A proton, a deuteron, and an alpha-particle h | vedclass

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Question

(A) The deflection $y$ of a charged particle moving with velocity $v$ in a uniform electric field $E$ over a distance $L$ is given by $y = \frac{1}{2}

Vedclass · Accepted Answer

$1 : 2 : 8$

Vedclass · Answer

(A) The deflection $y$ of a charged particle moving with velocity $v$ in a uniform electric field $E$ over a distance $L$ is given by $y = \frac{1}{2} a t^2$, where $a = \frac{qE}{m}$ and $t = \frac{L}{v}$.Substituting $t$, we get $y = \frac{1}{2} \left( \frac{qE}{m} ight) \left( \frac{L}{v} ight)^2 = \frac{qEL^2}{2mv^2}$.Since momentum $p = mv$, we have $v = \frac{p}{m}$. Substituting this, $y = \frac{qEL^2}{2m(p/m)^2} = \frac{qEL^2m}{2p^2}$.Since $E, L,$ and $p$ are constant for all particles, $y \propto qm$.For a proton $(p)$, deuteron $(d)$, and $\alpha$-particle $(\alpha)$:Charge ratio $q_p : q_d : q_\alpha = 1 : 1 : 2$.Mass ratio $m_p : m_d : m_\alpha = 1 : 2 : 4$.Therefore, the ratio of deflections is $y_p : y_d : y_\alpha = (q_p m_p) : (q_d m_d) : (q_\alpha m_\alpha) = (1 	imes 1) : (1 	imes 2) : (2 	imes 4) = 1 : 2 : 8$.

$A$ proton, a deuteron, and an $\alpha$-particle having the same momentum enter 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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