An $\alpha$-particle and a proton are accelerated from rest by a potential difference of 100 V . Aft

Question

An $\alpha$-particle and a proton are accelerated from rest by a potential difference of 100 V . After this, their de Broglie wavelengths are $\lambda_{\mathrm{a}}$ and $\lambda_{p}$ respectively. The ratio $\frac{\lambda_{p}}{\lambda_{\alpha}}$, to the nearest integer, is $\qquad$ .

Professor PC Thomas Classes · Answer

(3) : de Broglie wavelength,\ $\lambda=\frac{h}{\sqrt{2 m K}}$\ $\lambda=\frac{h}{\sqrt{2 m q V}}[\because$ Kinetic energy, $K=q V]$\ $\lambda_{p}=\frac{h}{\sqrt{2 m_{p} q_{p} V}}, \quad \lambda_{\alpha}=\frac{h}{\sqrt{2 m_{\alpha} q_{\alpha} V}}$\ $	herefore \frac{\lambda_{p}}{\lambda_{\alpha}}=\frac{h}{\sqrt{2 m_{p} q_{p} V}} 	imes \frac{\sqrt{2 m_{\alpha} q_{\alpha} V}}{h}$\ $=\frac{\sqrt{m_{\alpha} q_{\alpha}}}{\sqrt{m_{p} q_{p}}}=\frac{\sqrt{\left(4 m_{p}ight)\left(2 q_{p}ight)}}{\sqrt{m_{p} q_{p}}}=\sqrt{8}=2 \sqrt{2} \approx 3$\