Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A certain metal has $1.70 \times 10^{28}$ conduction electrons per cubic meter. A sample of that metal has a volume of $6.00 \times 10^{- 6} m^{3}$ and a temperature of 200K. How many occupied states are in the energy range of $3.20 x 10^{- 20} J$ that is centered on the energy $4.00 x 10^{- 19} J$ ? (Caution: Avoid round-off in the exponential.)

There are $7.98 \times 10^{19}$ occupied states in the given energy range.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

Number of conduction electrons per unit volume, $n = 1.70 \times 10^{28} m^{- 3}$
Volume of metal, $V = 6 \times 10^{- 6} m^{3}$
Temperature, $T = 200 K$
Given energy range, $∆ E = 3.20 \times 10^{- 20} J$
Energy in the center, $E = 4 x 10^{- 19} J$

Step 2: Understanding the concept of occupied states

At first, using the number of conduction electrons per unit volume, we can calculate the Fermi energy of the material. Then, we calculate the occupancy probability. Then, using Fermi-Dirac statistics we can calculate the value of the density of occupied states. Now, the value of the density of occupied states is calculated using the density of states at the energy level and the occupancy probability value is used in the formula of the number of occupied density states to get the required value.

Formulae:

The equation of Fermi energy $E_{F} = \frac{0.121 h^{2}}{m_{e}} n^{2 / 3}$ (i)

The occupancy probability is , $P (E) = \frac{1}{e^{(E - E_{F}) / k T} + 1}$ (ii)

The density of states associated with the conduction electrons of a material,

$N (E) = \frac{8 \sqrt{2} {πm}^{3 / 2}}{h^{3}} E^{1 / 2}$ (iii)

The density of occupied states, $N_{0} (E) = N (E) P (E)$ (iv)

The number of occupied states in the given energy range $∆ E$ ,

$N' = N_{0} (E) V ∆ E$ (v)

Step 3: Calculation of the number of occupied states

Using the value of conduction electrons per unit volume in equation (i), we can calculate the Fermi energy of the material as follows:

$E_{F} = \frac{0.121 {(6.26 \times 10^{- 34} J . s)}^{2}}{(9.11 \times 10^{- 31} k g)} {(1.70 \times 10^{28} m^{- 3})}^{2 / 3} = = 3.86 \times 10^{- 19} J$

Now,

$E - E_{F} = 4.00 x 10^{- 19} J - 3.86 \times 10^{- 19} J = 1.4 \times 10^{- 20} J$

and

role="math" localid="1661573166752" $\frac{E - E_{F}}{k T} = \frac{1.4 \times 10^{- 20} J}{(1.38 \times 10^{- 23} J / K) (200 K)} = 5.07$

Thus, using the above value in equation (ii), we can get the occupancy probability of the material as follows:

$P (E) = \frac{1}{e^{5.07} + 1} = 6.20 \times 10^{- 3}$

Now, the density of the states associated with the conduction electrons for the given energy level can be calculated using equation (iii) as follows:

$N (E) = \frac{8 \sqrt{2} π {(9.11 \times 10^{- 31} kg)}^{3 / 2}}{{(6.626 \times 10^{- 34} J . s)}^{3}} {(4.00 \times 10^{- 19} J)}^{1 / 2} = 6.717 \times 10^{46} / m^{3} J$

The density of the occupied states is given using equation (iv) as follows:

$N_{0} (E) = (6.717 \times 10^{46} / m^{3} . J) (6.20 \times 10^{- 3}) = 4.16 \times 10^{44} / m^{3} J$

With the given energy range of $∆ E = 3.20 \times 10^{- 20}$ and the given volume of material, we can get the number of occupied states using equation (v) as follows:

$N' = (4.16 \times 10^{44} / m^{3} J) (6 \times 10^{- 6} / m^{- 3}) (3.20 \times 10^{- 20} J) = 7.98 \times 10^{19}$

Hence, the value of the occupied states is $7.98 \times 10^{19}$ .

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