A proton accelerated through a potential difference of V volts has a de-Broglie wavelength λ associated with it. In order to get the same wavelength associated with an $α$-particle, the required accelerating potential is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → V/8 **
For a particle accelerated through potential $V$,
$\lambda = \frac{h}{\sqrt{2m q V}}$
For proton: $\lambda = \frac{h}{\sqrt{2 m_p e V}}$
For $\alpha$-particle: $\lambda = \frac{h}{\sqrt{2 m_\alpha (2e) V'}}$
For same $\lambda$,
$\sqrt{2 m_p e V} = \sqrt{2 m_\alpha (2e) V'}$
⟹ $m_p V = 2 m_\alpha V'$
Since $m_\alpha = 4 m_p$
$V' = \frac{V}{8}$
∴ Required accelerating potential for $\alpha$-particle is $\frac{V}{8}$