A radioactive nucleus with Z protons and N ne | vedclass

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(D) An $\alpha$-particle is a helium nucleus,represented as $_2^4	ext{He}$. When a nucleus emits an $\alpha$-particle,its atomic number $(Z)$ decreas

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$Z, N-4$

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(D) An $\alpha$-particle is a helium nucleus,represented as $_2^4	ext{He}$. When a nucleus emits an $\alpha$-particle,its atomic number $(Z)$ decreases by $2$ and its mass number $(A = Z+N)$ decreases by $4$. Since the mass number is the sum of protons and neutrons,the total number of nucleons decreases by $4$.$A$ $\beta^-$-particle is an electron,represented as $_{-1}^0	ext{e}$. When a nucleus emits a $\beta^-$-particle,its atomic number $(Z)$ increases by $1$,while the mass number remains unchanged.Given the emission of $1$ $\alpha$-particle and $2$ $\beta^-$-particles:Change in atomic number $(Z)$: $Z' = Z - 2 + (2 	imes 1) = Z$.Change in mass number $(A)$: $A' = A - 4 + (2 	imes 0) = A - 4$.Since $A = Z + N$,the new number of neutrons $N'$ is given by $N' = A' - Z' = (A - 4) - Z = (Z + N - 4) - Z = N - 4$.Therefore,the final number of protons is $Z$ and the final number of neutrons is $N - 4$.

$A$ radioactive nucleus with $Z$ protons and $N$ neutrons emits an $\alpha$-particle,$2\beta^-$-particles,and $2$ gamma rays. The number of protons and neutrons in the nucleus left after the decay,respectively,are:

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$37$ Rutherford equals

An element $X$ decays into element $Z$ by a two-step process:
$X \rightarrow Y + 4e$
$Y \rightarrow Z + 2e^{-}$
Then:

In a radioactive reaction $_{92}X^{232} \to _{82}Y^{204}$,the number of $\alpha$-particles emitted is:

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