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### Fundamentals Of Physics

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

# An electron (mass m) is contained in a cubical box of widths ${{\mathbf{L}}}_{{\mathbf{x}}}{\mathbf{=}}{{\mathbf{L}}}_{{\mathbf{y}}}{\mathbf{=}}{{\mathbf{L}}}_{{\mathbf{z}}}$. (a) How many different frequencies of light could the electron emit or absorb if it makes a transition between a pair of the lowest five energy levels? What multiple of ${\mathbf{h}}{\mathbf{/}}{\mathbf{8}}{{\mathbf{mL}}}^{{\mathbf{2}}}$ gives the (b) lowest, (c) second lowest, (d) third lowest, (e) highest, (f) second highest, and (g) third highest frequency?

(a)There are 7 different frequencies.

(b) The multiple of $\frac{\mathrm{h}}{8{\mathrm{mL}}^{2}}$ is 1.

(c) The multiple of $\frac{h}{8m{L}^{2}}$ is 2.

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

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

(f) The multiple of $\frac{h}{8m{L}^{2}}$ is 8.

(f) The multiple of $\frac{h}{8m{L}^{2}}$ is 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,

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

The frequency in terms of the energy is given by,

${\mathbf{f}}{\mathbf{=}}\frac{\mathbf{E}}{\mathbf{h}}$ ….. (2)

## Step 2: (a) Find the number of different frequencies of light that can electron emit or absorb:

The energies of lowest five levels are as follows.

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

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).

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

And similarly,

${\mathrm{E}}_{2,1,1}=6\left(\frac{{\mathrm{h}}^{2}}{8\mathrm{m}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{2,2,1}=9\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{1,1,3}=11\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{2,2,2}=12\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

From these energies, it can be seen that the non-iterated differences in the energy occurs between:

• The fourth excited state and the ground state
• The third excited state and the ground state
• The third excited state and the first excited state
• The fourth excited state and the third excited state
• The third excited state and the second excited state
• The second excited state and the ground state
• The second excited state and the first excited state

Therefore, there are 7 different frequencies.

## Step 3: (b) Find the multiple of h/8mL2 for the lowest frequency:

The transition between the energies above is simply the differences between the factors that are multiplied by $\frac{{h}^{2}}{8m{L}^{2}}$.

The lowest frequency occur at $\left({E}_{1,1,3}↔{E}_{2,2,2}\right)$.

Find the difference between ${\mathrm{E}}_{1,1,3}$ and ${E}_{2,2,2}$.

${\mathrm{E}}_{1,1,3}-{\mathrm{E}}_{2,2,2}=12\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)-11\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}$

Find the frequency by using equation (2).

${f}_{1,1,3↔2,2,2,}=\frac{\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}}{h}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}$

Therefore, the multiple of $\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}$ is 1.

## Step 4: (c) Find the multiple of for the second lowest frequency:

The second lowest frequency occur at $\left({E}_{1,1,3}↔{E}_{2,2,1}\right)$.

Find the difference between ${\mathrm{E}}_{1,1,3}$ and ${\mathrm{E}}_{2,2,1}$.

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

Find the frequency by using equation (2).

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

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

## Step 5: (d) Find the multiple of h/8mL2 for the third lowest frequency:

The third lowest frequency occur at $\left({E}_{2,1,1}↔{E}_{2,2,1,}\right)$.

Find the difference between ${\mathrm{E}}_{2,1,1,}$and ${\mathrm{E}}_{2,2,1}$.

${\mathrm{E}}_{2,2,1}-{\mathrm{E}}_{2,1,1}=9\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)-6\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=3\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

Find the frequency by using equation (2).

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

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

## Step 6: (e) Find the multiple of h/8mL2 for the highest frequency:

The third lowest frequency occur at $\left({E}_{1,1,1}↔{E}_{2,2,2}\right)$.

Find the difference between ${E}_{1,1,1}$ and ${E}_{2,,2,2}$.

${\mathrm{E}}_{2,2,2}-{\mathrm{E}}_{1,1,1}=12\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)-3\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=9\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

Find the frequency by using equation (2).

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

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

## Step 7: (f) Find the multiple of h/8mL2 for the second highest frequency:

The third lowest frequency occur at $\left({E}_{1,1,1}↔{E}_{1,1,3}\right)$.

Find the difference between ${\mathrm{E}}_{1,1,1}$ and ${\mathrm{E}}_{1,1,3}$.

role="math" localid="1661940120369" ${\mathrm{E}}_{1,1,3}-{\mathrm{E}}_{1,1,1}=11\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)-3\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=8\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

Find the frequency by using equation (2).

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

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

## Step 8: (g) Find the multiple of h/8mL2 for the third highest frequency:

The third lowest frequency occur at $\left({E}_{1,1,1}↔{E}_{2,2,1}\right)$.

Find the difference between ${\mathrm{E}}_{1,1,1}$ and ${\mathrm{E}}_{2,2,1}$.

${\mathrm{E}}_{2,2,1}-{\mathrm{E}}_{1,1,1}=9\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)-3\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=6\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

Find the frequency by using equation (2).

${\mathrm{f}}_{1,1,1↔2,2,1}=\frac{6\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)}{h}\phantom{\rule{0ex}{0ex}}=6\left(\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}\right)$

Therefore, the multiple of $\frac{{\mathrm{h}}^{2}}{8{\mathrm{mL}}^{2}}$ is 6.