Work function and the photoelectric effect

The work function, ϕ, is the least energy an electron needs to escape from the surface of a metal. It is the key to the photoelectric effect: light only releases electrons if each photon has at least ϕ of energy, and whatever is left over becomes the electron's kinetic energy. That gives the photoelectric equation, hf = ϕ + E_k(max), and the stopping potential is the voltage that just stops the fastest of those electrons.

What is the photoelectric effect?

Shine light of a high enough frequency on a clean metal surface and electrons are emitted from it. These are called photoelectrons. A classic demonstration, suggested in AQA's specification, puts a zinc plate on a charged electroscope: ultraviolet light from a UV lamp makes it discharge as electrons leave the zinc, but ordinary visible light does nothing.

OpenStax lists three features that a wave model of light can't explain:

1. There is a threshold frequency. Below a certain frequency, no electrons are released, however bright the light.
2. The maximum kinetic energy doesn't depend on intensity. Brighter light releases more electrons, not faster ones.
3. There is no delay. Electrons come out as soon as the light arrives, even when it is dim.

Einstein explained all three in 1905 by treating light as a stream of photons, each with energy E = hf. One photon gives all its energy to one electron, in one go. If that energy is less than the work function, the electron can't escape, so the frequency matters and the brightness doesn't. More on the wider argument is in is light a wave or a particle.

Work function and threshold frequency

Electrons are held in a metal by the attraction of the positive ions. The work function is the energy needed to pull out the least tightly held ones, those at the surface. Each metal has its own value. These are from OpenStax's table of typical work functions:

Measured work functions depend on the state of the surface, so tables differ a little. HyperPhysics, for example, lists sodium at 2.28 eV and, for polycrystalline samples, 2.36 eV, and notes that silver gives noticeably different values on different crystal faces. The examples below use the OpenStax values.

The threshold frequency, f₀, is the lowest frequency that releases electrons. At f₀ a photon has exactly enough energy to free an electron with nothing to spare, so:

Energies this small are often given in electronvolts. An electron accelerated through a potential difference of 1 V gains 1 electronvolt: 1 eV = 1.60 × 10⁻¹⁹ J. AQA expects you to convert between eV and J: multiply by 1.60 × 10⁻¹⁹ to go from eV to J, and divide to go back.

The photoelectric equation

It is conservation of energy for one photon and one electron: photon energy in, work function to escape, the rest as kinetic energy. It says maximum because electrons from deeper in the metal lose some extra energy getting out, so most leave slower. If you know the wavelength instead of the frequency, use hf = hc ÷ λ, with c = 3.00 × 10⁸ m s⁻¹.

Try a photon energy calculation, then press Mark to check it in the course.

Stopping potential

In a photocell, the emitted electrons cross a gap to a collecting electrode. If that electrode is made negative, it repels them. The stopping potential, V_s, is the smallest reverse potential difference that stops even the fastest photoelectrons, so the current falls to zero. The work done stopping them equals their maximum kinetic energy:

So a stopping potential of 1.5 V means E_k(max) = 1.5 eV, which is 1.5 × 1.60 × 10⁻¹⁹ = 2.4 × 10⁻¹⁹ J. Put it into the photoelectric equation and you get hf = ϕ + eV_s. AQA's specification notes that the experimental method for measuring the stopping potential is not required.

The graph of Ek(max) against frequency

Rearranging gives E_k(max) = hf − ϕ, which is y = mx + c. Plot maximum kinetic energy (up) against frequency (across) and you get a straight line:

• Gradient = h. OpenStax notes every metal gives a line with the same slope, the Planck constant.
• x-intercept = f₀, the threshold frequency. Below it there are no photoelectrons, so the line stops there.
• y-intercept = −ϕ when the line is extended back. A metal with a bigger work function gives a parallel line further to the right.

If the graph shows stopping potential instead, the line is V_s = (h ÷ e)f − ϕ ÷ e, so the gradient is h ÷ e and the y-intercept is −ϕ ÷ e.

Worked examples

Example 1: threshold frequency of zinc

Zinc has a work function of 4.31 eV. Find its threshold frequency and threshold wavelength.

1. Convert to joules: ϕ = 4.31 × 1.60 × 10⁻¹⁹ = 6.90 × 10⁻¹⁹ J.
2. f₀ = ϕ ÷ h = 6.90 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 1.04 × 10¹⁵ Hz.
3. λ₀ = c ÷ f₀ = 3.00 × 10⁸ ÷ 1.04 × 10¹⁵ = 2.88 × 10⁻⁷ m, which is 288 nm, in the ultraviolet.

That is why visible light, from about 400 nm to 700 nm, can't release electrons from zinc and the electroscope demonstration needs a UV lamp.

Example 2: kinetic energy and stopping potential for sodium

Ultraviolet light of wavelength 350 nm falls on sodium (ϕ = 2.46 eV). Find the maximum kinetic energy of the photoelectrons and the stopping potential.

1. Photon energy: E = hc ÷ λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ ÷ 3.50 × 10⁻⁷ = 5.68 × 10⁻¹⁹ J.
2. In electronvolts: 5.68 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = 3.55 eV.
3. E_k(max) = 3.55 − 2.46 = 1.09 eV, which is 1.75 × 10⁻¹⁹ J.
4. Since E_k(max) = eV_s, the stopping potential is 1.09 V.

Working in eV saves a step here: an energy in eV is numerically the stopping potential in volts.

Example 3: work function from a stopping potential

Radiation of frequency 1.5 × 10¹⁵ Hz falls on a metal and the stopping potential is 1.2 V. Find the work function and the threshold frequency.

1. Photon energy: hf = 6.63 × 10⁻³⁴ × 1.5 × 10¹⁵ = 9.95 × 10⁻¹⁹ J.
2. E_k(max) = eV_s = 1.60 × 10⁻¹⁹ × 1.2 = 1.92 × 10⁻¹⁹ J.
3. ϕ = hf − E_k(max) = 9.95 × 10⁻¹⁹ − 1.92 × 10⁻¹⁹ = 8.03 × 10⁻¹⁹ J, or 5.02 eV.
4. f₀ = ϕ ÷ h = 8.03 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 1.21 × 10¹⁵ Hz.

Common mistakes

• Mixing eV and J. Put every energy in the same unit before you add or subtract. h in J s gives energies in J.
• Saying brighter light gives faster electrons. Intensity changes the number of electrons per second, not their maximum energy.
• Thinking a long exposure helps below f₀. One photon frees one electron; photons can't save up energy, so no electrons come out at all.
• Leaving out "maximum". The equation gives the kinetic energy of the fastest electrons only.
• Using nanometres in hc ÷ λ. Convert to metres first: 350 nm = 3.50 × 10⁻⁷ m.
• Reading the gradient wrongly. On an Ek(max) against f graph the gradient is h; on a Vs against f graph it is h ÷ e.

The photoelectric effect is in the first year of A-level Physics, so practise it on today's A-level Physics: Year 1 short timed paper. The light and energy course has an interactive photoelectric experiment in its evidence for photons section, and the de Broglie wavelength guide covers the other half of wave and particle duality.

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