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
$\lambda = \frac{h}{\sqrt{2mqV}}$
$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha V_\alpha}{m_p q_p V_p}}$
$m_\alpha = 4m_p,\quad q_\alpha = 2e,\quad q_p = e$
$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{(4m_p)(2e)(64)}{(m_p)(e)(128)}}$
$= \sqrt{\frac{512\, m_p e}{128\, m_p e}}$
$= \sqrt{4}$
$\frac{\lambda_p}{\lambda_\alpha} = 2$ / 1
The required ratio is 2:1