Chapter 30, Problem 16P

To determine

The binding energy of ${}_{\text{1}}{}^{\text{2}}\text{H}$ (deuterium).

Answer to Problem 16P

Solution:

The binding energy of ${}_{\text{1}}{}^{\text{2}}\text{H}$ (deuterium)is $\text{2}\text{.224}\text{MeV}$.

Explanation of Solution

Formula used:

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

Solution:

Calculation:

The mass of ${}_{\text{1}}{}^{\text{2}}\text{H}$ is $2.014102\text{u}$. Mass of the constituents of ${}_{\text{1}}{}^{\text{2}}\text{H}$ is mass of proton + mass of neutron + mass of electron = $1.007276\text{u}+1.008665\text{u}+0.0005486\text{u}=\text{2}\text{.0164896}\text{u}$.

Mass defect is $\Delta \text{m}=\text{2}\text{.0164896}\text{u}-2.014102\text{u}=\text{0}\text{.0023876}\text{u}$.

Binding energy $\text{0}\text{.0023876}\text{u}\times \frac{931.5\text{MeV}}{\text{u}}=2.224\text{MeV}$

Conclusion:

The binding energy appears as radiation, kinetic energy etc.The binding energy of ${}_{\text{1}}{}^{\text{2}}\text{H}$ (deuterium)is $\text{2}\text{.224 MeV}$.