Question

Calculate the maximum kinetic energy of the $\beta$ particle emitted in the following decay equation in $\text{MeV}$.

${}_7{\text{N}}^{12}\to {}_6{\text{C}}^{{12}^{\ast }}+{}_{+1}{e}^0+\nu$
${}_6{\text{C}}^{{12}^{\ast }}\to {}_6{\text{C}}^{12}+\gamma \left(4.43\text{MeV}\right)$

The atomic mass of ${\text{N}}^{12}$ is $12.018613\text{u}$. The atomic mass of ${\text{C}}^{12}$ is $12\text{u}$.

• A. $11.9$
• B. $119$
• C. $12.9$
• D. $0.019$

Answer 1

The correct option is C $12.9$

Mass defect between ${}_7{\text{N}}^{12}$ and ${}_6{\text{C}}^{12}$

$\Delta m=12.018613\text{u}-12\text{u}$

$\Rightarrow \Delta m=0.018613\text{u}$

The energy released in the reaction is given by,

$E=\Delta m\times c^2$

We know that,

$1\text{u}\times c^2=931.5\text{MeV}$

$\Rightarrow E=0.018613\times 931.5\text{MeV}$

$\Rightarrow E=17.3380\text{MeV}$

Since, a $\gamma$ radiation of energy $4.43\text{MeV}$ is released,

The maximum $K.E$ of the $\beta$ particle emitted,

$K.E_{\beta }=17.3380\text{MeV}-4.43\text{MeV}$

$\Rightarrow K.E_{\beta }=12.9\text{MeV}$

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