Chapter 40, Problem 64P

(a)

To determine

The binding energy of a pair of $\alpha$ - particles from the atomic masses of ${}^{\text{4}}\text{He}$ and ${}^{\text{16}}\text{O}$ .

Explanation of Solution

Given:

The mass of one ${}^{\text{4}}\text{He}$ nucleus is $4.002603\text{u}$ .

The mass of one ${}^{\text{16}}\text{O}$ nucleus is $15.994915\text{u}$ .

The number of $\alpha$ - particles in ${}^{\text{16}}\text{O}$ is $4$ .

The number of $\alpha$ - particles at each vertex of the tetrahedron structure is $1$ .

Formula used:

Write the expression for the mass difference between $\alpha$ - particles and ${}^{\text{16}}\text{O}$ .

  $\Delta m=4m_{\alpha }-m_{{}^{\text{16}}\text{O}}$

Here, $\Delta m$ is the mass difference between $\alpha$ - particles and ${}^{\text{16}}\text{O}$ , $m_{\alpha }$ is the mass of $\alpha$ - particles and $m_{{}^{\text{16}}\text{O}}$ is the mass of one ${}^{\text{16}}\text{O}$ .

Write the expression for the binding energy of the pair of $\alpha$ - particles.

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

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

Substitute $4m_{\alpha }-m_{{}^{\text{16}}\text{O}}$ for $\Delta m$ in the above expression.

  $E_{\text{b}}=\left(4m_{\alpha }-m_{{}^{\text{16}}\text{O}}\right)c^2$ …… (1)

Write the expression for the total binding energy for the regular tetrahedron structure.

  $\begin{array}{c}{E}_{\text{b}\text{t}}=\frac{E_{\text{b}}}{\text{bond}}\\ =\frac{E_{\text{b}}}{\text{6}}\end{array}$ …… (2)

Here, $E_{\text{b}\text{t}}$ is the total binding energy for the regular tetrahedron structure.

Calculation:

Substitute $4.002603\text{u}$ for $m_{\alpha }$ and $15.994915\text{u}$ in equation (1).

  $\begin{array}{c}{E}_{\text{b}}=\left(4\left(4.002603\text{u}\right)-\left(15.994915\text{u}\right)\right)c^2\\ =\left(16.010412\text{u}-15.994915\text{u}\right)c^2\\ =\left(0.015497\text{u}\right)c^2\end{array}$

Substitute $\left(0.015497\text{u}\right)c^2$ for $E_{\text{b}}$ in equation (2).

  $\begin{array}{c}{E}_{\text{b}\text{t}}=\frac{\left(0.015497\text{u}\right)c^2}{\text{6}}\\ =\frac{\left(0.015497\text{u}\right)c^2}{\text{6}}\left(\frac{\frac{931.5\text{MeV}}{c^2}}{\text{u}}\right)\\ =2.406\text{MeV}\end{array}$

Conclusion:

Thus, the binding energy of a pair of $\alpha$ - particles from the atomic masses of ${}^{\text{4}}\text{He}$ and ${}^{\text{16}}\text{O}$ is $2.406\text{MeV}$ .

(b)

To determine

The binding energy of ${}^{\text{12}}\text{C}$ from the given model and from the atomic mass of ${}^{\text{12}}\text{C}$ and comparison between the two results.

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 one ${}^{\text{12}}\text{C}$ nucleus is $12.000000\text{u}$ .

The mass of one ${}^{\text{4}}\text{He}$ nucleus is $4.002603\text{u}$ .

The mass of one ${}^{\text{16}}\text{O}$ nucleus is $15.994915\text{u}$ .

The number of $\alpha$ - particles in ${}^{\text{16}}\text{O}$ is $4$ .

The number of $\alpha$ - particles at each vertex of the tetrahedron structure is $1$ .

Formula used:

Write the expression for the mass difference between $\alpha$ - particles and ${}^{\text{16}}\text{O}$ .

  $\Delta m=4m_{\alpha }-m_{{}^{\text{16}}\text{O}}$

Here, $\Delta m$ is the mass difference between $\alpha$ - particles and ${}^{\text{16}}\text{O}$ , $m_{\alpha }$ is the mass of $\alpha$ - particles and $m_{{}^{\text{16}}\text{O}}$ is the mass of one ${}^{\text{16}}\text{O}$ .

Write the expression for the binding energy of the pair of $\alpha$ - particles.

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

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

Substitute $4m_{\alpha }-m_{{}^{\text{16}}\text{O}}$ for $\Delta m$ in the above expression.

  $E_{\text{b}}=\left(4m_{\alpha }-m_{{}^{\text{16}}\text{O}}\right)c^2$ …… (1)

Write the expression for the total binding energy for the regular tetrahedron structure.

  $\begin{array}{c}{E}_{\text{b}\text{t}}=\frac{E_{\text{b}}}{\text{bond}}\\ =\frac{E_{\text{b}}}{\text{6}}\end{array}$ …… (2)

Here, $E_{\text{b}\text{t}}$ is the total binding energy for the regular tetrahedron structure.

Write the expression for the masses of protons and neutrons in ${}^{\text{4}}\text{He}$ .

  $m_{\text{p+a}}=2\left(m_{\text{p}}+m_{\text{a}}\right)$

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 ${}^{\text{4}}\text{He}$ .

Write the expression for the mass difference between total mass of both proton and neutron in ${}^{\text{4}}\text{He}$ and mass of ${}^{\text{4}}\text{He}$ alone.

  $\Delta m=2\left(m_{\text{p}}+m_{\text{a}}\right)-m_{{}^{\text{4}}\text{He}}$

Here, $\Delta m$ is the mass difference between total mass of both proton and neutron in ${}^{\text{4}}\text{He}$ and $m_{{}^{\text{4}}\text{He}}$ is the mass of ${}^{\text{4}}\text{He}$ .

Write the expression for the binding energy of ${}^{\text{4}}\text{He}$ .

  $E_{\text{b}}\left({}^{\text{4}}\text{He}\right)=\left(\Delta m\right)c^2$

Here, $E_{\text{b}}\left({}^{\text{4}}\text{He}\right)$ is the binding energy of ${}^{\text{4}}\text{He}$ and $c$ is the speed of light.

Substitute $2\left(m_{\text{p}}+m_{\text{a}}\right)-m_{{}^{\text{4}}\text{He}}$ for $\Delta m$ in the above expression.

  $E_{\text{b}}\left({}^{\text{4}}\text{He}\right)=\left(2\left(m_{\text{p}}+m_{\text{a}}\right)-m_{{}^{\text{4}}\text{He}}\right)c^2$ …… (3)

Write the expression for the binding energy of ${}^{\text{12}}\text{C}$ from the given model.

  $E_{\text{b}}\left({}^{\text{12}}\text{C}\right)=3\left(E_{\text{b}}\left({}^{\text{4}}\text{He}\right)\right)+3\left(E_{\text{b}\text{t}}\right)$ …… (4)

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

  $m_{\text{p+a}}=6\left(m_{\text{p}}+m_{\text{a}}\right)$

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.

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

Here, $\Delta m$ is the mass difference between total mass of both proton and neutron in ${}^{\text{4}}\text{He}$ and $m_{{}^{\text{12}}\text{C}}$

  $m_{{}^{\text{4}}\text{He}}$ is the mass of ${}^{\text{12}}\text{C}$ .

Write the expression for the binding energy of ${}^{\text{12}}\text{C}$ from the atomic masses of ${}^{\text{12}}\text{C}$ , proton and neutron.

  $E_{\text{b}}\left({}^{\text{12}}\text{C}\right)=\left(\Delta m\right)c^2$

Here, $E_{\text{b}}\left({}^{\text{12}}\text{C}\right)$ is the binding energy of ${}^{\text{12}}\text{C}$ and $c$ is the speed of light.

Substitute $6\left(m_{\text{p}}+m_{\text{a}}\right)-m_{{}^{\text{12}}\text{C}}$ for $\Delta m$ in the above expression.

  $E_{\text{b}}\left({}^{\text{12}}\text{C}\right)=\left(6\left(m_{\text{p}}+m_{\text{a}}\right)-m_{{}^{\text{12}}\text{C}}\right)c^2$ …… (5)

Calculation:

Substitute $4.002603\text{u}$ for $m_{\alpha }$ and $15.994915\text{u}$ in equation (1).

  $\begin{array}{c}{E}_{\text{b}}=\left(4\left(4.002603\text{u}\right)-\left(15.994915\text{u}\right)\right)c^2\\ =\left(16.010412\text{u}-15.994915\text{u}\right)c^2\\ =\left(0.015497\text{u}\right)c^2\end{array}$

Substitute $\left(0.015497\text{u}\right)c^2$ for $E_{\text{b}}$ in equation (2).

  $\begin{array}{c}{E}_{\text{b}\text{t}}=\frac{\left(0.015497\text{u}\right)c^2}{\text{6}}\\ =\frac{\left(0.015497\text{u}\right)c^2}{\text{6}}\left(\frac{931.5\text{MeV}/c^2}{\text{u}}\right)\\ =2.406\text{MeV}\end{array}$

Substitute $1.007825\text{u}$ for $m_{\text{p}}$ , $1.008665\text{u}$ for $m_{\text{a}}$ and $4.002603\text{u}$ for $m_{{}^{\text{4}}\text{He}}$ in equation (3).

  $\begin{array}{c}{E}_{\text{b}}\left({}^{\text{4}}\text{He}\right)=\left(2\left(1.007825\text{u}+1.008665\text{u}\right)-4.002603\text{u}\right)c^2\\ =\left(2\left(1.007825\text{u}+1.008665\text{u}\right)-4.002603\text{u}\right)c^2\left(\frac{931.5\text{MeV}/c^2}{\text{u}}\right)\\ =28.30\text{MeV}\end{array}$

Substitute $28.30\text{MeV}$ for $E_{\text{b}}\left({}^{\text{4}}\text{He}\right)$ and $2.406\text{MeV}$ for $E_{\text{b}\text{t}}$ in equation (4).

  $\begin{array}{c}{E}_{\text{b}}\left({}^{\text{12}}\text{C}\right)=3\left(28.30\text{MeV}\right)+3\left(2.406\text{MeV}\right)\\ =92.1\text{MeV}\end{array}$

Substitute $1.007825\text{u}$ for $m_{\text{p}}$ , $1.008665\text{u}$ for $m_{\text{a}}$ and $12.000000\text{u}$ for $m_{{}^{\text{12}}\text{C}}$ in equation (5).

  $\begin{array}{c}{E}_{\text{b}}\left({}^{\text{12}}\text{C}\right)=\left(6\left(1.007825\text{u}+1.008665\text{u}\right)-12.000000\text{u}\right)c^2\\ =\left(6\left(1.007825\text{u}+1.008665\text{u}\right)-12.000000\text{u}\right)c^2\left(\frac{931.5\text{MeV}/c^2}{\text{u}}\right)\\ =92.2\text{MeV}\end{array}$

Conclusion:

Thus, the binding energy of ${}^{\text{12}}\text{C}$ from the given model is $92.1\text{MeV}$ and from the atomic mass of ${}^{\text{12}}\text{C}$ is $92.2\text{MeV}$ . Hence the results are good agreement with the given model.