Question

A proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$ -particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

• A. $\frac{V}{8}$
• B. $\frac{V}{4}$
• C. $4 V$
• D. $8 V$

Answer 1

Answer: (A)

$\frac{V}{8}$

Answer 2

$\lambda_{p}=\lambda_{\alpha}$
$(m q V)_{p}=(m q V)_{\alpha}$
Potential difference
$V_{\alpha}=\frac{V}{8} [ \because m_{\alpha} =4 m_{p} q_{\alpha} =2 q_{p}]$