A radioactive nucleus (initial mass number A | vedclass

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(C) The initial nucleus is represented as $_{Z}X^{A}$.When an $\alpha$-particle $(_{2}He^{4})$ is emitted,the mass number decreases by $4$ and the ato

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The initial nucleus is represented as $_{Z}X^{A}$.When an $\alpha$-particle $(_{2}He^{4})$ is emitted,the mass number decreases by $4$ and the atomic number decreases by $2$.For $3$ $\alpha$-particles,the change is: $\Delta A = 3 	imes 4 = 12$ and $\Delta Z = 3 	imes 2 = 6$.After $3$ $\alpha$ emissions,the nucleus becomes $_{Z-6}Y^{A-12}$.When a positron $(_{+1}e^{0})$ is emitted,the mass number remains unchanged and the atomic number decreases by $1$.For $2$ positrons,the change is: $\Delta A = 2 	imes 0 = 0$ and $\Delta Z = 2 	imes 1 = 2$.After $2$ positron emissions,the final nucleus is $_{Z-6-2}Y^{A-12} = _{Z-8}Y^{A-12}$.Number of protons $(P)$ = $Z - 8$.Number of neutrons $(N)$ = Mass number - Atomic number = $(A - 12) - (Z - 8) = A - Z - 4$.The ratio of neutrons to protons is $\frac{N}{P} = \frac{A - Z - 4}{Z - 8}$.

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

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