The de Broglie wavelength of a proton
Question:

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is :

1. (1) $4: 2$

2. (2) $4: 1$

3. (3) $1: 4$

4. (4) $4: 3$

Correct Option: , 2

Solution:

(2)

From De-broglie’s wavelength : $\lambda=\frac{h}{m v}$

Given $\lambda_{\mathrm{p}}=\lambda_{\alpha}$

$v \alpha \frac{1}{m}$

$\frac{\mathrm{v}_{\mathrm{p}}}{\mathrm{v}_{\alpha}}=\frac{\mathrm{m}_{\alpha}}{\mathrm{m}_{p}}=\frac{4 \mathrm{~m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{p}}}=\frac{4}{1}$