A nucleus of mass M + Δ m is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c. The speed of daughter nuclei is (a) c(Δ m)/(M + Δ m) (b) c√((2Δ m)/(M)) (c) c√((Δ m)/(M)) (d) c√((Δ m)/(M + Δ m))

Answer: (b). Energy released Δ m c² becomes kinetic energy of the two equal fragments (equal speeds by momentum conservation). Δ m c² = 2 × 1/2M/2v² = (M v²)/(2) v = c√((2Δ m)/(M)) So the answer is (b).

A nucleus of mass M + Δ m is at rest and decays into two daughter… | JEE Main 2010 (AIEEE)

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Nuclear Reactions & Q-value

A nucleus of mass $M + \Delta m$ is at rest and decays into two daughter nuclei of equal mass $\frac{M}{2}$ each. Speed of light is $c.$

The speed of daughter nuclei is

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Answer: (b)

Energy released $\Delta m\,c^2$ becomes kinetic energy of the two equal fragments (equal speeds by momentum conservation).
$\Delta m\,c^2 = 2 \times \dfrac{1}{2}\dfrac{M}{2}v^2 = \dfrac{M v^2}{2}$
$v = c\sqrt{\dfrac{2\Delta m}{M}}$
So the answer is (b).

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