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Q. 7.21

Expert-verified
Found in: Page 276

### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

# An atomic nucleus can be crudely modeled as a gas of nucleons with a number density of $0.18{\mathrm{fm}}^{-3}$ (where $1\mathrm{fm}={10}^{-15}\mathrm{m}$ ). Because nucleons come in two different types (protons and neutrons), each with spin $1/2$, each spatial wavefunction can hold four nucleons. Calculate the Fermi energy of this system, in MeV. Also calculate the Fermi temperature, and comment on the result.

The Fermi energy of the system is$40.2\mathrm{MeV}$.

The Fermi temperature of the system is$4.6632×{10}^{11}\mathrm{K}$.

One would need lot of energy to excite those nucleons.

See the step by step solution

## Step 1. Given Information

We are given that the number density of gas of nucleons is $0.18{\mathrm{fm}}^{-3}$.

We have to find the fermi energy of given system and the Fermi temperature.

## Step 2. Atomic nucleous

An atomic nucleus can be modeled as a gas of nucleons with number density,

$\left(\frac{N}{V}\right)=0.18{\mathrm{fm}}^{-3}$

Since the nucleons are fourfold degenerate, then the total number of occupied states will be,

$N=4×\left(\text{Volume of eigth-sphere}\right)\phantom{\rule{0ex}{0ex}}=4\frac{1}{8}\left(\frac{4}{3}\pi {n}_{\mathrm{max}}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\pi {n}_{\mathrm{max}}^{3}$

Rearranging the equation for ${n}_{\mathrm{max}}$, we get

$\frac{3N}{2\pi }={n}_{\mathrm{max}}^{3}\phantom{\rule{0ex}{0ex}}{n}_{\mathrm{max}}={\left(\frac{3N}{2\pi }\right)}^{\frac{1}{3}}$

## Step 3. Finding the fermi energy of the system

The fermi energy of the system is given by,

${\epsilon }_{F}=\frac{{h}^{2}{n}_{\mathrm{max}}^{2}}{8m{L}^{2}}$

Putting ${n}_{max}={\left(\frac{3N}{2\pi }\right)}^{\frac{1}{3}}$, we get

${\epsilon }_{F}=\frac{{h}^{2}}{8m{L}^{2}}{\left(\frac{3N}{2\pi }\right)}^{\frac{2}{3}}\phantom{\rule{0ex}{0ex}}=\frac{{h}^{2}}{8m{\left({L}^{3}\right)}^{\frac{2}{3}}}{\left(\frac{3N}{2\pi }\right)}^{\frac{2}{3}}\phantom{\rule{0ex}{0ex}}=\frac{{h}^{2}}{8m}{\left(\frac{3}{2\pi }·\frac{N}{V}\right)}^{\frac{2}{3}}$

## Step 4. Finding the fermi energy of the system

The mass of nucleon is,

$m=\frac{1\mathrm{g}}{{N}_{A}}$

Here, ${N}_{A}$ is Avogadro's number.

Substituting $6.023×{10}^{23}\text{atoms}/\mathrm{mol}$, we get

$m=\frac{\left(1\mathrm{g}\right)\left(1\mathrm{kg}/1000\mathrm{g}\right)}{6.023×{10}^{23}}\phantom{\rule{0ex}{0ex}}=1.6603×{10}^{-27}\mathrm{kg}$

The number density of nucleons in the gas is $0.18{\mathrm{fm}}^{-3}$. Converting number density from fm to ${m}^{-3}$.

$\frac{N}{V}=\left(0.18{\mathrm{fm}}^{-3}\right){\left(\frac{{10}^{-15}\mathrm{m}}{\mathrm{fm}}\right)}^{-3}\phantom{\rule{0ex}{0ex}}=0.18×{10}^{45}{\mathrm{m}}^{-3}$

Substituting the values, we get

localid="1647858933038" ${\epsilon }_{F}=\frac{{\left(6.625×{10}^{-34}\mathrm{J}·\mathrm{s}\right)}^{2}}{8\left(1.6603×{10}^{-27}\mathrm{kg}\right)}{\left(\frac{3}{2\pi }\right)}^{\frac{2}{3}}{\left(0.18×{10}^{45}{\mathrm{m}}^{-3}\right)}^{2/3}\phantom{\rule{0ex}{0ex}}=\left(6.4355×{10}^{-12}\mathrm{J}\right)\left(\frac{1\mathrm{eV}}{1.6×{10}^{-19}\mathrm{J}}\right)\phantom{\rule{0ex}{0ex}}=\left(4.0223×{10}^{7}\mathrm{eV}\right)\left(\frac{1\mathrm{MeV}}{{10}^{6}\mathrm{eV}}\right)\phantom{\rule{0ex}{0ex}}=40.2\mathrm{MeV}$

Hence, the Fermi energy of the system is$40.2\mathrm{MeV}$.

## Step 5. Finding the fermi temperature

The Fermi temperature of the system is given by,

$\left({T}_{F}\right)=\frac{{E}_{F}}{{k}_{B}}$

Putting ${\epsilon }_{F}=4.0223×{10}^{7}\mathrm{eV}$ and

${k}_{B}=8.617×{10}^{-5}\mathrm{eV}/\mathrm{K}$, we get

${T}_{F}=\frac{4.0223×{10}^{7}\mathrm{eV}}{8.617×{10}^{-5}\mathrm{eV}/\mathrm{K}}\phantom{\rule{0ex}{0ex}}=4.6632×{10}^{11}\mathrm{K}$

Hence, the Fermi temperature of the system is$4.6632×{10}^{11}\mathrm{K}$ and lot of energy is required to excite those nucleons.

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