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Expert-verified Found in: Page 1273 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In Eq. 41-6 let, ${\mathbf{E}}{\mathbf{-}}{{\mathbf{E}}}_{{\mathbf{F}}}{\mathbf{=}}{\mathbf{∆}}{\mathbf{E}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{eV}}$ . (a) At what temperature does the result of using this equation differ by 1% from the result of using the classical Boltzmann equation ${\mathbf{P}}{\mathbf{\left(}}{\mathbf{E}}{\mathbf{\right)}}{\mathbf{=}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{∆}\mathbf{E}\mathbf{/}\mathbf{kT}}$ (which is Eq. 41-1 with two changes in notation)? (b) At what temperature do the results from these two equations differ by 10%?

1. The temperature at which the result of the given equation differs 1% by from the result is 2520 K .
2. The temperature at which the results from these two equations differ by 10% is $5.28×{10}^{3}\mathrm{K}$.
See the step by step solution

## Step 1: The given data

The given equation, $\mathrm{E}-{\mathrm{E}}_{\mathrm{F}}=∆\mathrm{E}=1.00\mathrm{eV}$

## Step 2: Occupancy Probability

The probability that an electron will occupy a certain level, according to Fermi-Dirac statistics and Boltzmann's statistics, is known as occupancy probability. Fo the Fermi level the occupancy probability is 0.5.

The occupancy probability due to Fermi-Dirac statistics,

${{\mathbf{P}}}_{{\mathbf{FD}}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{∆}\mathbf{E}\mathbf{/}\mathbf{kT}}\mathbf{+}\mathbf{1}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{\right)}}$

The Boltzmann occupation probability,

${{\mathbf{P}}}_{{\mathbf{B}}}{\mathbf{=}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{∆}\mathbf{E}\mathbf{/}\mathbf{kT}}\phantom{\rule{0ex}{0ex}}{\mathbf{where}}{\mathbf{}}{\mathbf{k}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{38}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{23}}{\mathbf{}}{\mathbf{J}}{\mathbf{/}}{\mathbf{K}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(}}{\mathbf{2}}{\mathbf{\right)}}$

## Step 3: a) Calculation of the temperature of result differing by 1%

Let the fractional difference between the probability differences be f.

Thus, using equations (1) and (2), the fractional difference can be given as follows:

$f=\frac{{P}_{B}-{P}_{FD}}{{P}_{B}}\phantom{\rule{0ex}{0ex}}=\frac{1/{e}^{∆E/kT}-1/\left({e}^{∆E/kT}+1\right)}{{e}^{∆E/kT}}\phantom{\rule{0ex}{0ex}}=\frac{\left({e}^{∆E/kT}+1\right)-{e}^{∆E/kT}}{\left({e}^{∆E/kT}+1\right){e}^{∆E/kT}.{e}^{∆E/kT}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\left({e}^{∆E/kT}+1\right)}$

The above equation can also be written as-

${e}^{∆E/kT}=\frac{1}{f}-1$

For f = 0.01,

${e}^{∆E/kT}=\frac{1}{0.01}-1\phantom{\rule{0ex}{0ex}}=99$

Taking logarithm on both sides

$\frac{∆\mathrm{E}}{\mathrm{kT}}=\mathrm{In}\left(99\right)\phantom{\rule{0ex}{0ex}}\mathrm{T}=\frac{∆\mathrm{E}}{\mathrm{k}\mathrm{In}\left(99\right)}\phantom{\rule{0ex}{0ex}}\mathrm{T}=\frac{1.0\mathrm{eV}×\left(1.6×{10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)×4.6}\phantom{\rule{0ex}{0ex}}\mathrm{T}=2520\mathrm{K}$

Hence, the value of the temperature is 2520 K.

## Step 4: b) Calculation of the temperature of result differing by 10%

If f = 0.1

Then, the temperature is given as-

$\mathrm{T}=\frac{∆\mathrm{E}}{\mathrm{k}\mathrm{In}\left(99\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1.00\mathrm{eV}\right)\left(1.6×{10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)\mathrm{In}\left(9\right)}\phantom{\rule{0ex}{0ex}}=5.28×{10}^{3}\mathrm{K}$

Hence, the value of the temperature is $5.28×{10}^{3}\mathrm{K}$ . ### Want to see more solutions like these? 