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

Options

A. `(4v)/(A - 4)`
B. `(2v)/(A - 4)`
C. `(4v)/(A + 4)`
D. `(2v)/(A + 4)`

Answer

A

Explanation

We assume that mass number of nucleus when it was at rest = A
`because` mass number of `alpha`- particle = 4
`therefore` mass number of remaining nucleus = A - 4
As there is no external force , so momentum of the system will remain conserved
`implies 0 = (A - 4) v. + 4 v implies v. = - (4v)/((A - 4))`
negative sign represents that direction is opposite to the direction of motion of `alpha` -particle .