Discussion
Three mutually non-interacting electrons are confined in a hollow spherical cavity of radius R with impenetrable walls. The average pressure exerted by the electrons on the walls of cavity when the system is in the lowest energy state is given by :
Answer: B
The keyed pressure is \(\frac{3\pi\hbar^{2}}{2mR^{5}}\).
The key treats the electrons in the s-wave levels of the spherical well, \(E_{n} = n^{2}\pi^{2}\hbar^{2}/2mR^{2}\). Two electrons (opposite spins) fill \(n = 1\) and the third goes to \(n = 2\):
\[ \begin{aligned} E &= (1 + 1 + 4)\frac{\pi^{2}\hbar^{2}}{2mR^{2}} \\ &= \frac{3\pi^{2}\hbar^{2}}{mR^{2}} \\ P &= -\frac{dE}{dV} = -\frac{dE/dR}{4\pi R^{2}} \\ &= \frac{6\pi^{2}\hbar^{2}}{mR^{3}}\cdot\frac{1}{4\pi R^{2}} \\ &= \frac{3\pi\hbar^{2}}{2mR^{5}} \end{aligned} \]
The pressure scales as \(R^{-5}\) because the energy goes as \(R^{-2}\) and the volume as \(R^{3}\), which rules out the \(R^{3}\) option; the factor 3/4 is half the correct coefficient and 7/2 does not come from these levels.
Note that in a full treatment the third electron would occupy the lowest \(l = 1\) level, whose energy (from the first zero of the spherical Bessel function \(j_{1}\), at 4.49) lies below the \(n = 2\), \(l = 0\) level; that gives a pressure not among the options, so the key rests on the s-wave picture.
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