Question

Determine the Fermi energy for a neutron star of radius 18 km and mass 2.1 times that of our Sun. Assume that the star is made entirely of neutrons and is of uniform density.

Answer 1

Answer:

$Fermi\ energy=7.49\times 10^{-12}J$

Explanation:

 We know that Fermi energy

       $F_e=\dfrac{h^2}{8m}\left(\dfrac{3N}{\pi V}\right)^{\dfrac{2}{3}}$

$Mass\ of\ neutron(m)=1.67 x 10^{-27}kg$

$Volume(V)=\dfrac{4}{3}\pi r^3$

   r=Radius of neutron star=18 km

$Volume(V)=\dfrac{4}{3}\pi \times (18\times 10^3)^3$

tex]V=2.44\times 10^{13}m^3[/tex]

Mass of neutron star(M)=2.1 x mass of sun

$Mass\ of\ sun(m_s)=1.98 x 10^{30}kg$

⇒$M=2.1\times 1.98\times 10^{30}kg$

$M=4.158\times 10^{30}kg$

$N=\dfrac{M}{m}$

$N=\dfrac{4.158\times 10^{30}}{1.67 x 10^{-27}}$

$N=2.8\times 10^{57}$

Now by putting all values in

 $F_e=\dfrac{h^2}{8m}\left(\dfrac{3N}{\pi V}\right)^{\dfrac{2}{3}}$

 $F_e=\dfrac{(6.62\times 10^{-34})^2}{8\times 1.674\times 10^{-27}}\left(\dfrac{2.8\times 10^{57}\times 3}{\pi \times 2.44\times 10^{13}}\right)^{\dfrac{2}{3}}$

$So\ Fermi\ energy=7.49\times 10^{-12} J$