A nucleus _n{X^m} emits one alpha and one bet | vedclass

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(D) The initial nucleus is $_n{X^m}$.When it emits one $\alpha$ particle $(_{2}He^{4})$,the atomic number decreases by $2$ and the mass number decreas

Vedclass · Accepted Answer

$_{n-1}Z^{m-4}$

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(D) The initial nucleus is $_n{X^m}$.When it emits one $\alpha$ particle $(_{2}He^{4})$,the atomic number decreases by $2$ and the mass number decreases by $4$. The nucleus becomes $_{n-2}X^{m-4}$.When this nucleus emits one $\beta$ particle $(_{-1}e^{0})$,the atomic number increases by $1$ while the mass number remains unchanged. The new atomic number is $(n-2) + 1 = n-1$.Thus,the resulting nucleus is $_{n-1}Z^{m-4}$.

$A$ nucleus $_n{X^m}$ emits one $\alpha$ and one $\beta$ particle. The resulting nucleus is:

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