If \( \lambda_0 \) is the de Broglie wavelength of a proton accelerated through a potential difference of 100 V, the de Broglie wavelength of an alpha particle accelerated through the same potential difference is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \lambda_0/2\sqrt{2} \)
A particle of charge \( q \) accelerated from rest through a potential difference \( V \) gains kinetic energy \( K=qV \), so its momentum is \( p=\sqrt{2mqV} \). Its de Broglie wavelength is therefore
\( \displaystyle \lambda=\frac{h}{\sqrt{2mqV}} \)
For the same \( V \), \( \lambda\propto1/\sqrt{mq} \). An alpha particle has about four times the mass of a proton (\( m_\alpha\approx4m_p \)) and twice its charge (\( q_\alpha=2e \)).
\( \begin{aligned} \frac{\lambda_\alpha}{\lambda_0} &= \sqrt{\frac{m_p\,e}{m_\alpha\,q_\alpha}} \\ &= \sqrt{\frac{m_p\,e}{4m_p\times2e}} \\ &= \frac{1}{\sqrt{8}} \\ &= \frac{1}{2\sqrt{2}} \end{aligned} \)
\( \lambda_\alpha=\lambda_0/2\sqrt{2} \)