Q136 / 160 · TS EAMCET 2012 · Physics

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

  • Correct answer
    A
    $\frac{V}{8}$
  • B
    $\frac{V}{4}$
  • C
    $4 \mathrm{~V}$
  • D
    $8 \mathrm{~V}$

Solution

$\lambda_p=\lambda_\alpha$ $(m q V)_p=(m q V)_\alpha$ Potential difference $V_\alpha=\frac{V}{8} \quad\left[\begin{array}{c}\because m_\alpha=4 m_p \\ q_\alpha=2 q_p\end{array}\right]$

Hence, the correct answer is option: 'A'.

Relevant Questions