Chapter 9, Problem 17P

(a)

To determine

The lowest energy of the system in which electrons occupy respective state.

Answer to Problem 17P

The lowest energy of the system in which electrons occupy respective state is $394.8\text{eV}$.

Explanation of Solution

The particle placed in a cubical box is a well-known model in the field of quantum mechanics which illustrates the motion of a particle confined to move inside a 3 dimensional box. The walls of this cubical box are impenetrable as the potential of these walls is infinite.

The particle in a box is a completely hypothetical model which illustrates the basic difference between the classical and quantum models. According to classical mechanics, the particle confined in a box can move with any velocity whereas in quantum mechanics the particles can occupy only particular states due to the presence of quantum effects.

Write the expression for the energy of the particle in cubical box.

  $E=\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(n_1^2+n_2^2+n_3^2\right)$        (I)

Here, $E$ is the energy of the system, $L$ is the side of the cubical box, $m$ is the mass of the confined particle and $n$ is the state occupied by the particle.

According to Pauli Exclusion Principle, no more than two fermions can occupy same state. The electros are also fermions and are identical due to which any two electrons can occupy any state.

The electrons will, first, occupy ground state and then they will occupy further states with two electrons filled in each state.

The minimum energy of the system is the sum of energies of electrons present in ground state and other states. The electrons can also occupy the degenerate energy state due to which there can be three possible combinations, of respective $n_1$, $n_2$ and $n_3$, $\left(1,1,2\right),\left(1,2,1\right)$ and $\left(2,1,1\right)$.

The energy of the electrons presents in the states $\left(1,1,2\right),\left(1,2,1\right)$ and $\left(2,1,1\right)$ is equal.

Write the expression for the minimum energy of the system of 8 electrons.

  $E_{\min }=2\left(E_{111}+E_{112}+E_{121}+E_{211}\right)$

Simplify the above expression.

  $E_{\min }=2\left(E_{111}+3E_{112}\right)$        (II)

Here, $E_{\min }$ is the minimum energy of the system, $E_{111}$ is the energy of electrons present in ground state and $E_{112}$ is the energy of the electrons present in $\left(1,1,2\right),\left(1,2,1\right)$ and $\left(2,1,1\right)$.

Conclusion:

Substitute $E_{111}$ for $E$, $1.054\times {10}^{-34}\text{Js}$ for $\hslash$, $9.11\times {10}^{-31}\text{kg}$ for $m$, $0.2\text{nm}$ for $L$,1 for $n_1$, 1 for $n_2$ and 1 for $n_3$ in equation (I).

  $\begin{array}{c}{E}_{111}=\frac{{\left(1.054\times {10}^{-34}\text{Js}\right)}^2{\pi }^2}{2\left(9.11\times {10}^{-31}\text{kg}\right){\left(0.2\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^2}\left(1^2+1^2+1^2\right)\\ =4.5\times {10}^{-18}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =28.2\text{eV}\end{array}$

Substitute $E_{112}$ for $E$, $1.054\times {10}^{-34}\text{Js}$ for $\hslash$, $9.11\times {10}^{-31}\text{kg}$ for $m$, $0.2\text{nm}$ for $L$,1 for $n_1$, 1 for $n_2$ and 2 for $n_3$ in equation (I).

  $\begin{array}{c}{E}_{112}=\frac{{\left(1.054\times {10}^{-34}\text{Js}\right)}^2{\pi }^2}{2\left(9.11\times {10}^{-31}\text{kg}\right){\left(0.2\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^2}\left(1^2+1^2+2^2\right)\\ =9\times {10}^{-18}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =56.4\text{eV}\end{array}$

Substitute $56.4\text{eV}$ for $E_{112}$ and $28.2\text{eV}$ for $E_{111}$ in equation (II).

  $\begin{array}{c}{E}_{\min }=2\left(28.2\text{eV}+3\left(56.4\text{eV}\right)\right)\\ =394.8\text{eV}\end{array}$

Thus, the lowest energy of the system in which electrons occupy respective state is $394.8\text{eV}$.

(b)

To determine

The lowest energy of the system of particles which have same mass as electrons but do not obey exclusion principle.

Answer to Problem 17P

The lowest energy of the system of particles which have same mass as electrons but do not obey exclusion principle is $225.6\text{eV}$.

Explanation of Solution

Since, the particles do not obey exclusion principle. Therefore all the particles can occupy same state that is ground state.

Write the expression for the minimum energy.

  $E_{\min }=8E_{111}$        (III)

Here, $E_{\min }$ is the minimum energy of the system of particles present in the same state.

Conclusion:

Substitute $E_{111}$ for $E$, $1.054\times {10}^{-34}\text{Js}$ for $\hslash$, $9.11\times {10}^{-31}\text{kg}$ for $m$, $0.2\text{nm}$ for $L$,1 for $n_1$, 1 for $n_2$ and 1 for $n_3$ in equation (I).

  $\begin{array}{c}{E}_{111}=\frac{{\left(1.054\times {10}^{-34}\text{Js}\right)}^2{\pi }^2}{2\left(9.11\times {10}^{-31}\text{kg}\right){\left(0.2\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^2}\left(1^2+1^2+1^2\right)\\ =4.5\times {10}^{-18}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =28.2\text{eV}\end{array}$

Substitute $28.2\text{eV}$ for $E_{111}$ in equation (III).

  $\begin{array}{c}{E}_{\min }=8\left(28.2\text{eV}\right)\\ =225.6\text{eV}\end{array}$

Thus, the lowest energy of the system of particles which have same mass as electrons but do not obey exclusion principle is $225.6\text{eV}$.