Question

The ratio of momentum of an electron and an alpha particle which are accelerated from rest by potential difference of 100 V is:

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

Answer 1

The correct option is D $\sqrt{\frac{m_e}{2m_{\alpha }}}$

Mass of proton $=1$ $u$

and, Mass of alpha particle $=4$ $u$

Charge of proton $=e$

And, charge of alpha particle $=2e$

For charge moving in a uniform potential field.

$K.E=$ Electric potential energy

$\frac{1}{2}m{v}^2=qV$

$v=\sqrt{\frac{2qV}{m}}$

OR,

Momentum $\left(P\right)=mv$

$\Rightarrow P=m\sqrt{\frac{2qV}{m}}$

$\Rightarrow P=\sqrt{m2qV}$

Momentum of proton,

$P_1=\sqrt{2qV{m}_1}=\sqrt{2eV{m}_e}$

Momentum of alpha particle

$P_2=\sqrt{2qV{m}_2}=\sqrt{2\times 2eV{m}_a}$

$\therefore \frac{P_1}{P_2}=\sqrt{\frac{2eV{m}_e}{4eV{m}_a}}$

$\therefore \frac{P_1}{P_2}=\sqrt{\frac{m_e}{2m_a}}$