Chapter 30, Problem 17P

To determine

The binding energy of last neutron in ${}_{\text{11}}{}^{\text{23}}\text{N}\text{a}$.

Answer to Problem 17P

Solution:

The binding energy of last neutron in ${}_{\text{11}}{}^{\text{23}}\text{N}\text{a}$ is $12.42\text{MeV}$.

Explanation of Solution

Formula used:

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

Solution:

When ${}_{\text{11}}{}^{\text{23}}\text{N}\text{a}$ loses one neutron it becomes ${}_{\text{11}}{}^{\text{22}}\text{N}\text{a}$.

Calculation:

Mass of ${}_{\text{11}}{}^{\text{22}}\text{N}\text{a}$ is $21.994437\text{u}$. Mass of neutron ${}_0{}^1\text{n}$ is $1.008665\text{u}$. Mass of ${}_{\text{11}}{}^{\text{23}}\text{N}\text{a}$ is $22.989769\text{u}$

Mass defect is $\Delta \text{m}=21.994437\text{u}+1.008665\text{u}-22.989769\text{u}=\text{0}\text{.013333}\text{u}$.

Binding energy $\text{0}\text{.013333}\text{u}\times \frac{931.5\text{MeV}}{\text{u}}=12.42\text{MeV}$

Conclusion:

The binding energy of last neutron in ${}_{\text{11}}{}^{\text{23}}\text{N}\text{a}$ is $12.42\text{MeV}$.The binding energy appears as radiation, kinetic energy etc.