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Q. 7.21

Expert-verifiedFound in: Page 276

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$4.6632\times {10}^{11}\mathrm{K}$.

One would need lot of energy to excite those nucleons.

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.

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,

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

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

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

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

${\epsilon}_{F}=\frac{{h}^{2}}{8m{L}^{2}}{\left(\frac{3N}{2\pi}\right)}^{\frac{2}{3}}\phantom{\rule{0ex}{0ex}}=\frac{{h}^{2}}{8m{\left({L}^{3}\right)}^{\frac{2}{3}}}{\left(\frac{3N}{2\pi}\right)}^{\frac{2}{3}}\phantom{\rule{0ex}{0ex}}=\frac{{h}^{2}}{8m}{\left(\frac{3}{2\pi}\xb7\frac{N}{V}\right)}^{\frac{2}{3}}$

The mass of nucleon is,

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

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

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

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

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

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

Substituting the values, we get

localid="1647858933038" ${\epsilon}_{F}=\frac{{\left(6.625\times {10}^{-34}\mathrm{J}\xb7\mathrm{s}\right)}^{2}}{8\left(1.6603\times {10}^{-27}\mathrm{kg}\right)}{\left(\frac{3}{2\pi}\right)}^{\frac{2}{3}}{\left(0.18\times {10}^{45}{\mathrm{m}}^{-3}\right)}^{2/3}\phantom{\rule{0ex}{0ex}}=\left(6.4355\times {10}^{-12}\mathrm{J}\right)\left(\frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right)\phantom{\rule{0ex}{0ex}}=\left(4.0223\times {10}^{7}\mathrm{eV}\right)\left(\frac{1\mathrm{MeV}}{{10}^{6}\mathrm{eV}}\right)\phantom{\rule{0ex}{0ex}}=40.2\mathrm{MeV}$

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

The Fermi temperature of the system is given by,

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

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

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

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

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

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