# A radioactive nucleus (initial mass number A and atomic number Z ) emits $\dpi{100} 3\alpha -$ particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be Option 1) $\frac{A-Z-4}{Z-2}$ Option 2) $\frac{A-Z-8}{Z-4}$ Option 3) $\frac{A-Z-4}{Z-8}$ Option 4) $\frac{A-Z-12}{Z-4}$

As we learnt

α -decay -

$^{A}_{Z}X\rightarrow ^{A-4}_{Z-2}Y+^{4}_{2}He+Q$

- wherein

$Q\: value = \left ( M_{X} -M_{Y}-M_{He}\right )c^{2}$

β plus decay -

$^{A}_{Z}X\rightarrow _{Z-1}^{A}Y+\beta ^{+}+ {\nu }+Q \ value$

- wherein

$\nu\rightarrow \: neutrino$

$Q\: value =\left [ M_{X}-M_{Y}-2M_{e} \right ]c^{2}$

Let's say original nuclei is $^{A}_{Z}X.$

$^{A}_{Z}X \;\overset{3\alpha}{\rightarrow}\;\ ^{A-12}_{Z-6}Y\; \overset{2\beta^{+}}{\rightarrow}\;^{A-12}_{Z-8}W$

Hence,  Number of proton = Z - 8.

Number of neutron = (A - 12) - (Z - 8) = A - Z - 4

Hence, Ratio $=\frac{A-Z-4}{Z-8}$

Correct option is 3.

Option 1)

$\frac{A-Z-4}{Z-2}$

This is an incorrect option.

Option 2)

$\frac{A-Z-8}{Z-4}$

This is an incorrect option.

Option 3)

$\frac{A-Z-4}{Z-8}$

This is the correct option.

Option 4)

$\frac{A-Z-12}{Z-4}$

This is an incorrect option.

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