de-Broglie wavelength is given by

$$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m q V}}$$

For same potential difference

$$\begin{gathered} \lambda \propto \frac{1}{\sqrt{m q}} \\ \frac{\lambda_\alpha}{\lambda_e}=\sqrt{\frac{m_e q_e}{m_\alpha q_\alpha}} \\ \because m_\alpha \gg m_e \\ \lambda_e>\lambda_\alpha \end{gathered}$$