Question

The ratio of momentum of an electron and $\alpha$ -particle which are accelerated from rest by a potential difference of $100\, V$ is

• A. $1$
• B. $\sqrt{\left(2m_{e} / m_{a}\right)}$
• C. $\sqrt{\left(m_{e} / m_{\alpha}\right)}$
• D. $\sqrt{\left(m_{e} / 2m_{\alpha}\right)}$

Answer 1

Answer: (D)

$ \frac{1}{2} m v^{2} =e V $
$ \Rightarrow v =\sqrt{\frac{2\, e V}{m}} $
$ p =m v=\sqrt{2 \,m e V} $
Now, $ p_{e} =\sqrt{2 \,m_{e} \times e \times 100} $
and $ p_{\alpha}=\sqrt{4 \,m_{\alpha} \times e \times 100} $
$\therefore \frac{p_{e}}{p_{\alpha}} =\sqrt{\frac{m_{e}}{2 m_{\alpha}}} $