A radioactive nucleus (initial mass number A and atomic number Z) emits 3 `alpha`-paticles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

Options

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

Answer

B

Explanation

`._(Z)X^(A)rarr3_(2)He^(4)+._(z-8)Y^(A-12)+2_(+1)e^(0)+2upsilon`
number of proton `=Z-8`
number of neutron `=(A-12)-(Z-8)=A-Z-4`
ratio is `=(A-Z-4)/(Z-8)`