Chapter 28, Problem 29P

(a)

To determine

The ground state energy of the electron

Answer to Problem 29P

The ground state energy of the electron is $0.40\text{eV}$

Explanation of Solution

The energy of the emitted photon during the transition from first excited state to the ground state is $1.2\text{ eV}$.

Write the expression for energy difference between ground state and first excited state

$\Delta E=E_2-E_1$                                                               (I)

Here, $\Delta E$ is the energy difference, $E_1$ is the ground state energy. $E_2$ is the energy of the first excited state.

Write the relation between $E_n$ and $E_1$

$E_n=n^2{E}_1$                                                               (II)

Substituting (II) in (I) and substituting $2$ for $n$

$\Delta E=\left(2^2-1\right)E_1$                                                         (III)

Rearranging (III)

$E_1=\frac{\Delta E}{3}$                                                     (IV)

Conclusion:

Substituting $1.2\text{ eV}$ for $\Delta E$ in (IV)  to find $E_1$

$\begin{array}{c}{E}_1=\frac{1.2\text{ eV}}{3}\\ =0.40\text{eV}\end{array}$

Thus, the ground state energy of the electron is $0.40\text{eV}$

(b)

To determine

The possible energies of the photon when a downward transition occur starting from the second excited state.

Answer to Problem 29P

The possible energies of the emitted photons are $3.2\text{ eV, }2.0\text{ eV},1.2\text{ eV}$.

Explanation of Solution

When the electron starts from the first excited state, the possible downward transitions are $3\to 1\text{, 3}\to 2\text{and 2}\to 1$.

The transitions are given in the energy level diagram given below

Write the expression for energy of the emitted photon during transition

$E_{n^{\prime }n}=E_{n^{\prime }}-E_n$                                                     (V)

Here, $E_{n^{\prime }n}$ is the energy of the emitted photon, $E_{n^{\prime }}$ is the energy of initial state and $E_n$ energy of final state.

Substituting the corresponding values of $n$ and $n^{\prime }$ for the above given transition.

$E_{32}=E_3-E_2$                                                  (VI)

$E_{31}=E_3-E_1$                                                   (VII)

$E_{21}=E_2-E_1$                                                   (VIII)

Thus, the possible energies of the emitted photons are $E_{31},E_{32},E_{21}$.

Conclusion

Substituting $3$ and $2$ for $n$ in (II)  to find $E_3$ and $E_2$ respectively

$E_3=3^2{E}_1=9E_1$

$E_2=2^2{E}_1=4E_1$

 Substituting $9E_1$ for $E_3$, $4E_1$ for $E_2$ and $0.40\text{eV}$ for $E_1$ in (VI)

$\begin{array}{c}{E}_{32}=\left(9-4\right)0.40\text{ eV}\\ =2.0\text{ ev}\end{array}$

Substituting $9E_1$ for $E_3$, and $0.40\text{eV}$ for $E_1$ in (VII)

$\begin{array}{c}{E}_{31}=\left(9-1\right)0.40\text{ eV}\\ =3.2\text{ ev}\end{array}$

Substituting $4E_1$ for $E_2$ and $0.40\text{eV}$ for $E_1$ in (VIII)

$\begin{array}{c}{E}_{21}=\left(4-1\right)0.40\text{ eV}\\ =1.2\text{ ev}\end{array}$

Thus, the possible energies of the emitted photons are $3.2\text{ eV, }2.0\text{ eV},1.2\text{ eV}$.

(c)

To determine

The length of the one dimensional box.

Answer to Problem 29P

The length of the box is $0.97\text{nm}$.

Explanation of Solution

The length of the box is $2a_0$. Here, $a_0$ is the Bohr radius

Write the expression for energy of a particle in a 1D box.

$E_1=\frac{h^2}{8m{L}^2}$                                                            (IX)

Here, $m$ is mass of the electron, $h$ is the Planck’s constant and $L$ is the length of the box.

Rearranging (IX)

$L=\sqrt{\frac{h^2}{8m{E}_1}}$                                                            (X)

Conclusion:

Substitute, $0.40\text{eV}$ for $E_1$, $9.109\times {10}^{-31}\text{kg}$ for $m$ and $6.626\times {10}^{-34}\text{ Js}$ for $h$ in equation (X)  to find $L$

$\begin{array}{c}L=\sqrt{\frac{{\left(6.626\times {10}^{-34}\text{ Js}\right)}^2}{8\times 9.109\times {10}^{-31}\text{kg}\times \left(0.40\text{eV}\right)}}\\ =\sqrt{\frac{{\left(6.626\times {10}^{-34}\text{ Js}\right)}^2}{8\times 9.109\times {10}^{-31}\text{kg}\times \left(\mathrm{0.401.602}\times {10}^{-19}\text{ J}\right)}}\\ =0.97\text{ nm}\end{array}$

Thus, the length of the box is $0.97\text{nm}$.