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CHEM 121 Studioby Learn4Less · UBC CHEM 121

5.2 · Light & matter

The photoelectric effect

The experiment that showed light comes in packets.

By the end you should be able to:

  • Explain and calculate the photoelectric effect

Key idea

What the experiment shows

Light shining on a clean metal surface can eject electrons. Four observations:

  1. Below a threshold frequency ν0\nu_0 no electrons are ejected, however intense the light.
  2. Above ν0\nu_0 electrons appear instantly, even in very dim light.
  3. The maximum kinetic energy of the electrons rises linearly with ν\nu and does not depend on intensity.
  4. Brighter light ejects more electrons (a larger current), not faster ones.

Classical wave theory predicted the opposite: the energy delivered should depend on intensity, and any frequency should work given enough time. Einstein (1905) explained the results with photons of energy hνh\nu, each ejecting at most one electron.

Formula

The photoelectric equation

KEmax⁡=hν−Φ\mathrm{KE}_{\max} = h\nu - \Phi

The work function Φ\Phi is the minimum energy needed to remove an electron from the metal's surface; it depends on the metal.

ν0=Φh,λ0=hcΦ\nu_0 = \frac{\Phi}{h}, \qquad \lambda_0 = \frac{hc}{\Phi}

Electrons are ejected only if ν>ν0\nu > \nu_0 (equivalently λ<λ0\lambda < \lambda_0). A plot of KEmax⁡\mathrm{KE}_{\max} against ν\nu is a straight line with slope hh, xx-intercept ν0\nu_0 and yy-intercept −Φ-\Phi. The fastest electrons move at v=2 KEmax⁡/mev = \sqrt{2\,\mathrm{KE}_{\max}/m_e}.

Method

Solving a photoelectric problem

  1. Put Φ\Phi and the photon energy in the same unit (1 eV = 1.602×10−191.602\times10^{-19} J).
  2. Photon energy: E=hc/λE = hc/\lambda (or hνh\nu).
  3. If E<ΦE < \Phi, no electrons are ejected; stop.
  4. Otherwise KEmax⁡=E−Φ\mathrm{KE}_{\max} = E - \Phi.
  5. If asked, the speed is v=2 KEmax⁡/mev = \sqrt{2\,\mathrm{KE}_{\max}/m_e} and the threshold wavelength is λ0=hc/Φ\lambda_0 = hc/\Phi.

Common mistake

Photoelectric traps

Wrong: "bright enough red light will eventually eject electrons." Right: below the threshold frequency no electrons are ejected at any intensity.

Wrong: "doubling the intensity doubles the kinetic energy." Right: intensity changes the number of electrons; their maximum kinetic energy depends only on ν\nu.

Wrong: reporting a negative kinetic energy when hν<Φh\nu < \Phi. Right: a negative result means no electrons are emitted.

Wrong: subtracting a work function in eV from a photon energy in J. Right: convert to the same unit first.

Worked example

Worked example: 300 nm light on sodium

Sodium has Φ=2.28\Phi = 2.28 eV =(2.28)(1.602×10−19)=3.65×10−19= (2.28)(1.602\times10^{-19}) = 3.65\times10^{-19} J.

Photon energy: E=1.986×10−253.00×10−7=6.62×10−19E = \dfrac{1.986\times10^{-25}}{3.00\times10^{-7}} = 6.62\times10^{-19} J, which exceeds Φ\Phi.

KEmax⁡=6.62×10−19−3.65×10−19=2.97×10−19 J (1.85 eV)\mathrm{KE}_{\max} = 6.62\times10^{-19} - 3.65\times10^{-19} = 2.97\times10^{-19}\ \mathrm{J}\ (1.85\ \mathrm{eV})

Speed: v=2(2.97×10−19)9.109×10−31=8.07×105v = \sqrt{\dfrac{2(2.97\times10^{-19})}{9.109\times10^{-31}}} = 8.07\times10^{5} m/s.

Threshold: λ0=1.986×10−253.65×10−19=5.44×10−7\lambda_0 = \dfrac{1.986\times10^{-25}}{3.65\times10^{-19}} = 5.44\times10^{-7} m = 544 nm.

Answer: KEmax⁡=2.97×10−19\mathrm{KE}_{\max} = 2.97\times10^{-19} J; light longer than 544 nm (yellow, orange, red) ejects nothing from sodium.

Worked example

Worked example: finding Φ from data

Light of 250 nm ejects electrons with KEmax⁡=3.00×10−19\mathrm{KE}_{\max} = 3.00\times10^{-19} J. Find Φ\Phi and λ0\lambda_0.

E=1.986×10−252.50×10−7=7.946×10−19 JE = \frac{1.986\times10^{-25}}{2.50\times10^{-7}} = 7.946\times10^{-19}\ \mathrm{J}

Φ=E−KEmax⁡=7.946×10−19−3.00×10−19=4.95×10−19 J (3.09 eV)\Phi = E - \mathrm{KE}_{\max} = 7.946\times10^{-19} - 3.00\times10^{-19} = 4.95\times10^{-19}\ \mathrm{J}\ (3.09\ \mathrm{eV})

λ0=hcΦ=1.986×10−254.946×10−19=4.02×10−7 m\lambda_0 = \frac{hc}{\Phi} = \frac{1.986\times10^{-25}}{4.946\times10^{-19}} = 4.02\times10^{-7}\ \mathrm{m}

Answer: Φ=4.95×10−19\Phi = 4.95\times10^{-19} J and λ0=402\lambda_0 = 402 nm.

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