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Energy & Momentum of a Photon (Cambridge (CIE) A Level Physics)

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Ashika

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Ashika

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Calculating photon energy

  • The energy of a photon can be calculated using the formula:

E space equals space h f

  • Where:

    • E = energy of a photon (J)

    • h = Planck's constant(J s)

    • f = frequency (Hz)

Photon representation

12-1-1-photons-ib-hl

A photon in a particle of light carrying discrete packets of energy

  • Using the wave equation f space equals space c over lambda, energy can also be equal to:

E space equals space fraction numerator space h c over denominator lambda end fraction

  • Where:

    • c = the speed of light (m s-1)

    • λ = wavelength (m)

  • This equation tells us:

    • The higher the frequency of EM radiation, the higher the energy of the photon

    • The energy of a photon is inversely proportional to the wavelength

    • Therefore, a long-wavelength photon of light has a lower energy than a shorter-wavelength photon

 

Worked Example

Light of wavelength 490 nm is incident normally on a surface, as shown in the diagram.

The power of the light is 3.6 mW. The light is completely absorbed by the surface.

Calculate the number of photons incident on the surface in 2.0 s.

Answer

Step 1: Write down the known quantities

  • Wavelength, λ = 490 nm = 490 × 10-9 m

  • Power, P = 3.6 mW = 3.6 × 10-3 W

  • Time, t = 2.0 s

Step 2: Write the equation for photon energy in terms of wavelength

E space equals fraction numerator space h c over denominator lambda end fraction

Step 3: Calculate the energy of one photon

E space equals space fraction numerator open parentheses 6.63 space cross times space 10 to the power of negative 34 end exponent close parentheses space cross times space open parentheses 3.0 space cross times space 10 to the power of 8 close parentheses space over denominator 490 space cross times space 10 to the power of negative 9 end exponent end fraction space equals space 4.06 space cross times space 10 to the power of negative 19 end exponent space straight J

Step 4: Calculate the number of photons hitting the surface every second

fraction numerator P o w e r space o f space l i g h t space s o u r c e over denominator E n e r g y space o f space o n e space p h o t o n end fraction space equals space fraction numerator 3.6 space cross times space 10 to the power of negative 3 end exponent space over denominator 4.06 space cross times space 10 to the power of negative 19 end exponent end fraction space equals space 8.9 space cross times space 10 to the power of 15 space straight s to the power of negative 1 end exponent

Step 5: Calculate the number of photons that hit the surface in 2 s

open parentheses 8.9 space cross times space 10 to the power of 15 close parentheses space cross times space 2 space equals space 1.8 space cross times space 10 to the power of 16

Examiner Tips and Tricks

The values of Planck’s constant and the speed of light are both given on your data sheet, however, it helps to memorise them to speed up calculation questions.

Remember this equation for E is only for the energy of a photon, not the energy for any other particle!

Photon momentum

  • Einstein showed that a photon travelling in a vacuum has momentum, despite it having no mass

  • The momentum of a photon is related to its energy by the equation:

p space equals space fraction numerator space E over denominator c end fraction

  • Where:

    • p = momentum (kg m s–1)

    • E = energy of a photon (J)

    • c = speed of light

Worked Example

A 5.0 mW laser beam is incident normally on a fixed metal plate. The cross-sectional area of the beam is 8.0 × 10-6 m2. The light from the laser has a frequency of 5.6 × 1014 Hz.

Assuming that all the photons are absorbed by the plate, calculate the momentum of the photon, and the pressure exerted by the laser beam on the metal plate.

Answer:

Step 1: Write down the known quantities

  • Power, P = 5.0 mW = 5.0 × 10-3 W

  • Frequency, f = 5.6 × 1014 Hz

  • Cross-sectional area, A = 8.0 × 10-6 m2

Step 2: Write the equations for photon energy and momentum

p space equals fraction numerator space E over denominator c end fraction space space space rightwards double arrow space space space p space equals fraction numerator space h f over denominator c end fraction

Step 3: Calculate the photon momentum

p space equals space fraction numerator space h f over denominator c end fraction space equals space fraction numerator open parentheses 6.63 space cross times space 10 to the power of negative 34 end exponent close parentheses space cross times space open parentheses 5.6 space cross times space 10 to the power of 14 close parentheses space over denominator 3.0 space cross times space 10 to the power of 8 end fraction space equals space 1.24 space cross times space 10 to the power of negative 27 end exponent space straight N space straight s

Step 4: Calculate the number of photons incident on the plate every second

fraction numerator P o w e r space o f space l i g h t space s o u r c e space over denominator E n e r g y space o f space o n e space p h o t o n end fraction space equals space fraction numerator 5.0 space cross times space 10 to the power of negative 3 end exponent space over denominator h f end fraction space equals space fraction numerator 5.0 space cross times space 10 to the power of negative 3 end exponent space over denominator open parentheses 6.63 space cross times space 10 to the power of negative 34 end exponent close parentheses space cross times space open parentheses 5.6 space cross times space 10 to the power of 14 close parentheses end fraction space equals space 1.35 space cross times space 10 to the power of 16 space straight s to the power of negative 1 end exponent

Step 5: Calculate the force exerted on the plate in a 1.0 s time interval

  • Force is the rate of change in momentum

F space equals space fraction numerator space increment p space over denominator increment t end fraction

F  = number of photons per second × momentum of each photon

F space equals space left parenthesis 1.35 space cross times space 10 to the power of 16 right parenthesis space cross times space left parenthesis 1.24 space cross times space 10 to the power of negative 27 end exponent right parenthesis space equals space 1.67 space cross times space 10 to the power of negative 11 end exponent space straight N

 Step 6: Calculate the pressure

  • Pressure (P) is the force per unit area

P space equals fraction numerator space F over denominator A end fraction space equals space fraction numerator 1.67 space cross times space 10 to the power of negative 11 end exponent over denominator 8.0 space cross times space 10 to the power of negative 6 end exponent end fraction space equals space 2.1 space cross times space 10 to the power of negative 6 end exponent space Pa

Examiner Tips and Tricks

It's not unusual for multiple equations to be required for a question involving the photon momentum. 

Always watch out of the units! For the momentum p to be in kg m s–1, the energy must be in J and the speed of light in m s–1.

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    Ashika

    Author: Ashika

    Expertise: Physics Project Lead

    Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.