Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2023 / Paper 2 / Section B / 21 / iv / Solution

The enlarged diamonds reach $|V_{sd}|=5.75\mathrm{mV}$ instead of $5.00\mathrm{mV}$. Subtract the unchanged charging contribution:Thus the two observed closed-orbital gaps are each approximately $0.75\mathrm{meV}$.

Define $A=h^2/(8m^{\ast }{L}_x^2)$ and $B=h^2/(8m^{\ast }{L}_y^2)$, and choose the labels so $A\le B$. With two spins per orbital, the first two closed orbitals correspond to occupancies $N=2$ and $N=4$. The lowest state is $(1,1)$ and the next is $(2,1)$, soThe next orbital is $(1,2)$, yielding the second observed gapThis ordering is self-consistent: $E_{11}=0.75$, $E_{21}=1.50$, $E_{12}=2.25$ and $E_{31}=2.75\mathrm{meV}$, as shown in the original level diagram.

Using $m^{\ast }=0.6m_e$ givesThe axes can be interchanged. Assigning the same $0.75\mathrm{meV}$ directly to both $3A$ and $3B$ would instead give a square box with degenerate $(2,1)$ and $(1,2)$ orbitals. That would require four electrons in that shell and would not give the stated second closed-orbital occupancy $N=4$.