Chapter 29, Problem 12P

(a)

To determine

The binding energy per nucleon of $\text{H}$.

Answer to Problem 12P

The binding energy per nucleon of $\text{H}$ is $1.11\text{MeV/nucleon}$.

Explanation of Solution

Formula to calculate the binding energy per nucleon is,

$\frac{BE}{A}=\frac{\left[Z{m}_p-N{m}_n-m\left(\text{H}\right)\right]c^2}{A}$

• $m_p$ is the mass of the proton.
• $m_n$ is the mass of the neutron.
• Z is the atomic number.
• N is the neutron number.
• $m\left(\text{H}\right)$ is the mass of $\text{H}$.
• c is the speed of light.
• A is the mass number.

Substitute 1 for Z, 1 for N,1.008625 u for $m_n$, 1.007825 u for $m_p$, 2.014102 u for $m\left(\text{H}\right)$, 931.5 MeV/u for $c^2$ and 2 for A in the above equation to find $BE/A$.

$\begin{array}{c}\frac{BE}{A}=\frac{\left[\left(1\right)\left(1.007825\text{u}\right)-\left(1\right)\left(1.008625\text{u}\right)-\left(2.014102\text{u}\right)\right]\left(931.5\text{MeV/u}\right)}{2}\\ =1.11\text{MeV/nucleon}\end{array}$

Conclusion:

The binding energy per nucleon of $\text{M}\text{g}$ is $1.11\text{MeV/nucleon}$.

(b)

To determine

The binding energy per nucleon of $\text{H}\text{e}$.

Answer to Problem 12P

The binding energy per nucleon of $\text{H}\text{e}$ is $\text{7}\text{.07}\text{MeV/nucleon}$.

Explanation of Solution

Formula to calculate the binding energy per nucleon is,

$\frac{BE}{A}=\frac{\left[Z{m}_p-N{m}_n-m\left(\text{H}\text{e}\right)\right]c^2}{A}$

• $m\left(\text{H}\text{e}\right)$ is the mass of $\text{H}\text{e}$.

Substitute 2 for Z, 2 for N,1.008625 u for $m_n$, 1.007825 u for $m_p$, 4.002603 u for $m\left(\text{H}\text{e}\right)$, 931.5 MeV/u for $c^2$ and 4 for A in the above equation to find $BE/A$.

$\begin{array}{c}\frac{BE}{A}=\frac{\left[\left(2\right)\left(1.007825\text{u}\right)-\left(2\right)\left(1.008625\text{u}\right)-\left(4.002603\text{u}\right)\right]\left(931.5\text{MeV/u}\right)}{4}\\ =\text{7}\text{.07}\text{MeV/nucleon}\end{array}$

Conclusion:

The binding energy per nucleon of $\text{R}\text{b}$ is $\text{7}\text{.07}\text{MeV/nucleon}$.

(c)

To determine

The binding energy per nucleon of $\text{F}\text{e}$.

Answer to Problem 12P

The binding energy per nucleon of $\text{F}\text{e}$ is $\text{8}\text{.79}\text{MeV/nucleon}$.

Explanation of Solution

Formula to calculate the binding energy per nucleon is,

$\frac{BE}{A}=\frac{\left[Z{m}_p-N{m}_n-m\left(\text{F}\text{e}\right)\right]c^2}{A}$

• $m\left(\text{F}\text{e}\right)$ is the mass of $\text{F}\text{e}$.

Substitute 26 for Z, 30 for N,1.008625 u for $m_n$, 1.007825 u for $m_p$, 55.934942 u for $m\left(\text{F}\text{e}\right)$, 931.5 MeV/u for $c^2$ and 56 for A in the above equation to find $BE/A$.

$\begin{array}{c}\frac{BE}{A}=\frac{\left[\left(26\right)\left(1.007825\text{u}\right)-\left(30\right)\left(1.008625\text{u}\right)-\left(55.934942\text{u}\right)\right]\left(931.5\text{MeV/u}\right)}{56}\\ =\text{8}\text{.79}\text{MeV/nucleon}\end{array}$

Conclusion:

The binding energy per nucleon of $\text{R}\text{b}$ is $\text{8}\text{.79}\text{MeV/nucleon}$.

(d)

To determine

The binding energy per nucleon of $\text{U}$.

Answer to Problem 12P

The binding energy per nucleon of $\text{U}$ is $\text{7}\text{.57}\text{MeV/nucleon}$.

Explanation of Solution

Formula to calculate the binding energy per nucleon is,

$\frac{BE}{A}=\frac{\left[Z{m}_p-N{m}_n-m\left(\text{U}\right)\right]c^2}{A}$

• $m\left(\text{U}\right)$ is the mass of $\text{U}$.

Substitute 92 for Z, 146 for N,1.008625 u for $m_n$, 1.007825 u for $m_p$, 238.050783 u for $m\left(\text{U}\right)$, 931.5 MeV/u for $c^2$ and 238 for A in the above equation.

$\begin{array}{c}\frac{BE}{A}=\frac{\left[\left(92\right)\left(1.007825\text{u}\right)-\left(146\right)\left(1.008625\text{u}\right)-\left(238.050783\text{u}\right)\right]\left(931.5\text{MeV/u}\right)}{238}\\ =\text{7}\text{.57}\text{MeV/nucleon}\end{array}$

Conclusion:

The binding energy per nucleon of $\text{U}$ is $\text{7}\text{.57}\text{MeV/nucleon}$