In a radioactive decay , an element `""_(Z) X^(A)` emits four `alpha`- particles , three `beta`- particles and eight gamma photons . The atomic number and mass number of the resulting final nucleus are

Options

A. Z- 11 , A - 16
B. Z - 5 , A - 13
C. Z - 5 , A - 16
D. Z - 8 , A - 13

Answer

C

Explanation

Emission of `alpha` - particle decreases the atomic number and mass number by 2 and 4 respectively . Emission of `beta` - particle increases the atomic number by 1 while the mass number remains unchanged . Emission of `alpha` - particle both the atomic number and the mass number remain unchanged .
`therefore` After the emission of four `alpha`- particles , three `beta`- particles and eight `gamma`- photons .
Decrease in atomic number = `4 xx 2 - 3 xx 1 = 5`
Decrease in mass number = `4 xx 4 - 0 = 16`
Thus , the atomic number and mass number of resulting final nucleus are Z -5 and A - 16 respectively .
`therefore` The resulting final nucleus is `""_(Z - 5) Y^(A -16)`