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Electrostatics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsElectrostatics3 Marks
Question
A proton and an $\alpha$-particle are accelerated from rest through the same potential difference. Compare their final speeds. Charge on an $\alpha$-particle $=2 \times$ charge on a proton, mass of an $\alpha$ particle $=4 \times$ mass of a proton.
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Answer

Lution: Let $q _1=$ charge on an $\alpha$-partic $1 e , q _2=$ charge on a proton, $m _1=$ mass of an $\alpha$-particle, $m_2=$ mass of a proton, $v_1=$ final speed of the $\alpha$-particle, $v_2=$ final speed of the proton
$
\therefore \frac{q_1}{q_2}=2 \text { and } \frac{m_1}{m_2}=4 \quad \therefore \frac{m_2}{m_1}=\frac{1}{4}
$
If the particles are accelerated from rest, through a potential difference $V$,
$
\begin{aligned}
& \frac{1}{2} m_1 v_1^2=q_1 V \text { and } \frac{1}{2} m_2 v_2^2=q_2 V \\
\therefore & \frac{m_1 v_1^2}{m_2 v_2^2}=\frac{q_1}{q_2} \quad \text {} \\
\therefore & \frac{v_1^2}{v_2^2}=\frac{q_1}{q_2} \times \frac{m_2}{m_1}=2 \times \frac{1}{4}=\frac{1}{2} \\
\therefore & \frac{v_1}{v_2}=\frac{1}{\sqrt{2}}=0.707
\end{aligned}
$

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