain the effect, if any, of this change on (i) the photoelectric current, and (ii) the maximum kinetic energy of the emitted photoelectrons. [2]

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Worked solution

Part (a): Work function

The work function ϕ\phi of a metal is the minimum energy needed to release an electron from the surface of that metal, leaving the electron with zero kinetic energy.

Part (b): Show that the threshold frequency is 5.16 × 10¹⁴ Hz

At the threshold frequency, an incident photon has just enough energy to release an electron with no kinetic energy left over, so hf0=ϕhf_0 = \phi, giving: f0=ϕhf_0 = \frac{\phi}{h}

First convert the work function to joules: ϕ=2.14×1.60×10−19=3.424×10−19 J\phi = 2.14\times1.60\times10^{-19} = 3.424\times10^{-19}\text{ J}

Then: f0=3.424×10−196.63×10−34=5.164×1014 Hzf_0 = \frac{3.424\times10^{-19}}{6.63\times10^{-34}} = 5.164\times10^{14}\text{ Hz}

Recompute as a check: 6.63×10−34×5.164×1014=3.424×10−19 J6.63\times10^{-34}\times5.164\times10^{14}=3.424\times10^{-19}\text{ J}, which matches ϕ\phi back again.

So f0=5.16×1014 Hzf_0 = 5.16\times10^{14}\text{ Hz} (3 s.f.), as required to show.

Part (c): Maximum kinetic energy at 380 nm

First find the frequency of the incident violet light: f=cλ=3.00×108380×10−9=7.895×1014 Hzf = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{380\times10^{-9}} = 7.895\times10^{14}\text{ Hz}

Then the photon energy: Ephoton=hf=6.63×10−34×7.895×1014=5.234×10−19 JE_{photon} = hf = 6.63\times10^{-34}\times7.895\times10^{14} = 5.234\times10^{-19}\text{ J}

Einstein’s photoelectric equation, hf=ϕ+12mvmax2hf = \phi + \tfrac12mv_{max}^2, rearranges to give the maximum kinetic energy of the emitted photoelectrons: KEmax=hf−ϕ=5.234×10−19−3.424×10−19=1.810×10−19 JKE_{max} = hf - \phi = 5.234\times10^{-19} - 3.424\times10^{-19} = 1.810\times10^{-19}\text{ J}

Converting to electron-volts: KEmax=1.810×10−191.60×10−19=1.13 eVKE_{max} = \frac{1.810\times10^{-19}}{1.60\times10^{-19}} = 1.13\text{ eV}

Recompute as a check, working entirely in eV: Ephoton=5.234×10−19/1.60×10−19=3.27 eVE_{photon}=5.234\times10^{-19}/1.60\times10^{-19}=3.27\text{ eV}, so KEmax=3.27−2.14=1.13 eVKE_{max}=3.27-2.14=1.13\text{ eV}, the same answer both ways.

Part (d): Maximum speed of the photoelectrons

KEmax=12mvmax2  ⟹  vmax=2KEmaxmKE_{max} = \tfrac12mv_{max}^2 \implies v_{max}=\sqrt{\frac{2KE_{max}}{m}}

vmax=2×1.810×10−199.11×10−31=3.974×1011=6.30×105 m s−1v_{max} = \sqrt{\frac{2\times1.810\times10^{-19}}{9.11\times10^{-31}}} = \sqrt{3.974\times10^{11}} = 6.30\times10^{5}\text{ m s}^{-1}

Recompute as a check: (6.30×105)2=3.97×1011(6.30\times10^{5})^2=3.97\times10^{11}, and 2×1.810×10−199.11×10−31=3.97×1011\dfrac{2\times1.810\times10^{-19}}{9.11\times10^{-31}}=3.97\times10^{11}, consistent.

Part (e): Effect of increasing the intensity

(i) The photocurrent increases. Increasing the intensity of light of the same frequency means more photons arrive at the surface per second, so more photoelectrons are released per second.

(ii) The maximum kinetic energy is unchanged. Each photon still carries the same energy hfhf (since the frequency has not changed), and each photoelectric emission event involves a single electron absorbing a single photon. The maximum kinetic energy hf−ϕhf-\phi therefore depends only on the photon frequency, not on how many photons arrive per second. This is exactly the observation a classical wave model cannot explain.

Final answers

  • (a) The work function is the minimum energy to release an electron from a metal surface with zero kinetic energy.
  • (b) f0=5.16×1014 Hzf_0=\boxed{5.16\times10^{14}}\text{ Hz}
  • (c) KEmax=1.81×10−19 J=1.13 eVKE_{max}=\boxed{1.81\times10^{-19}}\text{ J} = \boxed{1.13}\text{ eV}
  • (d) vmax=6.30×105 m s−1v_{max}=\boxed{6.30\times10^{5}}\text{ m s}^{-1}
  • (e) Photocurrent increases; maximum KE stays the same.