An
Question:

An $\alpha$ particle and a proton are accelerated from rest by a potential difference of $200 \mathrm{~V}$.

After this, their de Broglie wavelengths are $\lambda_{\alpha}$ and $\lambda_{\mathrm{p}}$ respectively. The ratio $\frac{\lambda_{\mathrm{p}}}{\lambda_{\alpha}}$ is :

  1. (1) 8

  2. (2) $2.8$

  3. (3) $3.8$

  4. (4) $7.8$

     


Correct Option: 2,

Solution:

(2)

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m q v}}$

$\frac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\frac{m_{\alpha} q_{\alpha}}{m_{p} q_{p}}}=\sqrt{\frac{4 \times 2}{1 \times 1}}$

$=2 \sqrt{2}=2.8$

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