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Found in: Page 1216

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A cubical box of widths ${{\mathbf{L}}}_{{\mathbf{x}}}{\mathbf{=}}{{\mathbf{L}}}_{{\mathbf{y}}}{\mathbf{=}}{{\mathbf{L}}}_{{\mathbf{z}}}{\mathbf{=}}{\mathbf{L}}$ contains an electron. What multiple of ,${{\mathbf{h}}}^{{\mathbf{2}}}{\mathbf{/}}{\mathbf{8}}{{\mathbf{mL}}}^{{\mathbf{2}}}$ where, m is the electron mass, is (a) the energy of the electron’s ground state, (b) the energy of its second excited state, and (c) the difference between the energies of its second and third excited states? How many degenerate states have the energy of (d) the first excited state and (e) the fifth excited state?

(a) The multiple of $\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}$ is 3 .

(b) The multiple of $\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}$ is 9.

(c) The difference is ${\mathrm{E}}_{1,1,3}-{\mathrm{E}}_{2,2,1}=2\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$.

(d) The number of degenerate states are 3.

(e) The number of degenerate states are 6.

See the step by step solution

## Step 1: The energy of three-dimensional electron traps:

The energy equation for an electron trapped in a three-dimensional cell is given by,

${{\mathbf{E}}}_{{\mathbf{n}}_{\mathbf{x}}\mathbf{,}{\mathbf{n}}_{\mathbf{y}}\mathbf{,}{\mathbf{n}}_{\mathbf{z}}}{\mathbf{=}}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{8}\mathbf{m}}\left(\frac{{n}_{x}^{2}}{{L}_{x}^{2}}+\frac{{n}_{y}^{2}}{{L}_{y}^{2}}+\frac{{n}_{z}^{2}}{{L}_{z}^{2}}\right)$ ….. (1)

Here, ${{\mathbf{n}}}_{{\mathbf{x}}}$ is quantum number for which the electron’s matter wave fits in well width ${{\mathbf{L}}}_{{\mathbf{x}}}$,${{\mathbf{n}}}_{{\mathbf{y}}}$ is quantum number for which the electron’s matter wave fits in well width ${{\mathbf{L}}}_{{\mathbf{y}}}$, ${{\mathbf{n}}}_{{\mathbf{z}}}$ is quantum number for which the electron’s matter wave fits in well width ${{\mathbf{L}}}_{{\mathbf{z}}}$, h is plank constant, and m is mass of the electron.

## Step 2: (a)  Find the multiple of  h2/8mL2 for the energy of the electron’s ground state:

The first excited state occur at ${\mathrm{n}}_{\mathrm{x}}=1,{\mathrm{n}}_{\mathrm{y}}=1,\mathrm{and}{\mathrm{n}}_{\mathrm{z}}=1$.

Substitute L for ${\mathrm{L}}_{\mathrm{x}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{y}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{z}},1\mathrm{for}{\mathrm{n}}_{\mathrm{x}},1\mathrm{for}{\mathrm{n}}_{\mathrm{y}},$and 1 for ${n}_{z}$ in equation (1).

${\mathrm{E}}_{1,1,1}=\frac{{\mathrm{h}}^{2}}{8\mathrm{m}}\left(\frac{{1}^{2}}{{\mathrm{L}}^{2}}+\frac{{1}^{2}}{{\mathrm{L}}^{2}}+\frac{{1}^{2}}{{\mathrm{L}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\left(1+1+1\right)\phantom{\rule{0ex}{0ex}}=3\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

Therefore, the multiple of $\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}$ is 3 .

## Step 3: (b) Find the multiple of h2/8mL2 for the energy of its second excited state:

The second excited state occur at ${\mathrm{n}}_{\mathrm{x}}=2,{\mathrm{n}}_{\mathrm{y}}=2,\mathrm{and}{\mathrm{n}}_{\mathrm{z}}=1$.

Substitute L for role="math" localid="1661931817535" ${\mathrm{L}}_{\mathrm{x}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{y}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{z}},2\mathrm{for}{\mathrm{n}}_{\mathrm{x}},2\mathrm{for}{\mathrm{n}}_{\mathrm{y}}$, and 1 for ${n}_{z}$ in equation (1).

${\mathrm{E}}_{2,2,1}=\frac{{\mathrm{h}}^{2}}{8\mathrm{m}}\left(\frac{{2}^{2}}{{\mathrm{L}}^{2}}+\frac{{2}^{2}}{{\mathrm{L}}^{2}}+\frac{{1}^{2}}{{\mathrm{L}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\left(\frac{4}{{\mathrm{L}}^{2}}+\frac{4}{{\mathrm{L}}^{2}}+\frac{1}{{\mathrm{L}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\left(4+4+1\right)\phantom{\rule{0ex}{0ex}}=9\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

Therefore, the multiple of $\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}$ is 9 .

## Step 4: (c) Find the multiple of h2/8mL2 for the difference between the energies of its second and third excited states:

The energy of the third exited state occurs at $\left({n}_{x},{n}_{y}\right)=\left(1,1\right)$ and ${n}_{z}=3$.

Substitute L for ${\mathrm{L}}_{\mathrm{x}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{y}},\mathrm{L}\mathrm{for}{\mathrm{L}}_{\mathrm{z}},1\mathrm{for}{\mathrm{n}}_{\mathrm{x}},1\mathrm{for}{\mathrm{n}}_{\mathrm{y}}$ and 3 for ${\mathrm{n}}_{\mathrm{z}}$ in equation (1).

${\mathrm{E}}_{1,1,3}=\frac{{\mathrm{h}}^{2}}{8\mathrm{m}}\left(\frac{{1}^{2}}{{\mathrm{L}}^{2}}+\frac{{1}^{2}}{{\mathrm{L}}^{2}}+\frac{{3}^{2}}{{\mathrm{L}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\left(\frac{1}{{\mathrm{L}}^{2}}+\frac{1}{{\mathrm{L}}^{2}}+\frac{9}{{\mathrm{L}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\left(1+1+9\right)\phantom{\rule{0ex}{0ex}}=11\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

The difference in energy between the ground state and the second exited state is calculated as follows.

${E}_{1,1,,3}-{E}_{2,2,1}=11\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)-9\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=2\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

Therefore, the difference in energy between the ground state and the second exited state is,

${E}_{1,1,,3}-{E}_{2,2,1}=2\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

## Step 5: (d) The number of degenerate states that have the energy of the first excited state:

The first exited state occur at $\left({n}_{x},{n}_{y},{n}_{z}\right)=\left(2, 1, 1\right)$, $\left({n}_{x},{n}_{y},{n}_{z}\right)=\left(1, 2, 1\right)$, and $\left({n}_{x},{n}_{y},{n}_{z}\right)=\left(1, 1, 2\right)$ .

Therefore, the number of degenerate states are 3.

## Step 6: (e) Find the number of degenerate states that have the energy of the fifth excited state:

The first exited state occur at .

$\left({n}_{x},{n}_{y},{n}_{z}\right)=\left(1, 2, 3\right), \left(1, 3, 2\right), \left(2, 3, 1\right), \left(2, 1, 3\right), \left(3, 1, 2\right), \left(3, 2, 1\right)$

Therefore, the number of degenerate states are 6.