Energy-Momentum Relation (Edexcel A Level Physics)

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Katie M


Katie M



Deriving the Energy-Momentum Relation

  • The equation for calculating the kinetic energy Ek of a particle m moving at velocity v is given by: 

E subscript k equals 1 half m v squared

  • The formula for the momentum p of the same particle is:

p space equals space m v

  • Combining these gives an equation that links kinetic energy to momentum, called the energy-momentum relation
    • Firstly, substituting the equation for velocity v equals p over minto the equation for kinetic energy gives:

E subscript k equals 1 half m open parentheses p over m close parentheses squared

    • Multiplying brackets out and simplifying gives: 

E subscript k equals 1 half m p squared over m squared equals 1 half p squared over m

  • Therefore the energy-momentum is presented as:

Ekfraction numerator p squared space over denominator 2 m end fraction

  • Where:
    • Ek = kinetic energy (J)
    • p = momentum (kg m s-1)
    • m = mass (kg)

Exam Tip

This is a common derivation, so make sure you're comfortable with deriving this from scratch! Think carefully about the algebra on each step.

Using the Energy-Momentum Relation

  • The energy-momentum relation is particularly useful for:
    • Calculations involving the kinetic energy of subatomic particles travelling at non-relativistic speeds (i.e. much slower than the speed of light)
    • Projectiles and collisions involving large masses 

Worked example

Calculate the kinetic energy, in MeV, of an alpha particle which has a momentum of 1.1 × 10–19 kg m s–1.

Use the following data:

  • Mass of a proton = 1.67 × 10–27 kg
  • Mass of a neutron = 1.67 × 10–27 kg

Step 1: Write the energy-momentum relation

    • The energy-momentum relation is given by Ek fraction numerator p squared over denominator 2 m end fraction

Step 2: Determine the mass of an alpha particle

    • An alpha particle is comprised of two protons and two neutrons
    • Therefore, the mass of an alpha particle mα = 2mp + 2mn, where mp and mn is the mass of a proton and neutron respectively
    • So mα = 2(1.67 × 10–27) + 2(1.67 × 10–27) = 6.68 × 10–27 kg

Step 3: Substitute the momentum and the mass of the alpha particle into the energy-momentum relation

Ek fraction numerator p squared over denominator 2 m end fraction

Ekfraction numerator left parenthesis 1.1 cross times 10 to the power of negative 19 end exponent right parenthesis squared over denominator 2 cross times left parenthesis 6.68 cross times 10 to the power of negative 27 end exponent right parenthesis end fraction= 9.1 × 10–13 J

Step 4: Convert the value of kinetic energy from J to MeV

1 MeV = 1.6 × 10–13 J

    • Therefore:

9.1 × 10–13 J = fraction numerator 9.1 cross times 10 to the power of negative 13 end exponent over denominator 1.6 cross times 10 to the power of negative 13 end exponent end fractionMeV = 5.7 MeV

Exam Tip

Calculations with the energy-momentum equation often require changing units, especially between eV and J due to it commonly being used for particles. Remember that 1 eV = 1.60 × 10-19 J. Therefore

eV → J = × (1.60 × 10-19)

J → eV = ÷ (1.60 × 10-19)

The prefix 'mega' (M) means × 106 therefore, 1 MeV = (1.60 × 10-19) × 10= 1.60 × 10-13

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Katie M

Author: Katie M

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.