Studyledge

Quantum Physics: Question 7

Syllabus 22.2

Structured A2 8 marks

A clean sodium surface has work function ϕ=2.30 eV\phi=2.30\text{ eV}.

(a) Show that the threshold wavelength for photoelectric emission from this surface is 540 nm540\text{ nm}. [2]

(b) The surface is illuminated with violet light of wavelength 400 nm400\text{ nm}. State, with a reason, whether photoelectrons are emitted from the surface. [1]

(c) Calculate the maximum kinetic energy of the photoelectrons emitted at 400 nm400\text{ nm}, giving your answer in both joules and electron-volts. [3]

(d) Calculate the stopping potential required to reduce the photoelectric current from this surface to zero. [2]

Show worked solution Hide worked solution

Worked solution

Part (a): Show that the threshold wavelength is 540 nm

At the threshold wavelength λ0\lambda_0, an incident photon has just enough energy to release an electron with zero kinetic energy left over, so hcλ0=ϕ\dfrac{hc}{\lambda_0}=\phi, giving: λ0=hcϕ\lambda_0 = \frac{hc}{\phi}

First convert the work function to joules: ϕ=2.30×1.60×10−19=3.68×10−19 J\phi = 2.30\times1.60\times10^{-19} = 3.68\times10^{-19}\text{ J}

Then, with h=6.63×10−34 J sh=6.63\times10^{-34}\text{ J s} and c=3.00×108 m s−1c=3.00\times10^{8}\text{ m s}^{-1}: λ0=6.63×10−34×3.00×1083.68×10−19=1.989×10−253.68×10−19=5.405×10−7 m\lambda_0 = \frac{6.63\times10^{-34}\times3.00\times10^{8}}{3.68\times10^{-19}} = \frac{1.989\times10^{-25}}{3.68\times10^{-19}} = 5.405\times10^{-7}\text{ m}

Recompute as a check: 5.405×10−7×3.68×10−19=1.989×10−255.405\times10^{-7}\times3.68\times10^{-19}=1.989\times10^{-25}, matching hchc found above.

So λ0=540 nm\lambda_0 = 540\text{ nm} (3 s.f.), as required to show.

Part (b): Are photoelectrons emitted at 400 nm?

Yes. The incident wavelength (400 nm400\text{ nm}) is shorter than the threshold wavelength (540 nm540\text{ nm}), which means the incident frequency is higher than the threshold frequency, so each photon carries enough energy to release a photoelectron.

Part (c): Maximum kinetic energy at 400 nm

First find the frequency of the incident light: f=cλ=3.00×108400×10−9=7.50×1014 Hzf = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{400\times10^{-9}} = 7.50\times10^{14}\text{ Hz}

Then the photon energy: Ephoton=hf=6.63×10−34×7.50×1014=4.973×10−19 JE_{photon} = hf = 6.63\times10^{-34}\times7.50\times10^{14} = 4.973\times10^{-19}\text{ J}

Einstein’s photoelectric equation, hf=ϕ+12mvmax2hf=\phi+\tfrac12mv_{max}^2, rearranges to give the maximum kinetic energy: KEmax=hf−ϕ=4.973×10−19−3.68×10−19=1.293×10−19 JKE_{max} = hf-\phi = 4.973\times10^{-19}-3.68\times10^{-19} = 1.293\times10^{-19}\text{ J}

Converting to electron-volts: KEmax=1.293×10−191.60×10−19=0.808 eVKE_{max} = \frac{1.293\times10^{-19}}{1.60\times10^{-19}} = 0.808\text{ eV}

Recompute as a check, working entirely in eV: Ephoton=4.973×10−19/1.60×10−19=3.11 eVE_{photon}=4.973\times10^{-19}/1.60\times10^{-19}=3.11\text{ eV}, so KEmax=3.11−2.30=0.81 eVKE_{max}=3.11-2.30=0.81\text{ eV}, the same answer (allowing for rounding) both ways.

Part (d): Stopping potential

The stopping potential VsV_s is the reverse potential difference that does just enough negative work on the fastest photoelectrons to bring them to rest, so: eVs=KEmax  ⟹  Vs=KEmaxeeV_s = KE_{max} \implies V_s = \frac{KE_{max}}{e}

Since KEmax=0.808 eVKE_{max}=0.808\text{ eV} (i.e. 0.808×e0.808\times e joules), dividing by ee leaves the stopping potential numerically equal to the kinetic energy expressed in eV: Vs=0.808 VV_s = 0.808\text{ V}

Recompute as a check, in joules: Vs=1.293×10−191.60×10−19=0.808 VV_s=\dfrac{1.293\times10^{-19}}{1.60\times10^{-19}}=0.808\text{ V}, consistent.

Final answers

  • (a) λ0=540 nm\lambda_0=\boxed{540}\text{ nm}
  • (b) Yes, 400 nm<540 nm400\text{ nm}<540\text{ nm}, so the photon energy exceeds the work function.
  • (c) KEmax=1.29×10−19 J=0.808 eVKE_{max}=\boxed{1.29\times10^{-19}}\text{ J} = \boxed{0.808}\text{ eV}
  • (d) Vs=0.808 VV_s=\boxed{0.808}\text{ V}