The ratio of the de-Broglie wavelengths associated with protons accelerated through a potential of 128 V and alpha particles accelerated through a potential of 64 V will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2 : 1
A charge \( q \) accelerated through a potential difference \( V \) gains kinetic energy \( K = qV \), so its momentum is \( p = \sqrt{2mqV} \) and its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m q V}} \)
For the ratio only the factor \( 1/\sqrt{mqV} \) matters. The alpha particle has four times the mass and twice the charge of the proton.
\( m_\alpha = 4m_p \)
\( q_\alpha = 2e,\ q_p = e \)
\( V_p = 128\,\mathrm{V},\ V_\alpha = 64\,\mathrm{V} \)
\( \displaystyle \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha\,q_\alpha\,V_\alpha}{m_p\,q_p\,V_p}} = \sqrt{\frac{4 \times 2 \times 64}{1 \times 1 \times 128}} = \sqrt{4} = 2 \)
So \( \lambda_p : \lambda_\alpha = 2 : 1 \).