Q122 / 180 · COMEDK 2020 · Physics
A proton accelerated through a potential $V$ has de-Broglie wavelength $\lambda$. Then, the de-Broglie wavelength of an $\alpha$-particle, when accelerated through the same potential $V$ is
- A$\frac{\lambda}{2}$
- B$\frac{\lambda}{\sqrt{2}}$
- Correct answerC$\frac{\lambda}{2 \sqrt{2}}$
- D$\frac{\lambda}{8}$
Solution
When a charged particle of charge $q$ is accelerated through a potential $V$, then de-Broglie wavelength is given as
$\lambda=\frac{h}{\sqrt{2 m q V}}$
$\Rightarrow \quad \lambda \propto \frac{1}{\sqrt{m q}}$
$\Rightarrow \quad \frac{\lambda_{\alpha}}{\lambda_{p}}=\sqrt{\frac{m_{p}}{m_{\alpha}}} \cdot \sqrt{\frac{q_{p}}{q_{\alpha}}}$
$\Rightarrow \quad \frac{\lambda_{\alpha}}{\lambda}=\sqrt{\frac{m_{p}}{4 m_{p}}} \sqrt{\frac{q_{p}}{2 q_{p}}} \quad\left[\begin{array}{c}\because m_{\alpha}=4 m_{p} \\ \text { and } q_{\alpha}=2 q_{p}\end{array}\right]$
$\Rightarrow \quad \frac{\lambda_{\alpha}}{\lambda}=\frac{1}{\sqrt{8}}$
$\Rightarrow \quad \lambda_{\alpha}=\frac{\lambda}{2 \sqrt{2}}$
Hence, the correct answer is option: 'C'.