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

Answer

C

Explanation

When an `alpha` particle is emitted , Z decreases by 2 and A decreases by 4 and when a positron is emitted, Z decreases by 1 but there is practically no change in A.
If the parent nucleus is `""_ZX^A` , then the transformation is represented as
`""_ZX^A overst(3alpha) (to) ""_(Z-6)Y_1^(A-12) overset(2e+)(to) ""_(Z-8)Y_2^(A-12)`
`therefore` In the final nucleus, no. of protons = Z - 8 `=N_P`
and the number of neutrons (A - 12) - (Z - 8)
`=A - Z - 4 = N_n`
`therefore (N_n)/(N_P) = (A-Z-4)/(Z - 8)`