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$H^{+}, {He}^{+},$ and $O^{2 +}$ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of $H^{+}, {He}^{+},$ and $O^{2 +}$ are $1 a m u, 4 a m u,$ and $16 a m u$ respectiely. Then,

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$H^{+}$ will be deflected most

$O^{2 +}$ will be deflected most

${He}^{+}$ and $O^{2 +}$ will be deflected equally

all will be deflected equally

answer is A, C.

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Detailed Solution

$r = \frac{m v}{B q} = \frac{p}{B q} = \frac{\sqrt{2 K m}}{B q}, i . e . r α \frac{\sqrt{m}}{q}$

If K and B are same.

i.e. $r_{H +} : r_{H e} : r_{O^{2 +}} = \frac{\sqrt{1}}{1} : \frac{\sqrt{4}}{1} : \frac{\sqrt{16}}{2} = 1 : 2 : 2$

Therefore, $H e^{+}$ and $O^{2 +}$ will be deflected equally but $H^{+}$ having the least radius will be deflected most.

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