Chapter 29, Problem 14P

To determine

The binding energy per nucleon of $\text{F}\text{e}$, $\text{M}\text{n}$ and $\text{C}\text{o}$.

Answer to Problem 14P

The binding energy per nucleon of $\text{F}\text{e}$ is greater than its neighbors.

Explanation of Solution

Section 1:

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

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

Explanation:

Formula to calculate the binding energy per nucleon is,

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

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

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.

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

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

Section 2:

To determine: The binding energy per nucleon of $\text{M}\text{n}$.

Answer: The binding energy per nucleon of $\text{M}\text{n}$ is $\text{8}\text{.77}\text{MeV/nucleon}$.

Explanation:

Formula to calculate the binding energy per nucleon is,

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

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

Substitute 25 for Z, 30 for N,1.008625 u for $m_n$, 1.007825 u for $m_p$, 54.938050 u for $m\left(\text{M}\text{n}\right)$, 931.5 MeV/u for $c^2$ and 55 for A in the above equation.

$\begin{array}{c}\frac{BE}{A}=\frac{\left[\left(26\right)\left(1.007825\text{u}\right)+\left(30\right)\left(1.008625\text{u}\right)-\left(55.938050\text{u}\right)\right]\left(931.5\text{MeV/u}\right)}{55}\\ =\text{8}\text{.77}\text{MeV/nucleon}\end{array}$

The binding energy per nucleon of $\text{M}\text{n}$ is $\text{8}\text{.77}\text{MeV/nucleon}$.

Section 3:

To determine: The binding energy per nucleon of $\text{C}\text{o}$.

Answer: The binding energy per nucleon of $\text{C}\text{o}$ is $\text{8}\text{.77}\text{MeV/nucleon}$.

Explanation:

Formula to calculate the binding energy per nucleon is,

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

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

Substitute 27 for Z, 32 for N,1.008625 u for $m_n$, 1.007825 u for $m_p$, 58.933200 u for $m\left(\text{C}\text{o}\right)$, 931.5 MeV/u for $c^2$ and 59 for A in the above equation.

$\begin{array}{c}\frac{BE}{A}=\frac{\left[\left(27\right)\left(1.007825\text{u}\right)+\left(32\right)\left(1.008625\text{u}\right)-\left(58.933200\text{u}\right)\right]\left(931.5\text{MeV/u}\right)}{59}\\ =\text{8}\text{.77}\text{MeV/nucleon}\end{array}$

The binding energy per nucleon of $\text{C}\text{o}$ is $\text{8}\text{.77}\text{MeV/nucleon}$.

Conclusion:

From Sections 1, 2 and 3, the binding energy per nucleon of $\text{F}\text{e}$ is greater than its neighbors.