Question

A proton, a deuteron and an $\alpha$-particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. If the electric field is perpendicular to the initial direction of the particles. Then, the ratio of deviations of these particles

• A. 1 : 2 : 8
• B. 1 : 2 : 4
• C. 1 : 2 : 2
• D. 1 : 1 : 2

Answer 1

Answer: (A)

Force acting on charge particle in normal direction of motion.
$F=Eq$
Acceleration $a=\frac{F}{m}=\frac{Eq}{m}$
If particle travels a distance $d$ in lime $t$,
then $t=\frac{d}{v}$ Motion in normal direction
$y=\frac{1}{2}a{t}^{2}y=\frac{1}{2}\times \frac{Eq}{m}\times \frac{d^2}{v^2}$ But $p=mv$
$\therefore y=\frac{Eq{d}^2\cdot m}{2p^2}$
$\therefore y\propto \frac{qm}{p^2}$
$\therefore y_p:y_d:y_{\alpha }=\frac{q_p{m}_p}{p_p^2}=\frac{q_d{m}_d}{p_d^2}$
$=\frac{q_{\alpha }{m}_{\alpha }}{p_{\alpha }^2}$
Given, $p_p=p_d=p_{\alpha }{q}_p=e,q_d=e$,
$q_{\alpha }=2e{m}_p=m,m_d=2m,m_{\alpha }=4m$
$\therefore y_p:y_d:y_{\alpha }=e\times m:e\times 2m:2e\times 4m$
$=1:2:8$