Question

The ratio of the momentum of a proton and a $\alpha$ - particle which is accelerated from rest by a potential difference of 200 V. $m_p$ and $m_{\alpha }$ are masses of the proton and $\alpha$- particles:

• A. $\sqrt{\frac{2m_p}{m_{\alpha }}}$
• B. $\sqrt{\frac{m_p}{2m_{\alpha }}}$
• C. $\frac{m_p}{2m_{\alpha }}$
• D. $\frac{2m_p}{m_{\alpha }}$

Answer 1

Answer: (B)

$\sqrt{\frac{m_p}{2m_{\alpha }}}$

Using the reltion $KE=\frac{p^2}{2m}=qV$, we get

$p=\sqrt{2mqV}$

Hence for same potential difference(V),

$\frac{p_p}{p_{\alpha }}=\frac{\sqrt{2m_p{q}_p}}{\sqrt{2m_{\alpha }{q}_{\alpha }}}$

SInce, $q_{\alpha }=2q_p$, hence

$\frac{p_p}{p_{\alpha }}=\frac{\sqrt{m_p}}{\sqrt{2m_{\alpha }}}$

Hence, B is the correct answer