Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A cubical box of widths $L_{x} = L_{y} = L_{z} = L$ contains an electron. What multiple of , $h^{2} / 8 {mL}^{2}$ 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 $\frac{h^{2}}{8 {mL}^{2}}$ is 3 .

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

(d) The number of degenerate states are 3.

(e) The number of degenerate states are 6.

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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 $L_{y}$ , $n_{z}$ 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.

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

The first excited state occur at $n_{x} = 1, n_{y} = 1, and n_{z} = 1$ .

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).

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

Therefore, the multiple of $\frac{h^{2}}{8 {mL}^{2}}$ 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 $n_{x} = 2, n_{y} = 2, and n_{z} = 1$ .

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

$E_{2, 2, 1} = \frac{h^{2}}{8 m} (\frac{2^{2}}{L^{2}} + \frac{2^{2}}{L^{2}} + \frac{1^{2}}{L^{2}}) = \frac{h^{2}}{8 {mL}^{2}} (\frac{4}{L^{2}} + \frac{4}{L^{2}} + \frac{1}{L^{2}}) = \frac{h^{2}}{8 {mL}^{2}} (4 + 4 + 1) = 9 (\frac{h^{2}}{8 {mL}^{2}})$

Therefore, the multiple of $\frac{h^{2}}{8 {mL}^{2}}$ 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 $(n_{x}, n_{y}) = (1, 1)$ and $n_{z} = 3$ .

Substitute L for $L_{x}, L for L_{y}, L for L_{z}, 1 for n_{x}, 1 for n_{y}$ and 3 for $n_{z}$ in equation (1).

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

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

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

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

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

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

The first exited state occur at $(n_{x}, n_{y}, n_{z}) = (2, 1, 1)$ , $(n_{x}, n_{y}, n_{z}) = (1, 2, 1)$ , and $(n_{x}, n_{y}, n_{z}) = (1, 1, 2)$ .

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 .

$(n_{x}, n_{y}, n_{z}) = (1, 2, 3), (1, 3, 2), (2, 3, 1), (2, 1, 3), (3, 1, 2), (3, 2, 1)$

Therefore, the number of degenerate states are 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 multiple of h2/8mL2 for the energy of the electron’s ground state:

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

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

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

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

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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 multiple of h2/8mL2 for the energy of the electron’s ground state:

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

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

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

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

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