Wave and Particle Nature of Light

Lessons

Senior 5 · Physics

Wave and Particle Nature of Light

5 lessons

Lesson 2 of 5

Photons and the Photoelectric Effect

25 min

A spark that liked ultraviolet

In 1887 Heinrich Hertz, testing Maxwell's theory with spark gaps, noticed that a spark jumped more easily when ultraviolet light from another spark fell on the gap. A year later Wilhelm Hallwachs showed the cause: a clean zinc plate, charged negatively and connected to an electroscope, loses its charge when ultraviolet light shines on it, but a positively charged plate does not, and a sheet of glass, which absorbs ultraviolet, stops the effect. Light was knocking electrons out of the metal. Ten years of careful measurement then produced results that the wave theory of light could not explain. Einstein's explanation, in 1905, treated light as a stream of particles and won him the 1921 Nobel Prize in Physics.

The photon

Einstein went further than Planck. Planck had quantised the energy of the oscillators in the walls of a cavity; Einstein proposed that the radiation itself travels as separate packets, later called photons. Each photon of light of frequency $f$ and wavelength $\lambda$ carries energy

E = hf = \frac{hc}{\lambda}

and, although it has no rest mass, a momentum

p = \frac{E}{c} = \frac{h}{\lambda}

A beam of light is a stream of photons. Its intensity (power per unit area) depends on how many photons arrive per second; the energy of each photon depends only on the frequency. Bright red light is many small packets; dim ultraviolet light is a few large ones.

Atomic-scale energies are usually given in electronvolts: $1\,\mathrm{eV} = 1.60 \times 10^{-19}\,\mathrm{J}$ , the energy gained by an electron accelerated through 1 V. A useful shortcut is $hc = 1240\,\mathrm{eV\,nm}$ , so a photon's energy in eV is $\frac{1240}{\lambda\text{ in nm}}$ : about 3.1 eV for violet (400 nm) and 1.8 eV for red (700 nm).

Definition

A photon is a quantum of electromagnetic radiation: an indivisible packet of energy $E = h f$ and momentum $p = \frac{h}{\lambda}$ , travelling at $c$ in a vacuum. Example: a photon of yellow-green light, $\lambda = 560$ nm, has $E = 3.55 \times 10^{-19}$ J $= 2.22$ eV and $p = 1.18 \times 10^{-27}\,\mathrm{kg\,m\,s^{-1}}$ .

Worked example 1: counting photons from a phone torch

The white LED of a phone torch emits about 0.20 W of visible light. Taking an average wavelength of 560 nm, how many photons does it emit each second?

Solution

Energy per photon: $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{560 \times 10^{-9}} = 3.55 \times 10^{-19}\,\mathrm{J}$ .

Photons per second: $N = \frac{P}{E} = \frac{0.20}{3.55 \times 10^{-19}} = 5.6 \times 10^{17}\,\mathrm{s^{-1}}$ .

The number is so large that the light looks perfectly continuous; its graininess shows only when the light is extremely faint or the detector extremely sensitive.

The photoelectric effect: what experiments show

The photoelectric effect is the emission of electrons from a surface, usually a metal, when electromagnetic radiation falls on it. The emitted electrons are called photoelectrons. It is studied with a vacuum photocell like the one below. Light falls on the emitter; photoelectrons cross the vacuum to the collector, and the microammeter measures the current. The potential of the collector can be made positive (to attract the electrons) or negative (to repel them).

Measurements with this apparatus show five facts:

Threshold frequency. For each metal there is a minimum frequency $f_{0}$ . Below it no electrons are emitted, however intense the light and however long it shines.
Instant emission. Above $f_{0}$ , emission begins without measurable delay (less than $10^{-9}$ s), even in very dim light.
Energy depends on frequency. The electrons leave with a range of kinetic energies up to a maximum $E_{k,\max}$ , which increases linearly with frequency and does not depend on the intensity.
Number depends on intensity. At a fixed frequency above $f_{0}$ , the photocurrent is proportional to the intensity.
Stopping potential. Making the collector negative reduces the current; it falls to zero at the stopping potential $V_{s}$ , which depends on the frequency and the metal but not on the intensity.

Why waves cannot explain it

Observation	Wave-theory prediction	Photon explanation
Threshold frequency	Any frequency works if the light is bright enough or shines long enough	One photon gives all its energy to one electron; if $hf \lt \phi$ , no single photon can free an electron
No time delay	In dim light an electron needs a long time to soak up enough energy	The first photon to arrive with $hf \gt \phi$ frees an electron at once
$E_{k,\max}$ independent of intensity	Brighter light means a bigger wave amplitude, so faster electrons	Brighter light means more photons, each with the same energy $h f$
$E_{k,\max}$ rises with frequency	No reason for frequency to matter	$E_{k,\max} = hf - \phi$ rises linearly with $f$

Einstein's photoelectric equation

An electron inside a metal is held by the attraction of the positive ions. The least energy needed to remove an electron from the surface is the work function $\phi$ of the metal. When a photon is absorbed, all its energy $h f$ goes to a single electron. Part of it, at least $\phi$ , is used to escape; the rest is kinetic energy. Electrons from just at the surface lose only $\phi$ and leave with the most kinetic energy, so energy conservation gives

hf = \phi + E_{k,\max} = \phi + \tfrac{1}{2} m_e v_{\max}^2

Electrons from deeper in the metal lose more energy on the way out, which is why there is a spread of energies below the maximum.

At the threshold frequency the electron just escapes with zero kinetic energy: $hf_0 = \phi$ , so

f_0 = \frac{\phi}{h}, \qquad \lambda_0 = \frac{hc}{\phi}

To stop even the fastest electrons, the collector must be negative enough that the work done against the field, $e V_{s}$ , equals their kinetic energy: $eV_s = E_{k,\max}$ . Substituting,

V_s = \frac{h}{e} f - \frac{\phi}{e}

A graph of $V_{s}$ against $f$ is a straight line of gradient $\frac{h}{e}$ , the same for every metal, which cuts the frequency axis at $f_{0}$ . Robert Millikan, who expected to prove Einstein wrong, measured these lines carefully in 1916 and found $h$ to within about 1%.

Metal	Work function $\phi$ / eV	Threshold wavelength $\lambda_0$ / nm
Caesium	2.1	590
Sodium	2.28	544
Calcium	2.9	430
Zinc	4.3	290
Copper	4.7	260

These are typical values; a measured work function depends on how clean the surface is. Only the alkali metals respond to visible light. Zinc and copper need ultraviolet, which is why Hallwachs's zinc plate needed an ultraviolet lamp.

Definition

The work function $\phi$ of a metal is the minimum energy needed to remove an electron from its surface. The threshold frequency $f_0 = \frac{\phi}{h}$ is the lowest frequency that can cause photoemission. The stopping potential $V_{s}$ is the potential difference that just stops the most energetic photoelectrons, so $eV_s = E_{k,\max}$ . Example: sodium, $\phi = 2.28$ eV, has $f_0 = 5.50 \times 10^{14}$ Hz.

Worked example 2: violet light on sodium

Light of wavelength 400 nm falls on a sodium surface ( $\phi = 2.28$ eV). Find (a) the photon energy in eV, (b) the maximum kinetic energy of the photoelectrons, (c) the stopping potential and (d) the maximum speed of the photoelectrons.

Solution

(a) $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{400 \times 10^{-9}} = 4.97 \times 10^{-19}\,\mathrm{J} = \frac{4.97 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.11\,\mathrm{eV}$ .

(b) $E_{k,\max} = hf - \phi = 3.11 - 2.28 = 0.83\,\mathrm{eV}$ .

(d) $E_{k,\max} = 0.83 \times 1.60 \times 10^{-19} = 1.32 \times 10^{-19}$ J, so $v_{\max} = \sqrt{\frac{2E_{k,\max}}{m_e}} = \sqrt{\frac{2(1.32 \times 10^{-19})}{9.11 \times 10^{-31}}} = 5.4 \times 10^{5}\,\mathrm{m\,s^{-1}}$ .

Worked example 3: finding $h$ and $\phi$ from measurements

In a school experiment the stopping potential for a metal is 0.82 V at $7.0 \times 10^{14}$ Hz and 1.65 V at $9.0 \times 10^{14}$ Hz. Find (a) Planck's constant, (b) the threshold frequency and (c) the work function in eV.

Solution

(a) The gradient of $V_{s}$ against $f$ is $\frac{h}{e} = \frac{1.65 - 0.82}{(9.0 - 7.0) \times 10^{14}} = 4.15 \times 10^{-15}\,\mathrm{V\,s}$ , so $h = (4.15 \times 10^{-15})(1.60 \times 10^{-19}) = 6.64 \times 10^{-34}\,\mathrm{J\,s}$ .

(b) At threshold $V_{s} = 0$ : $f_0 = 7.0 \times 10^{14} - \frac{0.82}{4.15 \times 10^{-15}} = 5.0 \times 10^{14}\,\mathrm{Hz}$ .

(c) $\phi = hf_0 = (6.64 \times 10^{-34})(5.02 \times 10^{14}) = 3.33 \times 10^{-19}\,\mathrm{J} = 2.1\,\mathrm{eV}$ , close to caesium's value.

Warning

Units trap. Mixing joules and electronvolts in $hf = \phi + E_{k,\max}$ is the commonest error in this topic. Convert everything to one unit first. And remember: $E_{k,\max}$ is the maximum; most photoelectrons leave with less.

Warning

Safety. Ultraviolet lamps used to show the photoelectric effect with zinc emit UV-C, which burns the eyes and skin within minutes. Never look at the lamp, shield it, and use it only under your teacher's direct supervision.

Tip

One photon, one electron. Frequency decides whether and how fast electrons come out; intensity decides how many.

Photoelectric Effect Demonstration, National STEM Centre. Watch the charged zinc plate: ultraviolet light discharges it only when it is negative, and a sheet of glass stops the effect.

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Practice

Find the energy, in eV, of a photon of wavelength 620 nm, and its momentum.
State two observations of the photoelectric effect that the wave theory of light cannot explain, and explain each with the photon model.
The work function of calcium is 2.9 eV. Find its threshold frequency and threshold wavelength.
Ultraviolet light of wavelength 250 nm falls on copper ( $\phi = 4.7$ eV). Find the stopping potential.
Light of wavelength 450 nm falls on a metal and the stopping potential is 0.66 V. Find the work function of the metal in eV.
A 2.0 mW beam of 480 nm light falls on a caesium surface. How many photons arrive each second?
A pupil doubles the intensity of the light on a photocell and also changes its colour from green to blue. Predict, with reasons, what happens to (a) the stopping potential, (b) the saturation current.
Explain why the $V_{s}$ – $f$ lines for all metals are parallel, and what the vertical intercept represents.

Show answersHide answers

$E = \frac{1240}{620} = 2.0$ eV; $p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{620 \times 10^{-9}} = 1.07 \times 10^{-27}\,\mathrm{kg\,m\,s^{-1}}$ .
Any two of: the threshold frequency (one photon frees one electron, so $h f$ must exceed $\phi$ ); no time delay (energy arrives in whole packets, not gradually); $E_{k,\max}$ independent of intensity (more photons, not bigger ones); $E_{k,\max}$ rising with frequency ( $E_{k,\max} = hf - \phi$ ).
$f_0 = \frac{(2.9)(1.60 \times 10^{-19})}{6.63 \times 10^{-34}} = 7.0 \times 10^{14}$ Hz; $\lambda_0 = \frac{1240}{2.9} = 430$ nm.
$E = \frac{1240}{250} = 4.96$ eV; $V_{s} = 4.96 - 4.7 = 0.26$ V.
$E = \frac{1240}{450} = 2.76$ eV; $\phi = 2.76 - 0.66 = 2.10$ eV.
$E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{480 \times 10^{-9}} = 4.14 \times 10^{-19}$ J; $N = \frac{2.0 \times 10^{-3}}{4.14 \times 10^{-19}} = 4.8 \times 10^{15}\,\mathrm{s^{-1}}$ .
(a) Increases: blue photons have higher $h f$ , so $E_{k,\max}$ and $V_{s}$ rise; intensity has no effect on it. (b) The saturation current depends on the number of photons per second. Doubling the intensity at a higher photon energy increases the photon rate, but by less than a factor of two, because each blue photon carries more energy; so the current increases, by somewhat less than double.
The gradient is $\frac{h}{e}$ , a universal constant, so it is the same for every metal. The vertical intercept (found by extending the line) is $-\frac{\phi}{e}$ , giving the work function.

Key points

Light travels as photons: $E = hf = \frac{hc}{\lambda}$ , $p = \frac{h}{\lambda}$ ; $h c = 1240$ eV nm.
Photoemission needs $f \ge f_0 = \frac{\phi}{h}$ ; it is instant; $E_{k,\max}$ depends on $f$ , the current on intensity.
Einstein: $hf = \phi + E_{k,\max}$ and $eV_s = E_{k,\max}$ .
$V_{s}$ against $f$ is a straight line of gradient $\frac{h}{e}$ with intercept $f_{0}$ on the frequency axis.

Sources and credits

Video: Photoelectric Effect Demonstration, National STEM Centre, embedded from YouTube.
Work functions: typical published values, rounded; they vary with surface condition.

Written by Ganzaa. The apparatus diagram, the graph (drawn to scale from Einstein's equation), the worked examples and the practice exercises are by Ganzaa.

1A spark that liked ultraviolet

2The photon

3The photoelectric effect: what experiments show

4Why waves cannot explain it

5Einstein's photoelectric equation

A spark that liked ultraviolet

The photon

The photoelectric effect: what experiments show

Why waves cannot explain it

Einstein's photoelectric equation