Question

Two identical nuclei $A$ and $B$ of the same radioactive element undergo ${\beta }^{-}$ decay.

$A$ emits a ${\beta }^{-}$ particle and changes to $A^{{}^{\prime }}$.

$B$ emits a ${\beta }^{-}$ particle and then a $\gamma$-photon and changes to $B^{{}^{\prime }}$

• A. $A^{{}^{\prime }}$ and $B^{{}^{\prime }}$ have different atomic number and different mass number.
• B. $A^{{}^{\prime }}$ and $B^{{}^{\prime }}$ have different atomic number but the same mass number.
  
• C. $A{}^{\prime }$ and $B^{{}^{\prime }}$ have the same atomic number and different mass numbers.
• D. $A^{{}^{\prime }}$ and $B^{{}^{\prime }}$ have the same atomic number and mass number.

Answer 1

The correct option is D $A^{{}^{\prime }}$ and $B^{{}^{\prime }}$ have the same atomic number and mass number.

The emission of ${\beta }^{-}-$particle, will lead to increase in $Z$ by $1$ and no change in$A$.

As both $A$ and $B$ emits ${\beta }^{-}-$ particle, $A^{{}^{\prime }}\text{and}{B}^{{}^{\prime }}$ will have the same atomic and mass number.

The emission of $\gamma -$ photon will not have any change in $A\text{and}Z$.

Therefore, $A^{{}^{\prime }}\text{and}{B}^{{}^{\prime }}$ will have the same atomic and mass number.

But the energy of $B^{{}^{\prime }}$ will be less than $A^{{}^{\prime }}$.

Hence, option $\left(D\right)$ is correct.