In the nuclear decay sequence given below:_Z{ | vedclass

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(B) The given decay sequence is: ${}_Z^AX 	o {}_{Z + 1}^AY 	o {}_{Z - 1}^{A - 4}K 	o {}_{Z - 1}^{A - 4}K$.$1$. In the first step,${}_Z^AX 	o {}_{Z

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$\beta, \alpha, \gamma$

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(B) The given decay sequence is: ${}_Z^AX 	o {}_{Z + 1}^AY 	o {}_{Z - 1}^{A - 4}K 	o {}_{Z - 1}^{A - 4}K$.$1$. In the first step,${}_Z^AX 	o {}_{Z + 1}^AY$: The atomic number increases by $1$ while the mass number remains the same. This corresponds to the emission of a $\beta^-$ particle.$2$. In the second step,${}_{Z + 1}^AY 	o {}_{Z - 1}^{A - 4}K$: The atomic number decreases by $2$ and the mass number decreases by $4$. This corresponds to the emission of an $\alpha$ particle.$3$. In the third step,${}_{Z - 1}^{A - 4}K 	o {}_{Z - 1}^{A - 4}K$: There is no change in the atomic number or mass number,which indicates the emission of a $\gamma$ ray (de-excitation of the nucleus).Therefore,the sequence of particles emitted is $\beta, \alpha, \gamma$.

In the nuclear decay sequence given below:
$_Z{X^A} \to {}_{Z + 1}{Y^A} \to {}_{Z - 1}{K^{A - 4}} \to {}_{Z - 1}{K^{A - 4}}$
the particles emitted in the sequence are:

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In the nuclear decay sequence given below:$_Z{X^A} \to {}_{Z + 1}{Y^A} \to {}_{Z - 1}{K^{A - 4}} \to {}_{Z - 1}{K^{A - 4}}$the particles emitted in the sequence are: