A proton is accelerated through a potential difference V. After acceleration, the de Broglie wavelength associated with it is λ. If the proton is…

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