The minimum kinetic energy required for ionization of a hydrogen atom is E₁ in case electron is collided with hydrogen atom. It is E₂ if the hydrogen ion is collided and E₃ when helium ion is collided. Then,

Assuming that ionization occurs as a result of a completely inelastic collision, we can write

$mv-0=\left(m+m_{\mathrm{H}}\right)u$

where m is the mass of incident particle, m_H the mass of hydrogen atom, v₀ the initial velocity of incident particle, and u the final common velocity of the particle after collision. Prior to collision, the KE of the incident particle was

$E_0=\frac{m{v}_0^2}{2}$

The total kinetic energy after collision

$E=\frac{\left(m+m_{\mathrm{H}}\right)u^2}{2}=\frac{m^2{v}_0^2}{2\left(m+m_{\mathrm{H}}\right)}$

The decrease in kinetic energy must be equal to ion ionization energy. Therefore,

$E_1=E_0-E=\left(\frac{m_{\mathrm{H}}}{m+m_{\mathrm{H}}}\right)E_0$

$\text{ i.e., }\frac{E_1}{E_0}=\frac{1}{1+\frac{m}{m_H}}$

i.e., the greater the mass m, the smaller the fraction of initial kinetic energy that be used for ionization.