(d) Momentum $p = \sqrt{2mk}$; where $K =$ kinetic energy $= Q.V$ $\Rightarrow p = \sqrt{2mQV} \Rightarrow p \propto \sqrt{mQ} \Rightarrow \frac{p_{e}}{p_{\alpha}} = \sqrt{\frac{m_{e}Q_{e}}{m_{\alpha}Q_{\alpha}}} = \sqrt{\frac{m_{e}}{2m_{\alpha}}}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.