91影视

The electrons present in different atoms of a star are placed in various shells. Thus, the distance between the two electrons is smaller relatively than the distance between the two protons or neutrons.

The interaction of proton and neutron with the neighbouring atom is little than the interaction of electron with the neighbouring atom. This is because the protons or neutrons of different atoms are so far apart from each other than the electrons of same atoms. Thus, electron interaction is considered first and becomes only first significant hindrance to the compression of the star.

Therefore, the electron becomes the first significant hindrance to compression of a star before protons and neutrons do.

The expression for the minimum energy of the neutron is given by:

${\mathbf{U}}_{\mathbf{\text{neutrons}}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{10}}{\left(\frac{3{蟺}^2{h}^3}{m_n^{\frac{3}{2}}V}\right)}^{\frac{\mathbf{2}}{\mathbf{3}}}{\mathbf{N}}^{\frac{\mathbf{5}}{\mathbf{3}}}$

Here, $m_n$ is the mass of neutron and $h$ is the plank's constant.

The expression for the total energy of the neutron star is given by,

$U_{\text{Total}}=U_{\text{neutron}}+U_{\text{g}}$

The expression for the gravitational energy of a sphere of mass$M$ and radius$R$ is given by,

$U_g=\frac{3G{M}^2}{5R}$

The minimum energy that is allowed by the exclusion principle is

$\begin{array}{c}{U}_{\text{neutrons}}=\frac{3}{10}{\left(\frac{3{蟺}^2{\mathrm{鈩�}}^3}{m_n^{\frac{3}{2}}V}\right)}^{\frac{2}{3}}{N}^{\frac{5}{3}}\\ =\frac{3{\mathrm{鈩�}}^2}{10m_n}{\left(\frac{3{蟺}^2}{V}\right)}^{\frac{2}{3}}{N}^{\frac{5}{3}}\\ =\frac{3{\mathrm{鈩�}}^2}{10m_n}{\left(\frac{3{蟺}^2}{\frac{4}{3}蟺R^3}\right)}^{\frac{2}{3}}{N}^{\frac{5}{3}}\\ =\frac{3{\mathrm{鈩�}}^2}{10m_n}{\left(\frac{9蟺}{4R^3}\right)}^{\frac{2}{3}}{N}^{\frac{5}{3}}\end{array}$

On further solving,

${\mathrm{U}}_{\text{electron}}=\frac{3{\mathrm{鈩�}}^2}{10{\mathrm{m}}_{\mathrm{n}}}{\left(\frac{81{\mathrm{蟺}}^2}{16}\right)}^{\frac{1}{3}}\frac{{\mathrm{N}}^{\frac{5}{3}}}{{\mathrm{R}}^2}$

${\mathrm{U}}_{\text{electron}}=\frac{3{\mathrm{鈩�}}^2}{10{\mathrm{m}}_{\mathrm{n}}{\mathrm{R}}^2}{\left(\frac{81{\mathrm{蟺}}^2{\mathrm{N}}^5}{16}\right)}^{\frac{1}{3}}$

Since the mass of neutron star $\left(\mathrm{M}\right)$ is because of number of neutrons $\left(\mathrm{N}\right)$ multiplied by the mass of single neutron $\left({\mathrm{m}}_{\mathrm{n}}\right)$ as:

$\begin{array}{l}\mathrm{M}={\mathrm{Nm}}_{\mathrm{n}}\\ \mathrm{N}=\frac{\mathrm{M}}{{\mathrm{m}}_{\mathrm{n}}}\end{array}$

Put the value of into equation (1),

$\begin{array}{c}{\mathrm{U}}_{\text{electron}}=\frac{3{\mathrm{鈩�}}^2}{10{\mathrm{m}}_{\mathrm{n}}{\mathrm{R}}^2}{\left(\frac{81{\mathrm{蟺}}^2{\left(\frac{\mathrm{M}}{{\mathrm{m}}_{\mathrm{n}}}\right)}^5}{16}\right)}^{\frac{1}{3}}\\ =\frac{3{\mathrm{鈩�}}^2}{10{\mathrm{m}}_{\mathrm{n}}{\mathrm{R}}^2}{\left(\frac{81{\mathrm{蟺}}^2\left(\frac{{\mathrm{M}}^5}{{\left({\mathrm{m}}_{\mathrm{n}}\right)}^5}\right)}{16}\right)}^{\frac{1}{3}}\\ =\frac{3{\mathrm{鈩�}}^2}{10{\mathrm{m}}_{\mathrm{n}}{\mathrm{R}}^2}{\left(\frac{81{\mathrm{蟺}}^2{\mathrm{M}}^5}{16{\left({\mathrm{m}}_{\mathrm{n}}\right)}^5}\right)}^{\frac{1}{3}}\\ =\frac{3{\mathrm{鈩�}}^2}{10{\mathrm{m}}_{\mathrm{n}}{\mathrm{R}}^2}\left(\frac{3}{2}\right){\left(\frac{3{\mathrm{蟺}}^2{\mathrm{M}}^5}{2{\left({\mathrm{m}}_{\mathrm{n}}\right)}^5}\right)}^{\frac{1}{3}}\end{array}$

On further solving,

${\mathrm{U}}_{\text{electron}}=\frac{9{\mathrm{鈩�}}^2}{20{\mathrm{R}}^2}{\left(\frac{3{\mathrm{蟺}}^2{\mathrm{M}}^5}{2{\left({\mathrm{m}}_{\mathrm{n}}\right)}^8}\right)}^{\frac{1}{3}}$

The total energy of the star is calculated as:

$\begin{array}{c}{\mathrm{U}}_{\text{Total}}={\mathrm{U}}_{\text{neutron}}+{\mathrm{U}}_{\text{g}}\\ =\frac{9{\mathrm{鈩�}}^2}{20{\mathrm{R}}^2}{\left(\frac{3{\mathrm{蟺}}^2{\mathrm{M}}^5}{2{\left({\mathrm{m}}_{\mathrm{n}}\right)}^8}\right)}^{\frac{1}{3}}鈭�\frac{3{\mathrm{GM}}^2}{5\mathrm{R}}\end{array}$

Here, $\mathrm{鈩�}$is the reduced Plank's constant and $G$ is the universal gravitational constant.

The minimum total energy is calculated by taking its derivative with respect radius equal to zero.

$\begin{array}{l}\frac{d}{dR}\left(U_{\text{Total}}\right)=0\\ \frac{d}{dR}\left[\frac{9{\mathrm{鈩�}}^2}{20R^2}{\left(\frac{3{蟺}^2{M}^5}{2{\left(m_n\right)}^8}\right)}^{\frac{1}{3}}鈭�\frac{3G{M}^2}{5R}\right]=0\\ (鈭�2)\frac{9{\mathrm{鈩�}}^2}{20R^3}{\left(\frac{3{蟺}^2{M}^5}{2{\left(m_n\right)}^8}\right)}^{\frac{1}{3}}鈭�(鈭�1)\frac{3G{M}^2}{5R^2}=0\\ \frac{鈭�9{\mathrm{鈩�}}^2}{10m_e{R}^3}{\left(\frac{3{蟺}^2{M}^5}{2{\left(m_n\right)}^8}\right)}^{\frac{1}{3}}+\frac{3G{M}^2}{5R^2}=0\end{array}$

On further solving, you get

$\begin{array}{l}\frac{3{\mathrm{GM}}^2}{5{\mathrm{R}}^2}=\frac{9{\mathrm{鈩�}}^2}{10{\mathrm{m}}_{\mathrm{e}}{\mathrm{R}}^3}{\left(\frac{3{\mathrm{蟺}}^2{\mathrm{M}}^5}{2{\left({\mathrm{m}}_{\mathrm{n}}\right)}^8}\right)}^{\frac{1}{3}}\\ {\mathrm{GM}}^2\mathrm{R}=\frac{3{\mathrm{鈩�}}^2}{2}{\left(\frac{3{\mathrm{蟺}}^2{\mathrm{M}}^5}{2{\left({\mathrm{m}}_{\mathrm{n}}\right)}^8}\right)}^{\frac{1}{3}}\end{array}$

$\begin{array}{c}\mathrm{R}=\frac{3{\mathrm{鈩�}}^2}{2{\mathrm{GM}}^2}{\left(\frac{3{\mathrm{蟺}}^2{\mathrm{M}}^5}{2{\left({\mathrm{m}}_{\mathrm{n}}\right)}^8}\right)}^{\frac{1}{3}}\\ =\frac{3{\mathrm{鈩�}}^2}{2\mathrm{G}}{\left(\frac{3{\mathrm{蟺}}^2}{2{\left({\mathrm{m}}_{\mathrm{n}}\right)}^8\mathrm{M}}\right)}^{\frac{1}{3}}\end{array}$

Therefore, the minimum total energy occurs when radius of star is .

Considered the known data as below.

The reduced planks constant, $\mathrm{鈩�}=1.055脳{10}^{鈭�34}\text{J}鈰�\text{s}$

The universal constant,$\mathrm{G}=6.67脳{10}^{鈭�11}\text{N}鈰�{\text{m}}^2/{\text{kg}}^2\text{鈭�}$

Mass of the single neutron, ${\mathrm{m}}_{\mathrm{n}}=1.67脳{10}^{鈭�27}\text{kg}$

The mass of neutron star, $\mathrm{M}=2脳{10}^{30}\text{kg}$

The radius of the star is calculated as:

$\mathrm{R}=\frac{3{\mathrm{鈩�}}^2}{2\mathrm{G}}{\left(\frac{3{\mathrm{蟺}}^2}{2{\left({\mathrm{m}}_{\mathrm{n}}\right)}^8\mathrm{M}}\right)}^{\frac{1}{3}}$

$\begin{array}{c}R=\frac{3{\left(1.055脳{10}^{鈭�34}\text{J}鈰�\text{s}\right)}^2}{2\left(6.67脳{10}^{鈭�11}\text{N}鈰�{\text{m}}^2/{\text{kg}}^2\right)}{\left(\frac{3{\left(3.14\right)}^2}{2{\left(1.67脳{10}^{鈭�27}\text{kg}\right)}^8\left(2\left(2脳{10}^{30}\text{kg}\right)\right)}\right)}^{\frac{1}{3}}\\ =0.2503脳{10}^{鈭�57}{\left(0.06112脳{10}^{186}\right)}^{\frac{1}{3}}\\ =0.2503脳{10}^{鈭�57}\left(0.394脳{10}^{62}\right)\\ =0.09862脳{10}^{5}m\end{array}$

$\begin{array}{c}R=9862\text{m}脳\left(\frac{1\text{km}}{1000\text{m}}\right)\\ =9.862\text{km}\\ 鈮�10\text{km}\end{array}$

Hence, the radius of neutron star whose mass is twice that of our sun is only about $10\text{km}$.