The minimum kinetic energy of an electron, hydrogen ion, helium ion required for ionization of a hydrogen atom is $E_1$ in case electron is collided with hydrogen atom. It is $E_2$ when hydrogen ion is collided and $E_3$ when helium ion is collided. Then

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

          $m{v}_0=\left(m+m_H\right)V$

where $m$ is the mass of incident particle, $m_H$ the mass of hydrogen atom, $v_0$ the initial velocity of incident particle and $V$ 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

          $E=\frac{\left(m+m_H\right)V^2}{2}=\frac{m^2{v}_0^2}{2\left(m+m_H\right)}$

The decrease in kinetic energy must be equal to ionization energy, so we have

$$\begin{gathered} E_i=E_0-E=\left(\frac{m_H}{m+m_H}\right)E_0 \\ \Rightarrow \frac{E_i}{E_0}=\frac{m_H}{m+m_H} \end{gathered}$$

i.e., the greater the mass $m$, the smaller the fraction of initial kinetic energy that will be used for ionization. Hence, the correct answer is (B).