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

`Fermi\ energy=7.49* 10^(-12)J`

Step-by-step explanation:

We know that Fermi energy

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

`Mass\ of\ neutron(m)=1.67 x 10^(-27)kg`

`Volume(V)=(4)/(3)\pi r^3`

r=Radius of neutron star=18 km

`Volume(V)=(4)/(3)\pi * (18* 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* 1.98* 10^(30)kg`

`M=4.158* 10^(30)kg`

`N=(M)/(m)`

`N=(4.158* 10^(30))/(1.67 x 10^(-27))`

`N=2.8* 10^(57)`

Now by putting all values in

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

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

`So\ Fermi\ energy=7.49* 10^(-12) J`