Chapter 40, Problem 15P

(a)

To determine

The binding energy and the binding energy per nucleon for ${}^{\text{12}}\text{C}$ .

Explanation of Solution

Given:

The mass of one proton is $1.007825\text{u}$ .

The mass of one neutron is $1.008665\text{u}$ .

The mass of ${}^{\text{12}}\text{C}$ is $12\text{u}$ .

Formula used:

Write the expression for the masses of protons and neutrons in ${}^{\text{12}}\text{C}$ .

  $m_{\text{p+a}}=6m_{\text{p}}+6m_{\text{a}}$

Here, $m_{\text{p}}$ is the mass of one proton, $m_{\text{a}}$ is the mass of one neutron and $m_{\text{p+a}}$ is the total mass of both proton and neutron in ${}^{\text{12}}\text{C}$ .

Write the expression for the mass difference between total mass of both proton and neutron in ${}^{\text{12}}\text{C}$ and mass of ${}^{\text{12}}\text{C}$ alone.

  $\Delta m=\left(6m_{\text{p}}+6m_{\text{a}}\right)-m_{{}^{\text{12}}\text{C}}$   ........ (1)

Here, $\Delta m$ is the mass difference between total mass of both proton and neutron in ${}^{\text{12}}\text{C}$ and $m_{{}^{\text{12}}\text{C}}$ is the mass of ${}^{\text{12}}\text{C}$ .

Write the expression for the binding energy.

  $E_{\text{b}}=\left(\Delta m\right)c^2$   ........ (2)

Here, $E_{\text{b}}$ is the binding energy and $c$ is the speed of light.

Write the expression for the binding energy per nucleon.

  ${E^{\prime }}_{\text{b}}=\frac{E_{\text{b}}}{A}$   ........ (3)

Here, ${E^{\prime }}_{\text{b}}$ is the binding energy per nucleon and $A$ is the number of neutrons.

Calculation:

Substitute $1.007825\text{u}$ for $m_{\text{p}}$ , $1.008665\text{u}$ for $m_{\text{a}}$ and $12\text{u}$ for $m_{{}^{\text{12}}\text{C}}$ in equation (1).

  $\begin{array}{c}\Delta m=\left(6\left(1.007825\text{u}\right)+6\left(1.008665\text{u}\right)\right)-\left(12\text{u}\right)\\ =0.098940\text{u}\end{array}$

Substitute $0.098940\text{u}$ for $\Delta m$ in equation (2).

  $\begin{array}{c}{E}_{\text{b}}=\left(0.098940\text{u}\right)c^2\left(\frac{\frac{931.5\text{MeV}}{c^2}}{\text{u}}\right)\\ =92.2\text{MeV}\end{array}$

Substitute $92.2\text{MeV}$ for $E_{\text{b}}$ and $12$ for $A$ in equation (3).

  $\begin{array}{c}{E^{\prime }}_{\text{b}}=\frac{\left(92.2\text{MeV}\right)}{12}\\ =7.68\text{MeV}\end{array}$

Conclusion:

Thus, the binding energy and the binding energy per nucleon for ${}^{\text{12}}\text{C}$ is $92.2\text{MeV}$ and $7.68\text{MeV}$ .

(b)

To determine

The binding energy and the binding energy per nucleon for ${}^{\text{56}}\text{F}$ .

Explanation of Solution

Given:

The mass of one proton is $1.007825\text{u}$ .

The mass of one neutron is $1.008665\text{u}$ .

The mass of ${}^{\text{56}}\text{F}$ is $56\text{u}$ .

Formula used:

Write the expression for the masses of protons and neutrons in ${}^{\text{56}}\text{F}$ .

  $m_{\text{p+a}}=26m_{\text{p}}+30m_{\text{a}}$

Here, $m_{\text{p}}$ is the mass of one proton, $m_{\text{a}}$ is the mass of one neutron and $m_{\text{p+a}}$ is the total mass of both proton and neutron in ${}^{\text{56}}\text{F}$ .

Write the expression for the mass difference between total mass of both proton and neutron in ${}^{\text{56}}\text{F}$ and mass of ${}^{\text{56}}\text{F}$ alone.

  $\Delta m=\left(26m_{\text{p}}+30m_{\text{a}}\right)-m_{{}^{\text{56}}\text{F}}$   ........ (4)

Here, $\Delta m$ is the mass difference between total mass of both proton and neutron in ${}^{\text{56}}\text{F}$ and $m_{{}^{\text{56}}\text{F}}$ is the mass of ${}^{\text{56}}\text{F}$ .

Write the expression for the binding energy.

  $E_{\text{b}}=\left(\Delta m\right)c^2$   ........ (2)

Here, $E_{\text{b}}$ is the binding energy and $c$ is the speed of light.

Write the expression for the binding energy per nucleon.

  ${E^{\prime }}_{\text{b}}=\frac{E_{\text{b}}}{A}$   ........ (3)

Here, ${E^{\prime }}_{\text{b}}$ is the binding energy per nucleon and $A$ is the number of neutrons.

Calculation:

Substitute $1.007825\text{u}$ for $m_{\text{p}}$ , $1.008665\text{u}$ for $m_{\text{a}}$ and $56\text{u}$ for $m_{{}^{\text{56}}\text{F}}$ in equation (4).

  $\begin{array}{c}\Delta m=\left(26\left(1.007825\text{u}\right)+30\left(1.008665\text{u}\right)\right)-\left(56\text{u}\right)\\ =0.528458\text{u}\end{array}$

Substitute $0.528458\text{u}$ for $\Delta m$ in equation (2).

  $\begin{array}{c}{E}_{\text{b}}=\left(0.528458\text{u}\right)c^2\left(\frac{\frac{931.5\text{MeV}}{c^2}}{\text{u}}\right)\\ =492\text{MeV}\end{array}$

Substitute $92.2\text{MeV}$ for $E_{\text{b}}$ and $56$ for $A$ in equation (3).

  $\begin{array}{c}{E^{\prime }}_{\text{b}}=\frac{\left(492\text{MeV}\right)}{56}\\ =8.79\text{MeV}\end{array}$

Conclusion:

Thus, the binding energy and the binding energy per nucleon for ${}^{\text{56}}\text{F}$ is $492\text{MeV}$ and $8.79\text{MeV}$ .

(c)

To determine

The binding energy and the binding energy per nucleon for ${}^{\text{238}}\text{U}$ .

Explanation of Solution

Given:

The mass of one proton is $1.007825\text{u}$ .

The mass of one neutron is $1.008665\text{u}$ .

The mass of ${}^{\text{238}}\text{U}$ is $238\text{u}$ .

Formula used:

Write the expression for the masses of protons and neutrons in ${}^{\text{238}}\text{U}$ .

  $m_{\text{p+a}}=92m_{\text{p}}+146m_{\text{a}}$

Here, $m_{\text{p}}$ is the mass of one proton, $m_{\text{a}}$ is the mass of one neutron and $m_{\text{p+a}}$ is the total mass of both proton and neutron in ${}^{\text{238}}\text{U}$ .

Write the expression for the mass difference between total mass of both proton and neutron in ${}^{\text{238}}\text{U}$ and mass of ${}^{\text{238}}\text{U}$ alone.

  $\Delta m=\left(92m_{\text{p}}+146m_{\text{a}}\right)-m_{{}^{\text{238}}\text{U}}$   ........ (5)

Here, $\Delta m$ is the mass difference between total mass of both proton and neutron in ${}^{\text{238}}\text{U}$ and $m_{{}^{\text{238}}\text{U}}$ is the mass of ${}^{\text{238}}\text{U}$ .

Write the expression for the binding energy.

  $E_{\text{b}}=\left(\Delta m\right)c^2$   ........ (2)

Here, $E_{\text{b}}$ is the binding energy and $c$ is the speed of light.

Write the expression for the binding energy per nucleon.

  ${E^{\prime }}_{\text{b}}=\frac{E_{\text{b}}}{A}$   ........ (3)

Here, ${E^{\prime }}_{\text{b}}$ is the binding energy per nucleon and $A$ is the number of neutrons.

Calculation:

Substitute $1.007825\text{u}$ for $m_{\text{p}}$ , $1.008665\text{u}$ for $m_{\text{a}}$ and $238\text{u}$ for $m_{{}^{\text{238}}\text{U}}$ in equation (5).

  $\begin{array}{c}\Delta m=\left(92\left(1.007825\text{u}\right)+146\left(1.008665\text{u}\right)\right)-\left(238\text{u}\right)\\ =1.934207\text{u}\end{array}$

Substitute $1.934207\text{u}$ for $\Delta m$ in equation (2).

  $\begin{array}{c}{E}_{\text{b}}=\left(1.934207\text{u}\right)c^2\left(\frac{\frac{931.5\text{MeV}}{c^2}}{\text{u}}\right)\\ =1802\text{MeV}\end{array}$

Substitute $1802\text{MeV}$ for $E_{\text{b}}$ and $238$ for $A$ in equation (3).

  $\begin{array}{c}{E^{\prime }}_{\text{b}}=\frac{\left(1802\text{MeV}\right)}{238}\\ =7.57\text{MeV}\end{array}$

Conclusion:

Thus, the binding energy and the binding energy per nucleon for ${}^{\text{238}}\text{U}$ is $1802\text{MeV}$ and $7.57\text{MeV}$ .