Chapter 40, Problem 63P

(a)

To determine

The lowest energy of a nucleon in the well.

Explanation of Solution

Given:

The mass of the nucleon is $1.0\text{u}$ .

The length of the square well is $3.0\text{fm}$ .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

  $E_n=n^2\left(\frac{h^2}{8m{L}^2}\right)$

Here, $E_n$ is the allowed energy for $n$ energy levels, $n$ are the allowed energy levels, $h$ is the Plank’s constant, $m$ is the mass of the particle and $L$ is the length of the square well.

For lowest energy level, $n$ is $1$ .

Substitute $1$ for $n$ in the above expression.

  $\begin{array}{c}{E}_1={\left(1\right)}^2\left(\frac{h^2}{8m{L}^2}\right)\\ =\frac{h^2}{8m{L}^2}\end{array}$   ........ (1)

Calculation:

Substitute $1.0\text{u}$ for $m$ , $3.0\text{fm}$ for $L$ and $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in equation (1).

  $\begin{array}{c}{E}_1=\frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(1.0\text{u}\right)\left(1.661\times {10}^{-27}\text{Kg/u}\right){\left(3.0\text{fm}\left(\frac{1\text{m}}{{10}^{15}\text{fm}}\right)\right)}^2}\\ =3.678\times {10}^{-12}\text{J}\left(\frac{1\text{MeV}}{1.602\times {10}^{-13}\text{J}}\right)\\ =23\text{MeV}\end{array}$

Conclusion:

Thus, the lowest energy of a nucleon in the well is $23\text{MeV}$ .

(b)

To determine

The ground state energy of $12$ nucleons in the well, all are neutrons.

Explanation of Solution

Given:

The mass of the nucleon is $1.0\text{u}$ .

The length of the square well is $3.0\text{fm}$ .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

  $E_n=n^2\left(\frac{h^2}{8m{L}^2}\right)$

Here, $E_n$ is the allowed energy for $n$ energy levels, $n$ are the allowed energy levels, $h$ is the Plank’s constant, $m$ is the mass of the particle and $L$ is the length of the square well.

For lowest energy level, $n$ is $1$ .

Substitute $1$ for $n$ in the above expression.

  $\begin{array}{c}{E}_1={\left(1\right)}^2\left(\frac{h^2}{8m{L}^2}\right)\\ =\frac{h^2}{8m{L}^2}\end{array}$   ........ (1)

Write the expression for $12$ neutrons which are fermions.

  $E=2\left(E_1+E_2+E_3+E_4+E_5+E_6\right)$

Write the above expression again for less than two neutrons per state.

  $\begin{array}{c}E=2\left(E_1+2^2{E}_1+3^2{E}_1+4^2{E}_1+5^2{E}_1+6^2{E}_1\right)\\ =182E_1\end{array}$   ........ (2)

Calculation:

Substitute $1.0\text{u}$ for $m$ , $3.0\text{fm}$ for $L$ and $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in equation (1).

  $\begin{array}{c}{E}_1=\frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(1.0\text{u}\right)\left(1.661\times {10}^{-27}\text{Kg/u}\right){\left(3.0\text{fm}\left(\frac{1\text{m}}{{10}^{15}\text{fm}}\right)\right)}^2}\\ =3.678\times {10}^{-12}\text{J}\left(\frac{1\text{MeV}}{1.602\times {10}^{-13}\text{J}}\right)\\ =23\text{MeV}\end{array}$

Substitute $23\text{MeV}$ for $E_1$ in equation (2).

  $\begin{array}{c}E=182\left(23\text{MeV}\right)\\ =4186\text{MeV}\left(\frac{1\text{GeV}}{{10}^3\text{MeV}}\right)\\ =4.186\text{GeV}\\ \approx 4.2\text{GeV}\end{array}$

Conclusion:

Thus, the ground state energy of $12$ nucleons in the well $4.2\text{GeV}$ .

(c)

To determine

The ground state energy of $12$ nucleons in the well, in which $6$ are neutrons and $6$ are protons.

Explanation of Solution

Given:

The mass of the nucleon is $1.0\text{u}$ .

The length of the square well is $3.0\text{fm}$ .

Formula used:

Write the expression for the allowed energy levels in one-dimensional infinite square well.

  $E_n=n^2\left(\frac{h^2}{8m{L}^2}\right)$

Here, $E_n$ is the allowed energy for $n$ energy levels, $n$ are the allowed energy levels, $h$ is the Plank’s constant, $m$ is the mass of the particle and $L$ is the length of the square well.

For lowest energy level, $n$ is $1$ .

Substitute $1$ for $n$ in the above expression.

  $\begin{array}{c}{E}_1={\left(1\right)}^2\left(\frac{h^2}{8m{L}^2}\right)\\ =\frac{h^2}{8m{L}^2}\end{array}$   ........ (1)

Write the expression for $12$ neutrons out of which $6$ are protons and $6$ are neutrons.

  $E=4\left(E_1+E_2+E_3\right)$

Write the above expression again for

  $4$ nucleons per state.

  $\begin{array}{c}E=4\left(E_1+2^2{E}_1+3^2{E}_1\right)\\ =56E_1\end{array}$   ........ (3)

Calculation:

Substitute $1.0\text{u}$ for $m$ , $3.0\text{fm}$ for $L$ and $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in equation (1).

  $\begin{array}{c}{E}_1=\frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(1.0\text{u}\right)\left(1.661\times {10}^{-27}\text{Kg/u}\right){\left(3.0\text{fm}\left(\frac{1\text{m}}{{10}^{15}\text{fm}}\right)\right)}^2}\\ =3.678\times {10}^{-12}\text{J}\left(\frac{1\text{MeV}}{1.602\times {10}^{-13}\text{J}}\right)\\ =23\text{MeV}\end{array}$

Substitute $23\text{MeV}$ for $E_1$ in equation (3).

  $\begin{array}{c}E=56\left(23\text{MeV}\right)\\ =1288\text{MeV}\left(\frac{1\text{GeV}}{{10}^3\text{MeV}}\right)\\ =1.288\text{GeV}\\ \approx 1.3\text{GeV}\end{array}$

Conclusion:

Thus, the ground state energy for $4$ protons and $4$ neutrons in the well is $1.3\text{GeV}$ .