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(C) In $\alpha$ decay,the mass number decreases by $4$ and the atomic number decreases by $2$.In $\beta$ decay,the mass number remains constant,but th

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$172, 69$

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(C) In $\alpha$ decay,the mass number decreases by $4$ and the atomic number decreases by $2$.In $\beta$ decay,the mass number remains constant,but the atomic number increases by $1$.In $\gamma$ decay,both the mass number and the atomic number remain constant.Given initial nucleus $X$ has mass number $A = 180$ and atomic number $Z = 72$.Step $1$: $X \xrightarrow{\alpha} X_1$$A_1 = 180 - 4 = 176$,$Z_1 = 72 - 2 = 70$.Step $2$: $X_1 \xrightarrow{\beta} X_2$$A_2 = 176$,$Z_2 = 70 + 1 = 71$.Step $3$: $X_2 \xrightarrow{\alpha} X_3$$A_3 = 176 - 4 = 172$,$Z_3 = 71 - 2 = 69$.Step $4$: $X_3 \xrightarrow{\gamma} X_4$$A_4 = 172$,$Z_4 = 69$.Thus,the mass number of $X_4$ is $172$ and the atomic number is $69$.

$A$ radioactive nucleus decays as follows:
$X \xrightarrow{\alpha} X_1 \xrightarrow{\beta} X_2 \xrightarrow{\alpha} X_3 \xrightarrow{\gamma} X_4$
If the atomic number and the mass number of $X$ are $72$ and $180$ respectively,then the mass number and atomic number of $X_4$ are:

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$A$ radioactive nucleus decays as follows:$X \xrightarrow{\alpha} X_1 \xrightarrow{\beta} X_2 \xrightarrow{\alpha} X_3 \xrightarrow{\gamma} X_4$If the atomic number and the mass number of $X$ are $72$ and $180$ respectively,then the mass number and atomic number of $X_4$ are: