Question

A proton is released from position near to positive plate. It reaches to opposite plate in time $t_1$, now one $\alpha$ particle is released & it reaches to opposite plate in time $t_2$, then $t_1:t_2$

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $1:\sqrt{2}$

Answer 1

The correct option is D $1:\sqrt{2}$

Here particles are released from rest under uniform acceleration
i.e. $a=\frac{qE}{m}$
$s=\frac{1}{2}a{t}^2$
$\Rightarrow t^2=\frac{2s}{a}$
$t^2=\frac{2s}{q\frac{E}{m}}$
$\Rightarrow t^2\propto \frac{m}{q}$
$\Rightarrow \frac{t_1}{t_2}=\frac{t_p}{t_{\alpha }}=\sqrt{\frac{\frac{m}{q}}{\frac{4m}{2q}}}=\sqrt{\frac{2}{4}}$
$=\frac{1}{\sqrt{2}}$