Part 5 · The old quantum theory

5.3 The photoelectric effect

Source: Postorino, week 4 slides, p. 22

Objectives

Picture. Light knocks electrons out of a metal. Their maximum energy depends on the light’s frequency, not its intensity; below a threshold frequency nothing comes out, however bright the light.
Theorem. Einstein: $K_{max} = h ν - ϕ$ .
Recipe. From the stopping potential, $K_{max} = e Δ V_{s}$ ; threshold $ν_{s} = ϕ / h$ , $λ_{s} = h c / ϕ$ .
Vocabulary. Photocurrent, stopping potential, work function, photon.

The experiment

The setup. An evacuated glass tube. Light falls on a metal surface (the cathode, −) and knocks electrons out of it. They fly to a detector (the collector, +), and an ammeter measures the resulting photocurrent. A battery sets the potential difference between the two.

The photocurrent against the potential difference, for high and low light intensity:

With a positive potential the current saturates: every emitted electron is collected. More intensity, more electrons, higher saturation current.
With a negative (retarding) potential fewer electrons make it. Both curves reach zero at the same potential, $- Δ V_{s}$ , whatever the intensity.

Definition 5.3.1Stopping potentialslides w4, p. 22

The stopping potential $Δ V_{s}$ is the (negative) potential that cancels the photocurrent. An electron leaving with kinetic energy $K$ climbs the barrier only if $K > e Δ V_{s}$ , so the stopping potential measures the maximum kinetic energy: $K_{max} = e Δ V_{s} .$

N.B. The electrons that come out have different energies, because those starting deeper below the surface spend energy getting there; only those from the surface leave with $K_{max}$ . A simulation is linked on the slide: javalab, photoelectric effect.

Key observation 5.3.2The observationsslides w4, p. 22

There is a threshold frequency $ν_{s}$ , specific to the metal. Below it no electrons come out, however intense the light.
The I–V curves converge at $- Δ V_{s}$ , and $Δ V_{s}$ depends linearly on the frequency and not on the intensity.
There is no delay in the emission of photoelectrons.
The kinetic energy depends on the frequency, not on the intensity.

WarningSlide slip

Item 2 says $Δ V_{s}$ depends linearly on $ν_{t}$ , the threshold frequency. The threshold is a fixed property of the metal; what $Δ V_{s}$ depends on linearly is the frequency $ν$ of the light, $e Δ V_{s} = h ν - ϕ$ . The slide also writes the threshold as $ν_{t}$ in the list, $ν_{s}$ in the box and $f_{c}$ in the plot: the same thing.

Non-example 5.3.3What classical physics expectsbeyond the lectures

Light is a wave whose energy is spread over the wavefront, with intensity $\propto E_{0}^{2}$ (Theorem 2.2.10):

more intense light shakes the electrons harder, so the kinetic energy should grow with intensity: against 2 and 4;
any frequency should eventually work if the light is intense enough: against 1;
with dim light an electron should need time to soak up enough energy to escape: against 3.

Einstein’s explanation

Theorem 5.3.4Einstein’s photon (1905)slides w4, p. 22

Electromagnetic radiation is quantised, and the photon is the quantum of energy. Each photon carries an energy $h ν$ and can only give it all at once, to a single electron. Then $K_{max} = h ν - ϕ \Rightarrow h ν_{s} = ϕ,$ where $ϕ$ is the work function: the energy needed to extract an electron from the metal, specific to each metal.

All four observations followbeyond the lectures

If $h ν < ϕ$ no single photon can free an electron, and photons don’t pool their energy.
$e Δ V_{s} = h ν - ϕ$ is linear in $ν$ . Intensity is the number of photons, which sets how many electrons come out (the saturation current), not their energy.
One photon, one electron, immediately: no accumulation.
Each electron’s energy comes from one photon, $h ν$ .

Work functions $ϕ$ in eV: Na 2.46, Al 4.08, Pb 4.14, Zn 4.31, Fe 4.50, Cu 4.70, Ag 4.73, Pt 6.35.

Their thresholds are $λ_{s} = h c / ϕ$ beyond the lectures: sodium, 504 nm, responds to visible light; platinum, 195 nm, only to ultraviolet.

The $K_{max}$ –frequency plot (the slide’s lower-left figure): $K_{max}$ against $ν$ for two metals.

Two parallel straight lines. Their slope is $h$ , the same for every metal.
Each line crosses the frequency axis at that metal’s threshold: metal 1 has the lower threshold $f_{c, 1}$ , metal 2 a higher one.
Extended back to $ν = 0$ , each line hits the vertical axis at $- ϕ$ : $- φ_{1}$ for metal 1, and $- φ_{2}$ , lower, for metal 2.

Measuring such lines (Millikan, 1916) gave $h$ independently of the blackbody, and the same value beyond the lectures.

WarningSlide slip

In the plot the labels are tangled. Both captions read “Soglia Metal 1”. The green one points to the label $ν_{s, 2}$ written over the metal 1 line; the dark-blue one points to the metal 2 crossing, labelled $ν_{s, 1}$ with $ν_{s, 2}$ printed over it. Read it as: metal 1 threshold $ν_{s, 1} = f_{c, 1}$ at the left crossing, metal 2 threshold $ν_{s, 2}$ at the right crossing.

Loading the demo…

Figure 5.3.1 The photoelectric effect. Choose a metal and drag the light’s

frequency

along the axis; set its intensity. Left:

K_{max}

against

ν

for the metals, parallel lines of slope

h

crossing the axis at their

thresholds

. Right: the photocurrent against the collector voltage. Intensity raises the saturation current; only the frequency moves the stopping potential.

Exercises

Exercise 5.3.1beyond the lectures

Sodium ( $ϕ = 2.46$ eV) is lit with 400 nm light. Find the photon energy, $K_{max}$ , the stopping potential and the threshold wavelength. What changes if the intensity doubles?

Solution

$h ν = 1240/400 = 3.10$ eV. $K_{max} = 3.10 - 2.46 = 0.64$ eV, so $Δ V_{s} = 0.64$ V. Threshold $λ_{s} = 1240/2.46 = 504$ nm. Doubling the intensity doubles the photocurrent; $K_{max}$ and $Δ V_{s}$ don’t change.

Exercise 5.3.2beyond the lectures

How many photons per second does a 1 mW red laser (633 nm) emit?

Solution

$E = 1240/633 = 1.96$ eV $= 3.14 \times 1 0^{- 19}$ J. $N = 1 0^{- 3} /3.14 \times 1 0^{- 19} = 3.2 \times 1 0^{15}$ photons/s: so many that the light looks continuous.

Exercise 5.3.3beyond the lectures

Stopping potentials of 1.0 V and 2.0 V are measured on one metal with light of $6.0 \times 1 0^{14}$ Hz and $8.4 \times 1 0^{14}$ Hz. Find $h$ and $ϕ$ .

Solution

$h = e Δ (Δ V_{s}) /Δ ν = 1.6 \times 1 0^{- 19} \cdot 1.0/2.4 \times 1 0^{14} = 6.7 \times 1 0^{- 34}$ J·s. Then $ϕ = h ν - e V_{s} = 6.7 \times 1 0^{- 34} \cdot 6 \times 1 0^{14} /1.6 \times 1 0^{- 19} - 1.0 = 2.5 - 1.0 = 1.5$ eV.

Exercise 5.3.4beyond the lectures

A classical estimate: light of intensity $1 0^{- 2}$ W/m² falls on sodium. If an atom (area $1 0^{- 19}$ m²) collected the energy falling on it, how long until it had 2.46 eV?

Solution

Power on one atom: $1 0^{- 2} \times 1 0^{- 19} = 1 0^{- 21}$ W. Time: $2.46 \times 1.6 \times 1 0^{- 19} /1 0^{- 21} = 394$ s, over six minutes. Electrons come out at once: classical energy accumulation is wrong.