A nucleus _Z{X^A} emits 3 alpha-particles and | vedclass

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(D) The initial nucleus is $_Z{X^A}$.When a nucleus emits an $\alpha$-particle $(_{2}^{4}He)$,the mass number decreases by $4$ and the atomic number d

Vedclass · Accepted Answer

$\frac{A - Z - 11}{Z - 1}$

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(D) The initial nucleus is $_Z{X^A}$.When a nucleus emits an $\alpha$-particle $(_{2}^{4}He)$,the mass number decreases by $4$ and the atomic number decreases by $2$.When it emits a $\beta$-particle $(_{-1}^{0}e)$,the mass number remains unchanged and the atomic number increases by $1$.After emitting $3 \alpha$-particles and $5 \beta$-particles,the new mass number $A'$ and atomic number $Z'$ are:$A' = A - 3(4) = A - 12$$Z' = Z - 3(2) + 5(1) = Z - 6 + 5 = Z - 1$Number of protons $P = Z' = Z - 1$Number of neutrons $N = A' - Z' = (A - 12) - (Z - 1) = A - Z - 11$Therefore,the ratio of neutrons to protons is $\frac{N}{P} = \frac{A - Z - 11}{Z - 1}$.

$A$ nucleus $_Z{X^A}$ emits $3 \alpha$-particles and $5 \beta$-particles. The ratio of total neutrons to protons in the final nucleus is:

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