Chapter 28, Problem 66P

(a)

To determine

The expression for the allowed energies of the electron in terms of the quantum numbers $n_x$ and $n_y$.

Answer to Problem 66P

The expression for the allowed energies of the electron in terms of the quantum numbers $n_x$ and $n_y$ is $\underset{\_}{\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2\right)}$.

Explanation of Solution

Write the expression for the allowed energy levels of the electron is given by,

    $E_n=\frac{h^2}{8m_e}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)$        (I)

Here, $E_n$ is the nth energy level, $h$ is the Planck’s constant, $m$ is the mass, $L$ is the length of the square well., $n$ varies as $1,2,\mathrm{3....}$.

Assuming $L_x=L_y=L$, equation (I) changes to,

    $E=\frac{h^2}{8m_e{L}^2}\left(n_x^2+n_y^2\right)$        (II)

Conclusion:

Therefore, the expression for the allowed energies of the electron in terms of the quantum numbers $n_x$ and $n_y$ is $\underset{\_}{\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2\right)}$.

(b)

To determine

The values of $n_x$ and $n_y$ correspond to the ground state.

Answer to Problem 66P

The values of $n_x$ and $n_y$ correspond to the ground state will be 1.

Explanation of Solution

An electron bound to an atom cannot have any value of energy , it can only occupy certain states which corresponds to certain energy levels.

According to quantum mechanics the particle can never be at rest. The least energy it have is corresponding to $n=1$. So the $x$ coordinate and the $y$ coordinate of $n$ will be 1.value $n_x$ and $n_y$ will be 1.

Conclusion:

Therefore, the values of $n_x$ and $n_y$ correspond to the ground state will be 1.

(c)

To determine

The energy of the ground state.

Answer to Problem 66P

The energy of the ground state is $\underset{\_}{\frac{h^2}{4m_e{L}^2}}$

Explanation of Solution

The allowed energy is given by,

    $E=\frac{h^2}{8m_e{L}^2}\left(n_x^2+n_y^2\right)$

Conclusion:

Substitute $1$ for $n_x$ and $1$ for $n_y$ in the above equation to find $E_{1,1}$.

    $\begin{array}{c}{E}_{1,1}=\frac{h^2}{8m_e{L}^2}\left(1^2+1^2\right)\\ =\frac{h^2}{4m_e{L}^2}\end{array}$

Therefore, the energy of the ground state is $\underset{\_}{\frac{h^2}{4m_e{L}^2}}$

(d)

To determine

The possible values of $n_x$ and $n_y$ for the second excited state.

Answer to Problem 66P

The possible values of $n_x$ and $n_y$ for the second excited state $\underset{\_}{n_x=2,n_y=1\text{or}{n}_x=1,n_y=2}$.

Explanation of Solution

The least energy it have is corresponding to $n=1$. So the $x$ coordinate and the $y$ coordinate of $n$ will be 1.value $n_x$ and $n_y$ will be 1. For the first excited state the value of $x$ or $y$ coordinate must be in the ground state. The possible values of $n$ will be $n_x=2,n_y=1\text{or}{n}_x=1,n_y=2$. This is a doubly degenerate state.

Conclusion:

Therefore, The possible values of $n_x$ and $n_y$ for the second excited state $\underset{\_}{n_x=2,n_y=1\text{or}{n}_x=1,n_y=2}$.

(e)

To determine

The possible values of $n_x$ and $n_y$ for the second excited state.

Answer to Problem 66P

The possible values of $n_x$ and $n_y$ for the second excited state is $\underset{\_}{n_x=2\text{and }{n}_y=2}$

Explanation of Solution

When the electron is the second excited state the possible values of $x$ and $y$ components of $n$ will be 2. Excited state is not the most stable state for an atom.

Conclusion:

Therefore, The possible values of $n_x$ and $n_y$ for the second excited state is $\underset{\_}{n_x=2\text{and }{n}_y=2}$

(f)

To determine

The energy of the second excited state.

Answer to Problem 66P

The energy of the second excited state is $\underset{\_}{\frac{h^2}{m_e{L}^2}}$

Explanation of Solution

The possible values of $n_x$ and $n_y$ for the second excited state is $n_x=2\text{and }{n}_y=2$.

The allowed energy is given by,

    $E=\frac{h^2}{8m_e{L}^2}\left(n_x^2+n_y^2\right)$

Conclusion:

Substitute $2$ for $n_x$ and $2$ for $n_y$ in the above equation to find $E_{1,1}$.

    $\begin{array}{c}{E}_{1,1}=\frac{h^2}{8m_e{L}^2}\left(2^2+2^2\right)\\ =\frac{h^2}{m_e{L}^2}\end{array}$

Therefore, the energy of the second excited state is $\underset{\_}{\frac{h^2}{m_e{L}^2}}$

(g)

To determine

The energy difference between the ground state and the second excited state.

Answer to Problem 66P

The energy difference between the ground state and the second excited state is $\underset{\_}{\frac{3h^2}{4m_e{L}^2}}$.

Explanation of Solution

The energy difference between the ground state and the second excited state is given by,

    $\Delta E=E_{2,2}-E_{1,1}$        (III)

Conclusion:

Substitute $\frac{h^2}{m_e{L}^2}$ for $E_{2,2}$ and $\frac{h^2}{4m_e{L}^2}$ for $E_{1,1}$ in equation (III) to find $\Delta E$.

    $\begin{array}{c}\Delta E=\frac{h^2}{m_e{L}^2}-\frac{h^2}{4m_e{L}^2}\\ =\frac{3h^2}{4m_e{L}^2}\end{array}$

Therefore, The energy difference between the ground state and the second excited state is $\underset{\_}{\frac{3h^2}{4m_e{L}^2}}$.

(h)

To determine

The wavelength of a photon which cause the transition between the ground state and the second excited state.

Answer to Problem 66P

The wavelength of a photon which cause the transition between the ground state and the second excited state is $\underset{\_}{\frac{4m_{e}c{L}^2}{3h}}$

Explanation of Solution

Write the expression for the energy in terms of wavelength.

    $\Delta E=\frac{hc}{\lambda }$        (IV)

Here, $h$ is the Planck’s constant, $\lambda$ is the wavelength.

Conclusion:

Substitute $\frac{3h^2}{4m_e{L}^2}$ for $\Delta E$ and solve equation (IV) to find $\lambda$.

    $\begin{array}{c}\frac{3h^2}{4m_e{L}^2}=\frac{hc}{\lambda }\\ \lambda =\frac{4m_{e}c{L}^2}{3h}\end{array}$

Therefore, the wavelength of a photon which cause the transition between the ground state and the second excited state is $\underset{\_}{\frac{4m_{e}c{L}^2}{3h}}$