A proton and an alpha particle accelerated through different potential differences have the same de-Broglie wavelength. The ratio of their respective accelerating potentials is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 8 : 1
A charge \( q \) accelerated through a potential difference \( V \) gains kinetic energy \( qV \), so its momentum is \( p = \sqrt{2mqV} \) and its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{\sqrt{2mqV}} \)
Equal wavelengths need equal values of \( mqV \).
\( m_\alpha = 4m_p \)
\( q_\alpha = 2e \)
\( q_p = e \)
\( \displaystyle m_p\,e\,V_p = (4m_p)(2e)\,V_\alpha \;\Rightarrow\; \frac{V_p}{V_\alpha} = 8 \)
So \( V_p : V_\alpha = 8 : 1 \).