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.
$F=E q$
Acceleration $a=\frac{F}{m}=\frac{E q}{m}$
If particle travels a distance $d$ in lime $t$,
then $t=\frac{d}{v}$ Motion in normal direction
$y=\frac{1}{2} a t^{2} y=\frac{1}{2} \times \frac{E q}{m} \times \frac{d^{2}}{v^{2}}$ But $p=m v$
$\therefore y=\frac{E q d^{2} \cdot m}{2 p^{2}}$
$\therefore y \propto \frac{q m}{p^{2}}$
$\therefore y_{p}: y_{d}: y_{\alpha}=\frac{q_{p} m_{p}}{p_{p}^{2}}=\frac{q_{d} m_{d}}{p_{d}^{2}}$
$=\frac{q_{\alpha} m_{\alpha}}{p_{\alpha}^{2}}$
Given, $p_{p}=p_{d}=p_{\alpha} q_{p}=e, q_{d}=e$,
$q_{\alpha}=2 e m_{p}=m, m_{d}=2 m, m_{\alpha}=4\, m$
$ \therefore y_{p}: y_{d}: y_{\alpha}=e \times m: e \times 2 m: 2 e \times 4\, m$
$=1: 2: 8$