A radioactive nucleus undergoes alpha-emissio | vedclass

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Question

(A) Let the mass of the parent radioactive nucleus be $A$. After $\alpha$-emission,the mass of the $\alpha$-particle is $4$ and the mass of the daught

Vedclass · Accepted Answer

$\frac{4V}{A - 4}$

Vedclass · Answer

(A) Let the mass of the parent radioactive nucleus be $A$. After $\alpha$-emission,the mass of the $\alpha$-particle is $4$ and the mass of the daughter nucleus is $(A - 4)$. Initially,the radioactive nucleus is at rest,so its initial momentum is $0$. According to the law of conservation of linear momentum: Initial momentum = Final momentum $0 = (A - 4)v' + 4V$ Here,$v'$ is the recoil velocity of the daughter nucleus and $V$ is the velocity of the $\alpha$-particle. $(A - 4)v' = -4V$ The negative sign indicates that the daughter nucleus moves in the opposite direction to the $\alpha$-particle. The magnitude of the recoil velocity is $v' = \frac{4V}{A - 4}$.

$A$ radioactive nucleus undergoes $\alpha$-emission to form a stable element. What will be the recoil velocity of the daughter nucleus if $V$ is the velocity of $\alpha$-emission and $A$ is the atomic mass of the radioactive nucleus?

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