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

Answer

A

Explanation

(a) The emission of an `alpha-`particle from the atom of an element reduces its atomic number by `2` and mass number by `4`.
Hence, the radioactive emission is as follows :
`._Z X^A overset (alpha-particl e)rarr ._(Z-2)Y^(A - 4) + ._ZHe^4 (alpha-partilce)`
Also from law of conservation of momentum,
`m xx 0 = m_y v_y + m_alpha v_alpha`
=`(A - 4) 4_y + 4v`
`rArr v_y = (4 v)/(A - 4)`.