Question

If the stationary proton and $\alpha$ - particle are accelerated through same potential difference, the ratio of de Broglie's wavelength will be

• A. $2$
• B. $1$
• C. $2 \sqrt{2}$
• D. $\sqrt{2}$

Answer 1

Answer: (C)

The gain in K.E. of a charged particle after moving through a potential difference of $V$ is given by $eV$, that is also equal to $\frac{1}{2} m v^{2}$ where $v$ is the velocity of the charged particle. Disregarding the relativistic effect,
$\frac{1}{2} m v^{2}=q V $
$ \Rightarrow v=\sqrt{\frac{2 q V}{m}} $
$ \Rightarrow m v=\sqrt{2 m q V} $
$\Rightarrow $ de Broglie wavelength $ \lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 m q V}}$
$\therefore \frac{\lambda_{p}}{\lambda_{a}}=\sqrt{\frac{m_{a} q_{\alpha} V_{a}}{m_{p} q_{p} V_{p}}}$
Putting $V_{\alpha}=V_{p}, \frac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\frac{(4)(2)}{(1)(1)}}=2 \sqrt{2}$