Chapter 30, Problem 23P

To determine

The decay nucleus ${}_6{}^{11}\text{C}\to {}_5{}^{10}\text{B}+\text{p}$ is not possible since energy would not be conserved.

Answer to Problem 23P

Solution:

Explanation of Solution

Formula used:

Binding energy is $\Delta m{c}^2$.

Solution:

Calculation:

Mass of ${}_6{}^{11}\text{C}$ is $11.011434\text{u}$. Mass of proton ${}_1{}^1\text{H}$ is $1.007825\text{u}$. Mass of ${}_5{}^{10}\text{B}$ is $10.013927\text{u}$.

Combined mass of ${}_5{}^{10}\text{B}$ and ${}_1{}^1\text{H}$ is $10.013927\text{u}+1.007825\text{u}=\text{11}\text{.021752}\text{u}$. Since this mass is greater than $11.011434\text{u}$, energy is not conserved, and the decay is not possible.

Conclusion:

Mass is equivalent to energy. Mass in a decay reaction cannot increase.