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A nucleus of mass $M + Δm$ is at rest and decays into two daughter nuclei of mass M/2 each. If the speed of light is c, then the speed of any one of the daughter nuclei is

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$c \sqrt{\frac{2 Δm}{M}}$

$c \sqrt{\frac{Δm}{M}}$

$c \sqrt{\frac{Δm}{M + Δm}}$

$c \frac{Δm}{M + Δm}$

answer is B.

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Detailed Solution

According to the conservation of momentum $0 = \frac{M}{2} V_{1} - \frac{M}{2} V_{2}$
$V_{1} = V_{2} = v$

The mass $∆ m$ is converted into kinetic energy of the daughter nuclei

${Δmc}^{2} = \frac{1}{2} \cdot \frac{M}{2} V_{1}^{2} + \frac{1}{2} \cdot \frac{M}{2} V_{2}^{2} \Rightarrow v = c \sqrt{\frac{2 Δm}{M}}$

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