A radioactive nucleus decays as follows:X xri | vedclass

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(A) Let the initial nucleus be represented as $_Z X^A$.$1$. An $\alpha$-decay reduces the mass number by $4$ and atomic number by $2$.$2$. $A$ $\beta$

Vedclass · Accepted Answer

$72, 180$

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(A) Let the initial nucleus be represented as $_Z X^A$.$1$. An $\alpha$-decay reduces the mass number by $4$ and atomic number by $2$.$2$. $A$ $\beta$-decay increases the atomic number by $1$ and keeps the mass number unchanged.$3$. $A$ $\gamma$-decay does not change the mass number or atomic number.Given the decay chain:$X(Z, A) \xrightarrow{\alpha} X_1(Z-2, A-4) \xrightarrow{\beta} X_2(Z-1, A-4) \xrightarrow{\alpha} X_3(Z-3, A-8) \xrightarrow{\gamma} X_4(Z-3, A-8)$Given $X_4$ has atomic number $69$ and mass number $172$:$Z - 3 = 69 \implies Z = 72$$A - 8 = 172 \implies A = 180$Thus, the atomic number and mass number of $X$ are $72$ and $180$ respectively.

$A$ radioactive nucleus decays as follows:
$X \xrightarrow{\alpha} X_1 \xrightarrow{\beta} X_2 \xrightarrow{\alpha} X_3 \xrightarrow{\gamma} X_4$
If the mass number and atomic number of $X_4$ are $172$ and $69$ respectively, then the atomic number and mass number of $X$ 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 mass number and atomic number of $X_4$ are $172$ and $69$ respectively, then the atomic number and mass number of $X$ are: