Chapter 30, Problem 19P

To determine

The comparison between average binding energy per nucleon in ${}_{\text{11}}{}^{\text{23}}\text{N}\text{a}$ to that in ${}_{\text{11}}{}^{\text{24}}\text{N}\text{a}$.

Answer to Problem 19P

Solution:

The binding energy per nucleon in ${}_{\text{11}}{}^{\text{23}}\text{N}\text{a}$ is greater than that in ${}_{\text{11}}{}^{\text{24}}\text{N}\text{a}$.

Explanation of Solution

Formula used:

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

Calculation:

Mass of ${}_{\text{11}}{}^{\text{23}}\text{N}\text{a}$ is $22.989769\text{u}$. Mass of neutron ${}_0{}^1\text{n}$ is $1.008665\text{u}$. Mass of proton ${}_1{}^1\text{H}$ is $1.007825\text{u}$. Mass defect is $\Delta \text{m}=12\left(1.008665\text{u}\right)+11\left(1.007825\text{u}\right)-22.989769\text{u}=\text{0}\text{.200286}\text{u}$.

Binding energy per nucleon is $\frac{\text{0}\text{.200286}\text{u}}{23}\times \frac{931.5\text{MeV}}{\text{u}}=8.113\text{MeV}$.

Mass of ${}_{\text{11}}{}^{\text{24}}\text{N}\text{a}$ is $23.990963\text{u}$. Mass of neutron ${}_0{}^1\text{n}$ is $1.008665\text{u}$. Mass of proton ${}_1{}^1\text{H}$ is $1.007825\text{u}$. Mass defect is $\Delta \text{m}=13\left(1.008665\text{u}\right)+11\left(1.007825\text{u}\right)-23.990963\text{u}=\text{0}\text{.207757}\text{u}$.

Binding energy per nucleon is $\frac{\text{0}\text{.207757}\text{u}}{24}\times \frac{931.5\text{MeV}}{\text{u}}=8.063\text{MeV}$.