The Photoelectric Equation (AQA A Level Physics)

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Katie M

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Katie M

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The Photoelectric Equation

  • The energy of a photon is given as:

E = hf

  • Photons of frequencies above the threshold frequency will have more energy than just the work function

    • An amount of energy equal to the work function is used to release the photoelectron from the metal

    • The remaining energy will be transferred as kinetic energy to the photoelectron

  • This equation is known as the photoelectric equation:

hf = Φ + Ek max

  • Which can also be written as:

E = hf = Φ + ½ mv2max

  • Where:

    • h = Planck's constant (J s)

    • f = the frequency of the incident radiation (Hz)

    • Φ = the work function of the material (J)

    • ½ mv2max Ek(max) = the maximum kinetic energy of the photoelectrons (J)

    • hf is equal to the energy of a single photon

 

  • 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

 hf0 = Φ

  • Where f0 = threshold frequency, photoelectric emission only just occurs

    • Ek(max) 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 Ek(max)

Graphical Representation of Work Function

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

y = mx + c

  • Comparing this to the photoelectric equation:

Ek(max) = hf - Φ

  •  A graph of maximum kinetic energy Ek(max) against frequency f can be obtained

  • The key elements of the graph:

    • 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

Worked Example

The graph below shows how the maximum kinetic energy Ek 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.

Answer:

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

E = hf = Φ + ½ mv2max         →    Ek(max) = hf - Φ

y = mx + c

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

When Ek = 0, f = f0

Therefore, the threshold frequency is f0 = 4 × 1014 Hz

Step 3: Calculate the work function

From the graph at f0, ½ mvmax2 = 0

Φ = hf0 = (6.63 × 10-34) × (4 × 1014) = 2.652 × 10-19 J

Step 4: Convert the work function into eV

1 eV = 1.6 × 10-19 J                 J → eV: divide by 1.6 × 10-19

straight E equals fraction numerator 2.652 cross times 10 to the power of negative 19 end exponent over denominator 1.6 cross times 10 to the power of negative 19 end exponent end fraction equals 1.66 space eV

Examiner Tips and Tricks

When using the photoelectric effect equation, hfΦ and Ek(max) must all have the same units (joules). Therefore make sure to convert any values given in eV into Joules (× (1.6 × 10-19))

Maximum Kinetic Energy

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 the rate of energy transferred per unit area and is related to the number of photons striking the metal plate

  • Increasing the number of photons striking the metal will not increase the kinetic energy of the photoelectrons; it will increase the number of photoelectrons emitted

Why is the Kinetic Energy a Maximum?

  • Each electron in the metal acquires the same amount of energy from the photons in the incident radiation for any given frequency

  • 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 as 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 a measure of the number of photoelectrons emitted per second

    • The value of the photoelectric current is calculated by the number of electrons emitted multiplied by the charge on one electron

  • Photoelectric current is proportional to the intensity of the radiation incident on the surface of the metal

  • This is because 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

2-4-3-ke-photocurrent-graphs

Sketch graphs showing the trends in the variation of electron KE with the frequency and intensity of the incident light and the variation of photocurrent with the intensity of the incident light

Examiner Tips and Tricks

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 space equals fraction numerator space P over denominator A end fraction space equals fraction numerator space E over denominator t A end fraction

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

I space equals fraction numerator space h f over denominator t A end fraction

So to account for n number of photons:

I space equals fraction numerator space n h f over denominator t A end fraction

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

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.