A proton, a deuteron and an alpha particle whose kinetic energies are same enter at right angles to a uniform magnetic field. Then the ratio of the radii of their circular paths is

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$\sqrt{2} : 1 : 1$

2 : 2 : 1

$1 : \sqrt{2} : 1$

1 : 2 : 1

answer is A.

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Detailed Solution

$r = \sqrt{2 mk} / Bq \Rightarrow rα \sqrt{m} / q$
$∴ r_{P} : r_{D} : r_{α} = \frac{\sqrt{m_{P}}}{q_{P}} : \frac{\sqrt{m_{D}}}{q_{d}} : \frac{\sqrt{m_{α}}}{q_{a}} = 1 : \sqrt{2} : 1$

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