A nucleus of mass M + Delta m at rest decays | vedclass

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Question

(C) The mass defect is $\Delta M = (M + \Delta m) - (M/2 + M/2) = \Delta m$.The energy released in the decay is $Q = \Delta M c^2 = \Delta m c^2$.Acco

Vedclass · Accepted Answer

$c\sqrt {\frac{{2\Delta m}}{M}} $

Vedclass · Answer

(C) The mass defect is $\Delta M = (M + \Delta m) - (M/2 + M/2) = \Delta m$.The energy released in the decay is $Q = \Delta M c^2 = \Delta m c^2$.According to the law of conservation of linear momentum,since the initial nucleus is at rest,the total initial momentum is $0$. Thus,the two daughter nuclei must move in opposite directions with equal speeds $v$. Let $v_1 = v_2 = v$.$0 = (M/2)v - (M/2)v$ (This confirms they move with equal speed in opposite directions).The energy released is converted into the kinetic energy of the two daughter nuclei:$Q = K_1 + K_2 = \frac{1}{2}(M/2)v^2 + \frac{1}{2}(M/2)v^2 = (M/2)v^2$.Equating the energy released to the kinetic energy:$\Delta m c^2 = (M/2)v^2$.Solving for $v$:$v^2 = \frac{2 \Delta m c^2}{M}$.Therefore,$v = c\sqrt{\frac{2 \Delta m}{M}}$.

$A$ nucleus of mass $M + \Delta m$ at rest decays into two daughter nuclei of equal mass $M/2$. If the speed of light is $c$,the speed of each daughter nucleus is:

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