A nucleus of an element _{84}X^{202} emits an | vedclass

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(C) The initial nucleus is $_{84}X^{202}$.$1$. Emission of an $\alpha$-particle $(_{2}He^{4})$: The atomic number decreases by $2$ and the mass number

Vedclass · Accepted Answer

$83$

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(C) The initial nucleus is $_{84}X^{202}$.$1$. Emission of an $\alpha$-particle $(_{2}He^{4})$: The atomic number decreases by $2$ and the mass number decreases by $4$.$_{84}X^{202} \rightarrow _{82}Y^{198} + _{2}He^{4}$.$2$. Emission of a $\beta$-particle $(_{-1}e^{0})$: The atomic number increases by $1$ and the mass number remains unchanged.$_{82}Y^{198} \rightarrow _{83}Z^{198} + _{-1}e^{0}$.$3$. Emission of a gamma photon: Gamma emission does not change the atomic number or the mass number of the nucleus.Therefore,the final nucleus is $_{83}Z^{198}$,which has an atomic number of $83$.

$A$ nucleus of an element $_{84}X^{202}$ emits an $\alpha$-particle first,a $\beta$-particle next,and then a gamma photon. The final nucleus formed has an atomic number:

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