Chapter 28, Problem 64P

(a)

To determine

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

Answer to Problem 64P

The energy difference is $\text{6 MeV}$.

Explanation of Solution

The length of the box is ${10}^{-14}\text{m}$.

Write the expression for energy of the quantized level.

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

Here, $E_1$ is the ground state energy, $h$ is the Planck’s constant, $m$ is the mass and $L$ is the length of the box.

Write the expression for energy of the quantized level in terms of the ground state energy.

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

Here, $E_n$ is the energy of the level $n$, $n$ is the quantum number and $E_1$ is the energy of the ground state.

Substituting $2$ for $n$ in (II)

$E_2=2^2{E}_1$                                                           (III)

Here, $E_2$ is the energy of the first excited state.

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

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

Here, $\Delta E$ is the energy difference.

Substituting (III) in (IV)

$\Delta E=2^2{E}_1-E_1=3E_1$                                                          (V)

Substituting (I) in (V)

$\Delta E=\frac{3h^2}{8m{L}^2}$                                                           (VI)

Substituting $6.626\times {10}^{-34}\text{ Js}$ for $h$, $1.673\times {10}^{-27}\text{kg}$ for $m$ and $0.20\text{ eV}$ for  $L$ in (VI) to find $\Delta E$

$\begin{array}{c}\Delta E=\frac{3\times {\left(6.626\times {10}^{-34}\text{ Js}\right)}^2}{8\times 1.673\times {10}^{-27}\text{kg}\times {\left({10}^{-14}\text{m}\right)}^2}\times \frac{1\text{ eV}}{1.602\times {10}^{-19}\text{}\text{J}}\\ =6\text{ MeV}\end{array}$

Thus, the energy difference is $\text{6 MeV}$.

(b)

To determine

The wavelength and frequency of the emitted photon.

Answer to Problem 64P

The wavelength of the emitted photon is $0.2\text{ pm}$, its frequency is $1.5\times {10}^{21}\text{ Hz}$ and it belongs to the gamma ray of the electromagnetic spectrum.

Explanation of Solution

Write the expression for the energy of a photon.

$E=\frac{hc}{\lambda }$                                                           (VII)

Here, $\lambda$ is the wavelength, $h$ is the Planck’s constant, $c$ is the velocity of light and $E$ is the energy of the photon.

Rearranging (I)

$\lambda =\frac{hc}{E}$                                                           (VIII)

Write the expression for frequency.

$\nu =\frac{c}{\lambda }$                                                          (IX)

 Here, $\nu$ is the frequency of the photon.

Substitute $6.626\times {10}^{-34}\text{ Js}$ for $h$, $3\times {10}^8{\text{ ms}}^{-1}$ for $c$ and  $6\text{ MeV}$ for $E$ in (VIII) to find $\lambda$

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}\text{ Js}\times 3\times {10}^8{\text{ ms}}^{-1}}{6\text{ MeV}}\\ =\frac{6.626\times {10}^{-34}\text{ Js}\times 3\times {10}^8{\text{ ms}}^{-1}}{6\times {10}^6\text{ eV}}\times \frac{1\text{ eV}}{1.602\times {10}^{-19}\text{ J}}\\ =0.2\times {10}^{-12}\text{ m}\\ =0.2\text{ pm}\end{array}$

Thus, the wavelength of the photon is $0.2\text{ pm}$.

Substitute $3\times {10}^8{\text{ ms}}^{-1}$ for $c$ and  $0.2\text{ pm}$ for $\lambda$ in (IX) to find $\nu$.

$\begin{array}{c}\nu =\frac{3\times {10}^8{\text{ ms}}^{-1}}{0.2\times {10}^{-12}\text{ m}}\\ =1.5\times {10}^{21}\text{ Hz}\end{array}$

Thus, the wavelength of the photon is $1.5\times {10}^{21}\text{ Hz}$.

This wavelength and frequency corresponds to the gamma ray of the electromagnetic spectrum.

(c)

To determine

The plot of the wave function of the electron in the ground state and first three excited states.

Answer to Problem 64P

The wave function is a sine wave.

Explanation of Solution

The length of the box is ${10}^{-14}\text{m}$ i.e. $10\text{fm}$.

Write the expression for the wave function of a particle in a box.

${\psi }_n=\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)$                                                     (X)

Here, ${\psi }_n$ is the wave function of the quantum state $n$, $L$ is the length of the box  and $x$ is the position co-ordinate.

Substitute $1,2,3\text{and 4}$ for $n$ in (VIII) to find their corresponding wavefunctions ${\psi }_1,{\psi }_2,{\psi }_3\text{and }{\psi }_4$ respectively.

${\psi }_1=\sqrt{\frac{2}{10\text{ fm}}}\sin \left(\frac{\pi x}{10\text{ fm}}\right)$                                                    (XI)

${\psi }_2=\sqrt{\frac{2}{10\text{ fm}}}\sin \left(\frac{2\pi x}{10\text{ fm}}\right)$                                                   (XII)

${\psi }_3=\sqrt{\frac{2}{10\text{ fm}}}\sin \left(\frac{3\pi x}{10\text{ fm}}\right)$                                                   (XIII)

${\psi }_4=\sqrt{\frac{2}{10\text{ fm}}}\sin \left(\frac{4\pi x}{10\text{ fm}}\right)$                                                   (XIV)

The plot of the wave function versus position is given below: