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  • A radioactive nucleus undergoes  emission to form a stable element. What will be the recoil velocity of

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?

Option: 1

\frac{4 v}{A-4}


Option: 2

\frac{2 v}{A-4}


Option: 3

\frac{4 \mathrm{v}}{\mathrm{A}+4}


Option: 4

None of these


Answers (1)

best_answer

We assume that mass number of nucleus when it was at rest = A

mass number of \alpha− particle = 4 \alpha

mass number of remaining nucleus = A – 4

As there is no external force, so momentum of the system will remain conserved.

\Rightarrow 0=(A-4) v^{\prime}+4 v \Rightarrow v^{\prime}=-\frac{4 v}{(A-4)}

Negative sign represents that direction is opposite to the direction of motion of \alpha− particle

Posted by

avinash.dongre

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