91Ó°ÊÓ

The minimum energy that is allowed by the exclusion principle is given by,

$U_{\mathrm{electrons}}=\frac{3}{10}{\left(\frac{3{\mathrm{Ï€}}^2{\mathrm{h}}^3}{{\mathrm{m}}_{\mathrm{e}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{V}}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{N}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Formula used:

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

${\mathit{U}}_{\mathbf{g}}\mathbf{=}\frac{\mathbf{3}\mathbf{G}{\mathbf{M}}^{\mathbf{2}}}{\mathbf{5}\mathbf{R}}$

The expression for the volume of the sphere is given by.

$\mathit{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{Ï€¸é}}^{\mathbf{3}}$

The minimum energy that is allowed by the exclusion principle is

$\begin{array}{rcl}{U}_{electrons}& =& \frac{3}{10}{\left(\frac{3{\mathrm{Ï€}}^2{\mathrm{h}}^3}{m_e^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}V}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{N}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& \frac{3{\mathrm{h}}^2}{10m_e}{\left(\frac{3{\mathrm{Ï€}}^2}{V}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{N}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& \frac{3{\mathrm{h}}^2}{10m_e}{\left(\frac{3{\mathrm{Ï€}}^2}{{\displaystyle \frac{4}{3}}{\mathrm{Ï€¸é}}^3}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{N}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& \frac{3{\mathrm{h}}^2}{10m_e}{\left(\frac{9\mathrm{Ï€}}{{\displaystyle 4R^3}}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{N}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\end{array}$

On further solving,

$$\begin{gathered} U_{electron}=\frac{3h^2}{10m_e}{\left(\frac{81{\mathrm{Ï€}}^2}{16}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\frac{N^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{R^2} \\ {U}_{electron}=\frac{3h^2}{10m_e{R}^2}{\left(\frac{81{\mathrm{Ï€}}^2{\mathrm{N}}^5}{16}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \end{gathered}$$

Since star is assumed electrically neutral thus the number of protons N that makes the half of dwarf's mass M is expressed as:

$\begin{array}{rcl}\frac{M}{2}& =& N{m}_P\\ N& =& \frac{M}{2m_P}\end{array}$

Put the value of N in equation (I),

role="math" localid="1659165527416" $\begin{array}{rcl}{U}_{electron}& =& \frac{3h^2}{10m_e{R}^2}{\left(\frac{81{\mathrm{Ï€}}^2{\left({\displaystyle \frac{\mathrm{M}}{2{\mathrm{m}}_{\mathrm{p}}}}\right)}^5}{16}\right)}^{\frac{1}{3}}\\ & =& \frac{3h^2}{10m_e{R}^2}{\left(\frac{81{\mathrm{Ï€}}^2\left({\displaystyle \frac{{\mathrm{M}}^5}{32{\left({\mathrm{m}}_{\mathrm{p}}\right)}^5}}\right)}{16}\right)}^{\frac{1}{3}}\\ & =& \frac{3h^2}{10m_e{R}^2}{\left(\frac{81{\mathrm{Ï€}}^2{\mathrm{M}}^5}{512{\left({\mathrm{m}}_{\mathrm{p}}\right)}^5}\right)}^{\frac{1}{3}}\\ & =& \frac{3h^2}{10m_e{R}^2}\left(\frac{3}{8}\right){\left(\frac{3{\mathrm{Ï€}}^2{\mathrm{M}}^5}{{\left({\mathrm{m}}_{\mathrm{p}}\right)}^5}\right)}^{\frac{1}{3}}\end{array}$

On further solving,

role="math" localid="1659165629679" $U_{electron}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{9h^2}{80m_e{R}^2}\end{array}\begin{array}{rcl}& & \left(\frac{3{\mathrm{Ï€}}^2{\mathrm{M}}^5}{{\left({\mathrm{m}}_{\mathrm{p}}\right)}^5}\right)\end{array}\begin{array}{rcl}& & {}^{\frac{1}{3}}\end{array}$

Therefore, the minimum electron energy is equal to $U_{electron}=\begin{array}{rcl}& & \frac{9h^2}{80m_e{R}^2}\end{array}\begin{array}{rcl}& & \left(\frac{3{\mathrm{Ï€}}^2{\mathrm{M}}^5}{{\left({\mathrm{m}}_{\mathrm{p}}\right)}^5}\right)\end{array}\begin{array}{rcl}& & {}^{\frac{1}{3}}\end{array}$.

Formula used:

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

${\mathit{U}}_{\mathbf{T}\mathbf{o}\mathbf{r}\mathbf{a}\mathbf{l}}\mathbf{=}{\mathit{U}}_{\mathbf{e}\mathbf{l}\mathbf{e}\mathbf{c}\mathbf{t}\mathbf{r}\mathbf{o}\mathbf{n}}\mathbf{+}{\mathit{U}}_{\mathbf{g}}$

The expression for the gravitational energy of a sphere of mass Mand radius Ris given by,

${\mathit{U}}_{\mathbf{g}}\mathbf{=}\frac{\mathbf{3}\mathbf{G}{\mathbf{M}}^{\mathbf{2}}}{\mathbf{5}\mathbf{R}}$

The total energy of the star is calculated as:

$\begin{array}{rcl}{U}_{Total}& =& U_{electron}+U_g\\ & =& \frac{9h^2}{80m_{\mathcal{l}}{R}^2}{\left(\frac{3{\mathrm{Ï€}}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}+\frac{3G{M}^2}{5R}\end{array}$

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

$\begin{array}{rcl}\frac{d}{dR}\left(U_{Total}\right)& =& 0\\ \frac{d}{dR}\left[\frac{9h^2}{80m_e{R}^2}{\left(\frac{3{\mathrm{Ï€}}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}-\frac{3G{M}^2}{5R}\right]& =& 0\\ \left(-2\right)\frac{9h^2}{80m_e{R}^3}{\left(\frac{3{\mathrm{Ï€}}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}-\left(-1\right)\frac{3G{M}^2}{5R^2}& =& 0\\ \frac{-9h^2}{80m_e{R}^3}{\left(\frac{3{\mathrm{Ï€}}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}+\frac{3G{M}^2}{5R^2}& =& 0\end{array}$

On further solving,

$\begin{array}{rcl}\frac{G{M}^2}{R^2}& =& \frac{3h^2}{8m_e{R}^3}{\left(\frac{3{\mathrm{Ï€}}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}\\ G{M}^2& =& \frac{3h^2}{8m_e{R}^3}{\left(\frac{3{\mathrm{Ï€}}^2{\mathrm{M}}^5}{{\left(m_p\right)}^5}\right)}^{\frac{1}{3}}\\ & =& \frac{3h^2}{8G}{\left(\frac{3{\mathrm{Ï€}}^2}{m_e^3{m}_p^{5}M}\right)}^{\frac{1}{3}}\end{array}$

Therefore, the minimum total energy occurs when radius of star role="math" localid="1659167163703" $R\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{3h^2}{8G}\end{array}\begin{array}{rcl}& & \left(\frac{3{\mathrm{Ï€}}^2}{m_e^3{m}_p^{5}M}\right)\end{array}\begin{array}{rcl}& & {}^{\frac{1}{3}}\end{array}$.

The mass of sun is m = 2x10³⁰kg.

Formula used:

The expression for the radius of the stars given by,

$R\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{3h^2}{8G}\end{array}\begin{array}{rcl}& & \left(\frac{3{\mathrm{Ï€}}^2}{m_e^3{m}_p^{5}M}\right)\end{array}\begin{array}{rcl}& & {}^{\frac{1}{3}}\end{array}$

The radius of the star is calculated as:

$\begin{array}{rcl}R& =& \frac{3h^2}{8G}\left(\frac{3{\mathrm{π}}^2}{m_e^3{m}_p^{5}M}\right){}^{\frac{1}{3}}\\ & =& \frac{3{\left(1.055×{10}^{-34}J·s\right)}^2}{8\left(6.67×{10}^{-11}N.m^2/kg\right)}\left(\frac{3{\left(3.14\right)}^2}{{\left(9.1×{10}^{-31}Kg\right)}^3{\left(1.67×{10}^{-27}kg\right)}^5\left(2×{10}^{30}kg\right)}\right)\frac{1}{3}\\ & =& 6.26×{10}^{-59}{\left(0.00151×{10}^{198}\right)}^{\frac{1}{3}}\\ & =& 6.26×{10}^{-59}×0.11T4×{10}^{66}\\ R& =& 0.713×{10}^{7}m\\ & =& 7.13×{10}^{6}m\\ & =& 7×{10}^{6}m\\ & & \end{array}$

Therefore, the radius of a star with mass as that of our sun is7x10⁶m .