The stability of ${ } _ { 6 } ^ { 11 } \mathrm { C }$ with respect to alpha, beta-plus and beta-minus decay is to be determined.
The following atomic masses are known:
${ } _ { 2 } ^ { 4 } \mathrm { He } : \quad 4.002603 \mathrm { u }$
${ } _ { 4 } ^ { 7 } \mathrm { Be } : \quad 7.016928 \mathrm { u }$
${ } _ { 5 } ^ { 11 } \mathrm {~B} : \quad 11.009305 \mathrm { u }$
${ } _ { 6 } ^ { 11 } \mathrm { C } : \quad 11.011433 \mathrm { u }$
${ } _ { 7 } ^ { 11 } \mathrm {~N} : \quad 11.026742 \mathrm { u }$
The ${ } _ { 6 } ^ { 11 } \mathrm { C }$ nucleus is

Question

Quizplus · Accepted Answer

To determine the stability of ${ } _ { 6 } ^ { 11 } \mathrm { C }$ with respect to different decay modes, we compare its mass with the masses of possible decay products:- **Alpha Decay**: This would produce ${ } _ { 4 } ^ { 7 } \mathrm { Be }$ and ${ } _ { 2 } ^ { 4 } \mathrm { He }$. The combined mass of these products is $7.016928 \mathrm { u } + 4.002603 \mathrm { u } = 11.019531 \mathrm { u }$, which is greater than the mass of ${ } _ { 6 } ^ { 11 } \mathrm { C }$, making alpha decay energetically unfavorable.- **Beta-minus Decay ($\beta^-$)**: This would convert a neutron in the carbon nucleus into a proton, forming ${ } _ { 7 } ^ { 11 } \mathrm { N }$. However, the mass of ${ } _ { 7 } ^ { 11 } \mathrm { N }$ is $11.026742 \mathrm { u }$, which is greater than the mass of ${ } _ { 6 } ^ { 11 } \mathrm { C }$, making $\beta^-$ decay energetically unfavorable.- **Beta-plus Decay ($\beta^+$)**: This would convert a proton in the carbon nucleus into a neutron, forming ${ } _ { 5 } ^ { 11 } \mathrm { B }$. The mass of ${ } _ { 5 } ^ { 11 } \mathrm { B }$ is $11.009305 \mathrm { u }$, which is less than the mass of ${ } _ { 6 } ^ { 11 } \mathrm { C }$, making $\beta^+$ decay energetically favorable.Therefore, ${ } _ { 6 } ^ { 11 } \mathrm { C }$ is subject to beta-plus decay only.

The Stability Of 611C{ } _ { 6 } ^ { 11 } \mathrm { C }611​C

The Stability Of ${ } _ { 6 } ^ { 11 } \mathrm { C }$