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A proton, a deutron and an $\alpha$-particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is
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$1: 2: 8$
The deflection suffered by charged particle in an electric field is
$y=\frac{q E L D}{m \mu^2}=\frac{q E L D}{p^2 / m} \quad(p=m u)$
$\Rightarrow \quad y \propto \frac{q m}{p^2}$
$\Rightarrow y_p: y_d: y_\alpha=\frac{q_p m_p}{p_p^2}: \frac{q_d m_d}{p_d^2}: \frac{q_\alpha m_\alpha}{p_a^2}$
Since, $\quad p_\alpha=p_d=p_p \quad$ (given)
$m_p: m_d: m_\alpha=1: 2: 4$
and $\quad q_1: q_d: q_\alpha=1: 1: 2$
$\Rightarrow \quad y_b: y_d: y_\alpha=1 \times 1: 1 \times 2: 2 \times 4$
$=1: 2: 8$
$y=\frac{q E L D}{m \mu^2}=\frac{q E L D}{p^2 / m} \quad(p=m u)$
$\Rightarrow \quad y \propto \frac{q m}{p^2}$
$\Rightarrow y_p: y_d: y_\alpha=\frac{q_p m_p}{p_p^2}: \frac{q_d m_d}{p_d^2}: \frac{q_\alpha m_\alpha}{p_a^2}$
Since, $\quad p_\alpha=p_d=p_p \quad$ (given)
$m_p: m_d: m_\alpha=1: 2: 4$
and $\quad q_1: q_d: q_\alpha=1: 1: 2$
$\Rightarrow \quad y_b: y_d: y_\alpha=1 \times 1: 1 \times 2: 2 \times 4$
$=1: 2: 8$
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