The Photoelectric Equation (AQA AS Physics)

• 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 + E_{k max}

• Which can also be written as:

E = hf = Φ + ½ mv²_max

• Where:
  • h = Planck's constant (J s)
  • f = the frequency of the incident radiation (Hz)
  • Φ = the work function of the material (J)
  • ½ mv²_max = E_k(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
  • E_k(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 E_k(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:

E_k(max) = hf - Φ

•  A graph of maximum kinetic energy E_k(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

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

       straight line

E = hf = Φ + ½ mv²_max         →    E_k(max) = hf - Φ

y = mx + c

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

When E_k = 0, f = f₀

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

Step 3: Calculate the work function

From the graph at f₀, ½ mv_max² = 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

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

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