The Photoelectric Equation (AQA A Level Physics): Revision Note

The graph below shows how the maximum kinetic energy E_k of electrons emitted from the surface of sodium metal varies with the frequency f of the incident radiation.

Calculate the work function of sodium in eV.

[3]

Answer:

Step 1: Write out the photoelectric equation and rearrange to fit the equation of a straight line

$hf=\varphi +E_{k\left(max\right)}$

$E_{k\left(max\right)}=hf-\varphi$

$y=mx+c$

 Step 2: Identify the threshold frequency from the x-axis of the graph

• When $E_k=0$, $f=f_0$

• Therefore, the threshold frequency is $f_0=4\times {10}^{14}\mathrm{Hz}$ [1 mark]

Step 3: Calculate the work function

$0=h{f}_0-\varphi$

$\varphi =h{f}_0=\left(6.63\times {10}^{-34}\right)\times \left(4\times {10}^{14}\right)$

$\varphi =2.652\times {10}^{-19}\mathrm{J}$ [1 mark]

Step 4: Convert the work function into eV

1 eV = 1.6 × 10^-19 J

J → eV: divide by 1.6 × 10^-19

$\varphi =\frac{2.652\times {10}^{-19}}{1.6\times {10}^{-19}}=1.66\mathrm{eV}$ [1 mark]

When using the photoelectric effect equation, hf, Φ and E_k(max) must all have the same units (joules). Therefore make sure you convert any values given in eV into joules (× (1.6 × 10^-19))

Monochromatic light is incident on a metal surface in a vacuum, and photoelectrons are emitted from the surface. The photoelectric current $I$ is the rate of flow of charge from the surface. The maximum kinetic energy of the photoelectrons is $E_{k\left(max\right)}$. $E_{k\left(max\right)}$ and $I$ are measured.

The frequency of the light is then increased. There is no change to the rate at which energy is incident on the surface.

What happens to $E_{k\left(max\right)}$ and $I$ when the frequency is increased?

[1]

Answer:

Step 1: Determine the effect of increasing frequency on maximum kinetic energy

• The maximum kinetic energy of the photoelectrons is given by:

$hf=\Phi +E_{k\left(\text{max}\right)}$ → $E_{k\left(\text{max}\right)}=hf-\Phi$

• The work function $\Phi$ is a property of the metal and does not change

• So, increasing the frequency $f$ increases each photon's energy $hf$, and therefore $E_{k\left(max\right)}$ increases

Therefore, C and D can be ruled out

Step 2: Determine the effect of increasing frequency on photoelectric current

• Keeping the rate at which energy is incident on the surface the same means the intensity of the light is kept constant

• Each photon now carries more energy, so fewer photons arrive at the surface each second

• Each photon ejects at most one photoelectron, so the number of photoelectrons emitted per second falls, and therefore $I$ decreases

Therefore, the correct answer is A [1 mark]

Note: B is the trap — it relies on the shortcut "intensity controls current, frequency controls KE", which only works when intensity is varied by adding more photons of the same frequency. Here, intensity is kept constant while frequency changes, so the shortcut does not apply.

If you change the frequency of the incident light whilst keeping the number of photons emitted from the light source constant, then the photoelectric current will remain constant.

This is because changing the frequency will change the energy of the emitted photons, but the number of photons will remain the same.

If you change the frequency of the incident light whilst keeping the intensity constant, then the photoelectric current will change.

This is because intensity is power per unit area which is equal to the rate of energy transfer per unit area

$I=\frac{P}{A}=\frac{E}{tA}$

The energy transferred comes from the photons, where the energy of a single photon is hf

$I=\frac{hf}{tA}$

So to account for n number of photons:

$I=\frac{nhf}{tA}$

If the frequency f is increased and the intensity $I$ remains constant, then the number of photons n must decrease.

Planck's constant h and the area A of the metal plate do not change.

This is because at higher frequencies, each photon has a higher energy, so fewer photons are required to maintain the intensity.