A proton is accelerated through a potential difference V. After acceleration, the de Broglie wavelength associated with it is $\displaystyle \lambda$. If the proton is replaced by an alpha particle, then find the de Broglie wavelength associated with it if it were accelerated through the same potential difference V. What will be the momentum of the alpha particle?

$$\begin{array}{|ll|} \hline \text { - } & \text { Finding de- Broglie wavelength } \text { - } & \text { Finding the momentum of } \alpha \text { particles } \end{array} \quad 1_{2}^{1} ⿱ 12$\displaystyle \lambda_{p}=\frac{h}{\sqrt{2 q_{p} m_{p} V}}$, $\displaystyle \lambda_{\alpha}=\frac{h}{\sqrt{2 \cdot 2 q_{p} \cdot 4 m_{p} V}}$ $\displaystyle \frac{\lambda_{\alpha}}{\lambda_{p}}=\frac{1}{2 \sqrt{2}}$ $\displaystyle \lambda_{\alpha}=\frac{h}{p_{p}}, \quad \lambda_{\alpha}=\frac{h}{p_{\alpha}}$ $\displaystyle \frac{p_{\alpha}}{p_{p}}=\frac{1}{2 \sqrt{2}}$,