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A proton, a deuteron and an alpha particle are accelerated through potentials of $V,\: 2V\: and\: 4V$ respectively. Their velocities will bear a ratio

$\begin {array} {1 1} (a)\;1:1:1 & \quad (b)\;1: \sqrt 2 : 1 \\ (c)\;\sqrt 2 : 1 : 1 & \quad (d)\;1 : 1: \sqrt 2 \end {array}$

 

1 Answer

Mass of the proton $m_p = m_p$
Mass of the deutron $m_d = 2m_p$
Mass of the alpha particles $m_{\alpha} = 4 m_p$
Change of the proton $=q_p $
Change of deutron $= 2q_p$
$\begin{align*}v = \frac{\sqrt{2qV}}{m} \end{align*}$
$\begin{align*} v_p = \frac{2 \times q_p \times V}{m_p}; \; \; \; V_q V_d = \frac{\sqrt{2 \times 2q_p \times 2V}}{2m_p} \end{align*}$
$ \begin{align*}V_{\alpha}=\frac{\sqrt{2 \times 2q_p \times 4V}}{4 m_p} \end{align*}$
Hence the velocities bear the ratio = $1 : 1 : \sqrt{2}$
answered Dec 27, 2013 by thanvigandhi_1
edited Nov 27, 2017 by priyanka.c
 

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