Question:
The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is :
Correct Option: , 2
Solution:
(2)
From De-broglie's wavelength : $\lambda=\frac{h}{m v}$
Given $\lambda_{\mathrm{p}}=\lambda_{\alpha}$
$v \alpha \frac{1}{m}$
$\frac{\mathrm{v}_{\mathrm{p}}}{\mathrm{v}_{\alpha}}=\frac{\mathrm{m}_{\alpha}}{\mathrm{m}_{p}}=\frac{4 \mathrm{~m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{p}}}=\frac{4}{1}$