A radiaoactive nucleus (initial mass number `A` and atomic number `Z` emits `3 alpha`- particles and 2 positrons The ratio of number of neutrons to that of proton in the final nucleus will be

Options

A. `(A-Z-4)/(Z-8)`
B. `(A-Z-12)/(Z-4)`
C. `(A-Z-4)/(Z-2)`
D. `(A-Z-8)/(Z-4)`

Answer

A

Explanation

As per question, the nuclear reaction is
`._(Z)X^(A)to._(Z-8)Y^(A-12)+3._(2)He^(4)+2._(+1)e^(0)`
Charge number of final nucleus =(Z-8)
Mass number of final nucleus =(A-12)
`:.` Number of protons in final nucleus, `n_(p)=(Z=8)`
Number of neutron in final nucleus,
`n_(n)=(A-12)-(Z-8)=A-Z-4`
`(n_(n))/(n_(p))=((A-Z-4))/(Z-8)`