= Solution
As an estimate, model the confinement by an <infinite square well> of width $L$ and use the stipulated <electron> <mass>. Its <energy eigenvalues> are $E_j=j^2\pi^2\hbar^2/(2mL^2)$, $j=1,2,\ldots$, so the gap controlling <thermal occupation of a quantum well> is
$$
\Delta E=E_2-E_1=\frac{3\pi^2\hbar^2}{2mL^2}
=1.81\times10^{-21}\,\mathrm J=11.3\,\mathrm{meV},
\qquad \boxed{\frac{\Delta E}{k_B}\simeq131\,\mathrm K}.
$$
Thus \b[$T\ll130\,\mathrm K$] suppresses excitation: $p_2/p_1=e^{-\Delta E/(k_BT)}$. For example $T\lesssim28\,\mathrm K$ makes this ratio below one percent in this model. There is no sharp cutoff. The <ground state> <energy> itself corresponds to about $44\,\mathrm K$, but it is the excitation gap that enters the <Boltzmann factor>. Finite barrier height and a different <effective mass> would change the numerical scale; neither is supplied.
Back to article page
