Semiconductor Sub-band Occupancy

[Matt atkinson]

Homework Statement

a) A quantum well contains electrons at a sheet carrier density of $n_s =2 \times 10^{16}m^{-2}$. The electron effective mass is $0.1m_e^*$. Calculate the Fermi energy of the carrier distribution in the well. You may assume the spacings between sub-bands in the quantum well is very much greater than the Fermi energy.
b) If instead the spacing between the lowest two sub-bands is $25meV$, deduce the resultant occupancy in meV of each of the two sub-bands.

Homework Equations

For a) I used
$$\varepsilon_F=\frac{\hbar^2\pi}{m_e^*}n_s$$
which was derived from the 2D density of states.

The Attempt at a Solution

So I understand part a), but part b) has me confused.
I interpreted the problem as shown in the image below.
I assume the thing that I need to find is the energies I denoted as $\Delta E_{1_o}$ and $\Delta E_{2_o}$, and when I asked my lecturer about the question he said to look at it as a geometry problem but I just can't see, would I literally just need to calculate the energys $E_1$ and $E_2$ and use that and the fermi energy to determine the width?

Using;
$$25meV=\frac{\hbar^2\pi^2}{2m_e^*d^2}\left( 2^2-1^2 \right)$$

I found the QW width and then used the following relations to get the occupancy;
$$\Delta E_{1_o}=\varepsilon_F -E_1$$
$$\Delta E_{2_o}=\varepsilon_F -E_2$$

 

[Asim Riaz]

I then substituted the corresponding values for the Fermi energy, E1 and E2 to get;\Delta E_{1_o}=13.92meV\Delta E_{2_o}=11.08meVWould this be correct? I'm not 100% sure on this.