X-rays and the Photoelectric Effect

Photons and the photoelectric effect

Shine light on a clean metal surface and, under the right conditions, electrons are emitted. This is photoelectric emission. The surprising part is which light works.

• Ultraviolet light on zinc frees electrons, however dim it is.

• Red light on the same zinc frees none, however bright it is.

There is a minimum frequency below which nothing happens.

The wave model of light cannot explain this. A wave spreads its energy over the surface, so a bright enough red beam should eventually give an electron enough energy to escape. It does not. The particle model does explain it: light arrives in packets called photons, and each photon has energy

where $h = 6.63\times10^{-34}\ \text{J s}$ is Planck's constant and $f$ is the frequency. One photon gives all its energy to one electron, or none at all. The electron cannot save up energy from several weak photons.

If $hf$ is bigger than $\Phi$, the extra energy becomes kinetic energy of the electron. This is conservation of energy:

The electron has maximum kinetic energy when it comes from the surface. Electrons from deeper in the metal lose some energy on the way out.

Energies of electrons are very small in joules, so we often use the electronvolt: $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.

Stopping potential and finding h

How do we measure the maximum kinetic energy of the electrons? We make the collecting plate negative so that it repels them. The smallest p.d. that stops even the fastest electrons reaching it is the stopping potential, $V_s$. The electrons lose their kinetic energy doing work against the field, so

Rearranged as $V_s = \dfrac{h}{e}f - \dfrac{\Phi}{e}$, this is a straight line, $y = mx + c$.

• The line meets the frequency axis at $f_0$, the threshold frequency.

• The slope is $\dfrac{h}{e}$, so $h = e\times\text{slope}$.

• The line extended back to $f = 0$ meets the $V_s$ axis at $-\dfrac{\Phi}{e}$.

The slope is the same for every metal. Only $f_0$ changes, so different metals give parallel lines.

You can verify the photoelectric effect without a lab: take published $V_s$ and $f$ data for a metal from a secondary source (a textbook table or a university website), plot $V_s$ against $f$ and check that the points lie on a straight line. Then compare the slope $\times\, e$ with the accepted value of $h$.

Photocells

A photocell (photoelectric cell) uses the photoelectric effect. Light falls on a metal cathode inside an evacuated tube. Emitted electrons are collected by the anode, so a small current flows in the circuit. This current, the photocurrent, is proportional to the light intensity as long as $f > f_0$.

Uses of photocells:

• automatic doors, which open when a beam is broken

• burglar alarms and counters on a production line

• light meters in cameras

• reading the sound track on old cinema film

To trigger a door or alarm, the circuit detects the photocurrent dropping to zero when something blocks the beam.

Producing X-rays

X-rays are electromagnetic radiation with very high frequency (wavelengths of about $10^{-11}$ to $10^{-9}\ \text{m}$). They are produced when fast electrons are stopped by a metal target.

In an X-ray tube:

1. A heated filament (the cathode) emits electrons by thermionic emission.

2. A very high p.d. (tens of kV) accelerates the electrons towards the metal target (the anode), in a vacuum.

3. The electrons strike the target and are rapidly stopped.

Energy transformation at the target. The kinetic energy of the electrons becomes:

• X-ray photons (only about 1%)

• heat (about 99%)

This is why the target is made of a metal with a high melting point (such as tungsten) and is cooled. Modern machines use a rotating anode, so the electron beam strikes a different part of the target all the time and the heat is spread over a larger area.

Maximum photon energy. An electron accelerated through a p.d. $V$ gains kinetic energy $eV$. The most energetic photon is made when an electron gives all of its energy to one photon:

So a higher accelerating voltage gives a higher maximum frequency and a shorter minimum wavelength, which means more penetrating X-rays.

The X-ray spectrum

If we plot intensity against wavelength we see two features:

• A sharp cut-off at $\lambda_{min}$. No photon can have more energy than the electron that made it, $eV$. So there is a shortest wavelength, set only by the voltage.

• A continuous background. Electrons slow down in many stages, in collisions of different strength, so photons have all energies up to the maximum.

• Sharp peaks. An incoming electron can knock an inner electron out of a target atom. When an outer electron drops into the gap, it emits a photon of a definite energy. These peaks depend on the target metal, not on the voltage.

Uses of X-rays. In medicine: imaging bones (X-ray radiographs) and CT scans, and radiotherapy for cancer. In industry: checking welds and cracks in metal, and airport security scanners. X-rays are ionising, so exposure is kept as low as possible.

Two inverse processes

The two effects mirror each other:

In the photoelectric effect a photon's energy is turned into an electron's kinetic energy (after paying $\Phi$). In an X-ray tube the electron's kinetic energy is turned back into a photon. Both are explained only by the particle model of light, where energy is exchanged in whole photons of $hf$. The work function is negligible for X-rays, because $eV$ is thousands of times bigger.

Summary

• Light comes in photons of energy $E = hf$. One photon frees at most one electron.

• The work function is $\Phi = hf_0$. Photoelectric emission follows $hf = \Phi + \tfrac12mv^2_{max}$.

• Frequency controls the electrons' maximum kinetic energy, and intensity controls how many are emitted. Below $f_0$ nothing is emitted. The wave model cannot explain this.

• $eV_s = hf - \Phi$. A $V_s$ against $f$ graph has slope $h/e$ and meets the frequency axis at $f_0$.

• A photocell turns light into a photocurrent, which is useful in alarms, doors and light meters.

• In an X-ray tube, $eV = hf_{max} = hc/\lambda_{min}$. About 1% of the beam power becomes X-rays and the rest is heat, so the target is cooled.

• The X-ray spectrum has a continuous background, a sharp cut-off $\lambda_{min}$ and characteristic peaks.