Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

An electron (mass m) is contained in a cubical box of widths $L_{x} = L_{y} = L_{z}$ . (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 $h / 8 {mL}^{2}$ 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 $\frac{h}{8 {mL}^{2}}$ is 1.

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

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

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

(f) The multiple of $\frac{h}{8 m L^{2}}$ is 6.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The energy of three dimensional electron traps:

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

$E_{n_{x}, n_{y}, n_{z}} = \frac{h^{2}}{8 m} (\frac{n_{x}^{2}}{L_{x}^{2}} + \frac{n_{y}^{2}}{L_{y}^{2}} + \frac{n_{z}^{2}}{L_{z}^{2}})$ ….. (1)

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

The frequency in terms of the energy is given by,

$f = \frac{E}{h}$ ….. (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.

$E_{n_{x}, n_{y}, n_{z}} = \frac{h^{2}}{8 m} (\frac{n_{x}^{2}}{L_{x}^{2}} + \frac{n_{y}^{2}}{L_{y}^{2}} + \frac{n_{z}^{2}}{L_{z}^{2}})$

Substitute L for $L_{x}, L for L_{y}, L for L_{z}, 1 for n_{x}, 1 for n_{y},$ and 1 for $n_{z}$ in equation (1).

$E_{1, 1, 1} = (\frac{h^{2}}{8 m}) (\frac{1}{L^{2}} + \frac{1}{L^{2}} + \frac{1}{L^{2}}) = 3 (\frac{h^{2}}{8 {mL}^{2}})$

And similarly,

$E_{2, 1, 1} = 6 (\frac{h^{2}}{8 m}) E_{2, 2, 1} = 9 (\frac{h^{2}}{8 {mL}^{2}}) E_{1, 1, 3} = 11 (\frac{h^{2}}{8 {mL}^{2}}) E_{2, 2, 2} = 12 (\frac{h^{2}}{8 {mL}^{2}})$

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 $\frac{h^{2}}{8 m L^{2}}$ .

The lowest frequency occur at $(E_{1, 1, 3} \leftrightarrow E_{2, 2, 2})$ .

Find the difference between $E_{1, 1, 3}$ and $E_{2, 2, 2}$ .

$E_{1, 1, 3} - E_{2, 2, 2} = 12 (\frac{h^{2}}{8 {mL}^{2}}) - 11 (\frac{h^{2}}{8 {mL}^{2}}) = \frac{h^{2}}{8 {mL}^{2}}$

Find the frequency by using equation (2).

$f_{1, 1, 3 \leftrightarrow 2, 2, 2,} = \frac{\frac{h^{2}}{8 {mL}^{2}}}{h} = \frac{h^{2}}{8 {mL}^{2}}$

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

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

The second lowest frequency occur at $(E_{1, 1, 3} \leftrightarrow E_{2, 2, 1})$ .

Find the difference between $E_{1, 1, 3}$ and $E_{2, 2, 1}$ .

$E_{1, 1, 3} - E_{2, 2, 1} = 11 (\frac{h^{2}}{8 {mL}^{2}}) - 9 (\frac{h^{2}}{8 {mL}^{2}}) = 2 (\frac{h^{2}}{8 {mL}^{2}})$

Find the frequency by using equation (2).

$E_{1, 1, 3 \leftrightarrow 2, 2, 1} = \frac{2 (\frac{h^{2}}{8 {mL}^{2}})}{h} 1 = 2 (\frac{h^{2}}{8 {mL}^{2}})$

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

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

The third lowest frequency occur at $(E_{2, 1, 1} \leftrightarrow E_{2, 2, 1,})$ .

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

$E_{2, 2, 1} - E_{2, 1, 1} = 9 (\frac{h^{2}}{8 {mL}^{2}}) - 6 (\frac{h^{2}}{8 {mL}^{2}}) = 3 (\frac{h^{2}}{8 {mL}^{2}})$

Find the frequency by using equation (2).

$f_{2, 1, 1 \leftrightarrow 2, 2, 1} = \frac{3 (\frac{h^{2}}{8 {mL}^{2}})}{h} 1 = 3 (\frac{h^{2}}{8 {mL}^{2}})$

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

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

The third lowest frequency occur at $(E_{1, 1, 1} \leftrightarrow E_{2, 2, 2})$ .

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

$E_{2, 2, 2} - E_{1, 1, 1} = 12 (\frac{h^{2}}{8 {mL}^{2}}) - 3 (\frac{h^{2}}{8 {mL}^{2}}) = 9 (\frac{h^{2}}{8 {mL}^{2}})$

Find the frequency by using equation (2).

$f_{1, 1, 1 \leftrightarrow 2, 2, 2} = \frac{9 (\frac{h^{2}}{8 {mL}^{2}})}{h} = 9 (\frac{h^{2}}{8 {mL}^{2}})$

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

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

The third lowest frequency occur at $(E_{1, 1, 1} \leftrightarrow E_{1, 1, 3})$ .

Find the difference between $E_{1, 1, 1}$ and $E_{1, 1, 3}$ .

role="math" localid="1661940120369" $E_{1, 1, 3} - E_{1, 1, 1} = 11 (\frac{h^{2}}{8 {mL}^{2}}) - 3 (\frac{h^{2}}{8 {mL}^{2}}) = 8 (\frac{h^{2}}{8 {mL}^{2}})$

Find the frequency by using equation (2).

$f_{1, 1, 1 \leftrightarrow 1, 1, 3} = \frac{8 (\frac{h^{2}}{8 {mL}^{2}})}{h} = 8 (\frac{h^{2}}{8 {mL}^{2}})$

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

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

The third lowest frequency occur at $(E_{1, 1, 1} \leftrightarrow E_{2, 2, 1})$ .

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

$E_{2, 2, 1} - E_{1, 1, 1} = 9 (\frac{h^{2}}{8 {mL}^{2}}) - 3 (\frac{h^{2}}{8 {mL}^{2}}) = 6 (\frac{h^{2}}{8 {mL}^{2}})$

Find the frequency by using equation (2).

$f_{1, 1, 1 \leftrightarrow 2, 2, 1} = \frac{6 (\frac{h^{2}}{8 {mL}^{2}})}{h} = 6 (\frac{h^{2}}{8 {mL}^{2}})$

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

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The energy of three dimensional electron traps:

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

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

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

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

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

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

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

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The energy of three dimensional electron traps:

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

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

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

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

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

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

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

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