The Photoelectric Equation (HL) | HL IB Physics Revision Notes 2025

The Photoelectric Equation

The energy possessed by a photon is equal to

$E = h f$

The fraction of the energy transferred to the electron to release it from the metal is equal to the work function $Φ$
The remaining amount is given as kinetic energy to the emitted photoelectron

$E_{K m a x} = \frac{1}{2} m_{e} {v_{m a x}}^{2}$

Since energy is always conserved, the energy of a photon incident on the surface of a metal is equal to:

The work function + the maximum kinetic energy of the photoelectron

This equation is known as the photoelectric equation:

$E = h f = Φ + \frac{1}{2} m_{e} {v_{m a x}}^{2}$

Where:
- h = Planck's constant (J s)
- f = frequency of the incident radiation (Hz)
- $Φ$ = work function of the metal (J)
- $E_{K m a x}$ = maximum kinetic energy of a photoelectron (J)
- $m_{e}$ = mass of an electron (kg)
- $v_{m a x}$ = maximum velocity of a photoelectron (m s⁻¹)

This equation demonstrates:
- If the incident photons do not have a high enough frequency and energy to overcome the work function (Φ), then no electrons will be emitted
- When $E = h f_{0} = Φ$ , where $f_{0}$ = threshold frequency, photoelectric emission only just occurs
- The maximum kinetic energy $E_{K m a x}$ depends only on the frequency of the incident photon, and not the intensity of the radiation
- The majority of photoelectrons will have kinetic energies less than $E_{K m a x}$

Graphical Representation of Work Function

The photoelectric equation can be rearranged into the straight-line equation:

$y = m x + c$

Comparing this to the photoelectric equation:

$E_{K m a x} = h f - Φ$

A graph of maximum kinetic energy $E_{K m a x}$ against frequency $f$ can be obtained

The key elements of the graph are:
- The work function Φ is the y-intercept
- The threshold frequency f₀ is the x-intercept
- The gradient is equal to Planck's constant h
- There are no electrons emitted below the threshold frequency f₀

Kinetic Energy & Intensity

The kinetic energy of the photoelectrons is independent of the intensity of the incident radiation
- This is because each electron can only absorb one photon
- Kinetic energy is only dependent on the frequency of the incident radiation
Intensity is a measure of the number of photons incident on the surface of the metal
- So, increasing the number of electrons striking the metal will not increase the kinetic energy of the electrons, it will increase the number of photoelectrons emitted

Why Kinetic Energy is a Maximum

Each electron in the metal acquires the same amount of energy from the photons in the incident monochromatic radiation.
However, the energy required to remove an electron from the metal varies because some electrons are on the surface whilst others are deeper in the metal
- The photoelectrons with the maximum kinetic energy will be those on the surface of the metal since they do not require much energy to leave the metal
- The photoelectrons with lower kinetic energy are those deeper within the metal since some of the energy absorbed from the photon is used to approach the metal surface (and overcome the work function)
- There is less kinetic energy available for these photoelectrons once they have left the metal

Photoelectric Current

The photoelectric current is the number of photoelectrons emitted per second
Photoelectric current is proportional to the intensity of the radiation incident on the surface of the metal
This is because the intensity is proportional to the number of photons striking the metal per second
Since each photoelectron absorbs a single photon, the photoelectric current must be proportional to the intensity of the incident radiation

KE & Photocurrent Graphs, downloadable AS & A Level Physics revision notes

Graphs showing the variation of electron KE and photocurrent with the frequency of the incident light

Stopping Voltage

Stopping voltage $V_{S}$ is defined as:

The voltage required to stop photoelectron emission from occurring

The photons arriving at the metal plate cause photoelectrons to be emitted
- This is called the emitter plate
The electrons that cross the gap are collected at the other metal plate
- This is called the collector plate

Stopping Potential, downloadable AS & A Level Physics revision notes

This setup can be used to determine the maximum kinetic energy of the emitted photoelectrons

The flow of electrons across the gap sets up an e.m.f. between the plates that allows a current to flow around the rest of the circuit
- Effectively, it becomes a photoelectric cell which produces a photoelectric current

If the e.m.f. of the variable power supply is initially zero, the circuit operates only on the photoelectric current
As the supply is turned up, the emitter plate becomes more positive
- This is because it is connected to the positive terminal of the supply

As a result, electrons leaving the emitter plate are attracted back towards it
- This is because the p.d. across the tube opposes the motion of the electrons between the plates

If any electrons escape with high enough kinetic energy, they can overcome this attraction and cross to the collector plate
- And if they don't have enough energy, they can't cross the gap

By increasing the e.m.f. of the supply, eventually, a p.d. will be reached at which no electrons will be able to cross the gap
- This value of e.m.f. is equal to the stopping voltage $V_{S}$

At this point, the energy needed to cross the gap is equal to the maximum kinetic energy $E_{K m a x}$ of the electrons

$E_{K m a x} = e V_{S}$

Where:
- $E_{K m a x}$ = maximum kinetic energy of the electrons (J)
- $e$ = elementary charge (C)
- $V_{S}$ = stopping voltage (V)

Worked example

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.

5-2-2-work-function-graph-worked-example
Calculate the work function of sodium in eV.

Answer:

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

$h f = Φ + E_{K m a x}$

$E_{K m a x} = h f - Φ$

$y = m x + c$

Therefore, when $E_{K} = 0$ , $h f = Φ$ and $f = f_{0}$

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

From the graph:
- When $E_{K} = 0$ , threshold frequency: $f = f_{0}$ = 4 × 10¹⁴ Hz

Step 3: Calculate the work function

$Φ = h f_{0} = (6.63 \times 10^{- 34}) \times (4 \times 10^{14})$

Work function: $Φ$ = 2.652 × 10⁻¹⁹ J

Step 4: Convert the work function into eV

To convert from J to eV: divide by 1.6 × 10⁻¹⁹ J

$E = \frac{2.652 \times 10^{- 19}}{1.6 \times 10^{- 19}}$ = 1.66 eV

Worked example

Monochromatic light of wavelength $λ_{1}$ is incident on the surface of a metal. The stopping voltage for this light is $V_{1}$ .

When another monochromatic light of wavelength $λ_{2}$ is incident on the same surface, the stopping voltage is $V_{2}$ .

What is the quantity $\frac{(V_{2} - V_{1})}{(\frac{1}{λ_{2}} - \frac{1}{λ_{1}})}$ equal to?

A. $\frac{h}{2 π}$ B. $\frac{h}{c}$ C. $\frac{h c}{e}$ D. $\frac{h}{e}$

Answer: C

The photoelectric equation for light 1 is

$h f_{1} = Φ + E_{K m a x (1)}$

Where $E_{K m a x (1)} = e V_{1}$ and $E_{1} = h f_{1} = \frac{h c}{λ_{1}}$

The photoelectric equation for light 2 is

$h f_{2} = Φ + E_{K m a x (2)}$

Where $E_{K m a x (2)} = e V_{2}$ and $E_{2} = h f_{2} = \frac{h c}{λ_{2}}$

Since the metal is the same, the work function $Φ$ is the same for both, so:

$Φ = \frac{h c}{λ_{1}} - e V_{1} = \frac{h c}{λ_{2}} - e V_{2}$

Collecting the terms together and simplifying gives

$e V_{2} - e V_{1} = \frac{h c}{λ_{2}} - \frac{h c}{λ_{1}}$

$e (V_{2} - V_{1}) = h c (\frac{1}{λ_{2}} - \frac{1}{λ_{1}})$

$\frac{(V_{2} - V_{1})}{(\frac{1}{λ_{2}} - \frac{1}{λ_{1}})} = \frac{h c}{e}$

Exam Tip

When using the photoelectric effect equation, hf, Φ and E_k(max) must all have the same units; Joules.

All values given in eV need to be converted into Joules by multiplying by 1.6 × 10⁻¹⁹. Do this right away, in the same way as you would convert into SI units before calculating.

It is important to note that the stopping voltage actually holds a negative value, but since we use it to determine the maximum kinetic energy of the emitted electrons, its sign is not important in calculations, it's acceptable to just quote its magnitude.

The Photoelectric Equation (HL) (HL IB Physics)

Revision Note