Answer:
The number of available energy states per unit volume is [tex]4.01\times10^{48}[/tex]
Explanation:
Given that,
Average energy [tex]E=2\times10^{-4}\ eV[/tex]
Photon = [tex]4\times10^{-5}\ eV[/tex]
We need to calculate the number of available energy states per unit volume
Using formula of energy
[tex]g(\epsilon)d\epsilon=\dfrac{8\pi E^2dE}{(hc)^3}[/tex]
Where, E = energy
h = Planck constant
c = speed of light
Put the value into the formula
[tex]g(\epsilon)d\epsilon=\dfrac{8\times\pi\times2\times10^{-4}\times4\times10^{-5}\times1.6\times10^{-19}}{(6.67\times10^{-34}\times3\times10^{8})^3}[/tex]
[tex]g(\epsilon)d\epsilon=4.01\times10^{48}[/tex]
Hence, The number of available energy states per unit volume is [tex]4.01\times10^{48}[/tex]