A radioactive nucleus (initial mass number A | vedclass

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(C) Initial nucleus: Mass number $= A$, Atomic number $= Z$, Number of neutrons $(N) = A - Z$.Emission of $3 \,\alpha-$ particles:Each $\alpha-$ parti

Vedclass · Accepted Answer

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

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(C) Initial nucleus: Mass number $= A$, Atomic number $= Z$, Number of neutrons $(N) = A - Z$.Emission of $3 \,\alpha-$ particles:Each $\alpha-$ particle emission decreases mass number by $4$ and atomic number by $2$.New mass number $A' = A - (3 	imes 4) = A - 12$.New atomic number $Z' = Z - (3 	imes 2) = Z - 6$.Emission of $2$ positrons ($\beta^+$ decay):Each positron emission leaves mass number unchanged and decreases atomic number by $1$.Final mass number $A'' = A' = A - 12$.Final atomic number $Z'' = Z' - 2 = (Z - 6) - 2 = Z - 8$.Final number of protons $p = Z'' = Z - 8$.Final number of neutrons $n = A'' - Z'' = (A - 12) - (Z - 8) = A - Z - 4$.Ratio of neutrons to protons $= \frac{n}{p} = \frac{A - Z - 4}{Z - 8}$.

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

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