A heavy nucleus N,at rest,undergoes fission N | vedclass

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(A,C,D) The fission reaction is $N \rightarrow P + Q$. The energy released in the fission process is given by the mass defect $\delta$ multiplied by $

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$A, C, D$

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(A,C,D) The fission reaction is $N \rightarrow P + Q$. The energy released in the fission process is given by the mass defect $\delta$ multiplied by $c^2$,which is $E = \delta c^2$.Since the initial nucleus $N$ is at rest,the total energy released is converted into the kinetic energies of the product nuclei $P$ and $Q$. Thus,$E_P + E_Q = \delta c^2$. Statement $(A)$ is correct.By the law of conservation of linear momentum,the initial momentum is zero,so the magnitudes of the momenta of $P$ and $Q$ must be equal: $p_P = p_Q = p$. Thus,$M_P v_P = M_Q v_Q$,which implies $\frac{v_P}{v_Q} = \frac{M_Q}{M_P}$. Statement $(C)$ is correct.The kinetic energy is given by $E = \frac{p^2}{2M}$. Since $p_P = p_Q = p$,we have $E_P = \frac{p^2}{2M_P}$ and $E_Q = \frac{p^2}{2M_Q}$.Substituting these into the energy equation: $\frac{p^2}{2M_P} + \frac{p^2}{2M_Q} = \delta c^2$.$\frac{p^2}{2} \left( \frac{M_Q + M_P}{M_P M_Q} \right) = \delta c^2$.Using the reduced mass $\mu = \frac{M_P M_Q}{M_P + M_Q}$,we get $\frac{p^2}{2\mu} = \delta c^2$,which gives $p = c \sqrt{2 \mu \delta}$. Statement $(D)$ is correct.Statement $(B)$ is incorrect because $E_P = \frac{p^2}{2M_P} = \frac{2\mu \delta c^2}{2M_P} = \frac{M_Q}{M_P+M_Q} \delta c^2$.

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