In the given reaction _Z{X^A} to _{Z+1}{Y^A} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$\beta, \alpha, \gamma$

Vedclass · Answer

(B) The given reaction sequence is: $_Z{X^A} 	o _{Z+1}{Y^A} 	o _{Z-1}{K^{A-4}} 	o _{Z-1}{K^{A-4}}$.$1$. In the first step,$_Z{X^A} 	o _{Z+1}{Y^A}$,the atomic number increases by $1$ while the mass number remains the same. This corresponds to the emission of a $\beta^-$-particle $(_{-1}e^0)$.$2$. In the second step,$_{Z+1}{Y^A} 	o _{Z-1}{K^{A-4}}$,the atomic number decreases by $2$ and the mass number decreases by $4$. This corresponds to the emission of an $\alpha$-particle $(_{2}He^4)$.$3$. In the third step,$_{Z-1}{K^{A-4}} 	o _{Z-1}{K^{A-4}}$,there is no change in atomic number or mass number,which indicates the emission of a $\gamma$-ray (electromagnetic radiation).Therefore,the sequence of emissions is $\beta, \alpha, \gamma$.

In the given reaction $_Z{X^A} \to _{Z+1}{Y^A} \to _{Z-1}{K^{A-4}} \to _{Z-1}{K^{A-4}}$,radioactive radiations are emitted in the sequence:

Similar Questions

In the given nuclear reaction,the element $X$ is:
${ }_{11}^{22} Na \rightarrow X + e ^{+} + \nu$

The radioactive nucleus $_7N^{13}$ decays to $_6C^{13}$ through the emission of

Which of the following can pass through $20 \, cm$ thickness of steel?

${}^{238}U$ has $92$ protons and $238$ nucleons. It decays by emitting an alpha particle and becomes:

Curie is a unit of

Vedclass Test Series

Exam Paper Generator

Online Exam Module