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Question: A nucleus \({}_Z^AX\)emits an \(\alpha - \)particle with velocity\(v\). The recoil speed of the daug...

A nucleus ZAX{}_Z^AXemits an α−\alpha - particle with velocityvv. The recoil speed of the daughter nucleus is:
A. A−44v\dfrac{{A - 4}}{{4v}}
B. 4vA−4\dfrac{{4v}}{{A - 4}}
C. vv
D. v4\dfrac{v}{4}

Explanation

Solution

Nuclear fission is a process in nuclear physics in which the nucleus of an atom splits into two or more smaller nuclei as fission products, and usually some by-product particles. The total momentum of the interacting particles before and after a reaction are the same. Conservation of energy. Energy, including rest mass energy, is conserved in nuclear reactions.
For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision. For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision ∴⇒pivi=pfvf\therefore \Rightarrow pivi = pfvf

Complete step by step solution:
A nucleus ZAX{}_Z^AXemits an α−\alpha - particle with velocityvv.
Mass of α−\alpha - particle = 4
Velocity of α−\alpha - particle = vv
Let the daughter particle produced by particle B and its velocity be xx ms−1m{s^{ - 1}}
∵\because Mass is conserved and alpha particle is emitted from the parent atom
∴\therefore Mass of particle B will be = A−4A - 4
Now,
By applying momentum conservation we can say that since initial momentum is zero therefore final momentum will must be zero.
Therefore both particles produced in fission will have same magnitude of momentum (but direction of momentum will be opposite)
∴⇒4v+(−(A−4)x)=0 ⇒4v=(A−4)x ⇒x=4v(A−4) \begin{gathered} \therefore \Rightarrow 4v + ( - (A - 4)x) = 0 \\\ \Rightarrow 4v = (A - 4)x \\\ \Rightarrow x = \dfrac{{4v}}{{(A - 4)}} \\\ \end{gathered}

Hence the recoil speed of the daughter nucleus will be = 4v(A−4)\dfrac{{4v}}{{(A - 4)}}

Note: In some questions of these types sometimes there may be change of mass into energy so be careful during these cases and also in case of momentum handle signs with care.