A proton accelerated through a potential difference of \( V \) volts has a de-Broglie wavelength \( \lambda \) associated with it. In order to get the same wavelength associated with an alpha-particle, the required accelerating potential is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( V/8 \)
A charged particle accelerated through a potential difference \( V \) gains kinetic energy \( qV \), so its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2m\,qV}} \)
For the same wavelength, the product \( m\,q\,V \) must be the same for both particles.
\( \displaystyle m_{\alpha}\,q_{\alpha}\,V' = m_{p}\,q_{p}\,V \)
An alpha particle has \( m_{\alpha} = 4m_p \) and \( q_{\alpha} = 2q_p \), so \( m_\alpha q_\alpha = 8\,m_p q_p \).
\( \displaystyle 8\,V' = V \;\Rightarrow\; V' = \frac{V}{8} \)