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Q.

A proton, a deuteron and an α particle having same momentum enter a uniform magnetic field at right angles to the field. Then the ratio of their angular momenta during their motion in the magnetic field is

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a

2 : 2 : 1

b

4 : 1 : 2

c

2 : 1 : 3

d

4 : 2 : 1

answer is A.

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Detailed Solution

$\begin{array}{l} 1^{H^{1}} 1^{H^{2}} 2^{{He}^{4}} \\ \vec{L} = \vec{r} \times \vec{P} \Rightarrow L = rpsin 90^{0} \Rightarrow L = rP \\ L = \frac{P^{2}}{qB} (∵ r = \frac{mv}{Bq} and P = mv) \\ L_{1} : L_{2} : L_{3} = \frac{1}{q_{1}} : \frac{1}{q_{2}} : \frac{1}{q_{3}}; \frac{1}{1} : \frac{1}{1} : \frac{1}{2} = 2 : 2 : 1 \end{array}$

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