Q105 / 180 · AIIMS 2018 (27 May) · Physics

A deutron has an energy of $4 \mathrm{MeV}$ when accelerated through a potential difference $V$. What will be the energy of an $\alpha$-particle accelerated by source $V$ ?

  • A
    $4 \mathrm{MeV}$
  • Correct answer
    B
    $8 \mathrm{MeV}$
  • C
    $12 \mathrm{MeV}$
  • D
    $20 \mathrm{MeV}$

Solution

Energy of a charged particle when accelerated through V $\begin{gathered}E=q V \\ \text { So, } \quad \frac{E_d}{E_\alpha}=\frac{q_d}{q_\alpha} \times \frac{V_d}{V_\alpha}=\frac{(e) V}{(2 e) V}=\frac{1}{2} \\ E_\alpha=2 E_d=2 \times 4 \mathrm{MeV}=8 \mathrm{MeV}\end{gathered}$

Hence, the correct answer is option: 'B'.

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