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A proton, a deuteron and an $ \alpha$ - particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If $r_p, r_d$ and $r_{\alpha}$ denote respectively the radii of the trajectories of these particles, then

$\begin {array} {1 1} (a)\;r_{\alpha} = r_p < r_d & \quad (b)\;r_{\alpha} = r_d < r_p \\ (c)\;r_{\alpha} = r_p = r_d & \quad (d)\;r_{\alpha} < r_p < r_d \end {array}$

 

1 Answer

$\begin{align*}r = \frac{mV}{qB} \end{align*}$
$mp = 1; \; \; \; qp = 1$
$md = 2; \; \; \; qd =1 $
$m_2 = 4; \; \; \; q_2 = 2$
$\begin{align*} \therefore r_p = \frac{1m \times V}{1q \times B}; \; \; \; r_d = \frac{2m \times V}{1q \times B} \end{align*}$
$\begin{align*}r_{\alpha} = \frac{4m \times V}{2q \times B} \end{align*}$
$\begin{align*}K.E_p = \frac{1}{2} m \times v^2 \;\; \implies v = \sqrt{\frac{2 K E}{m} } \end{align*}$
$\begin{align*}K.E_{\alpha} = \frac{1}{2} \times 2m \times v^2 \implies v = \sqrt{\frac{K.E}{m}}\end{align*}$
$\begin{align*}K.E_{\alpha} = \frac{1}{2} \times 4m \times v^2 \implies v = \sqrt{\frac{K.E}{2m}}\end{align*}$
Now substituting the values in $r_p$, $r_d$ and $r_{\alpha}$ we get,
$r_{\alpha} = r_p < r_d$
answered Feb 23, 2014 by thanvigandhi_1
edited Nov 27 by priyanka.c
 

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