Pressure exerted by a single electron in an spherical bubble

Question

This is from APhO 2010, Problem 3, Part (A)

When an electron is planted inside liquid helium, it can repel atoms of liquid helium and form what is called an electron bubble. The bubble contains nothing but the electron itself. We shall be interested mainly in its size and stability.

We use $\Delta f$ to denote the uncertainty of a quantity $f$. The components of an electron’s position vector $\vec{q}=(x,y,z)$ and momentum vector $\vec{p} = (p_x, p_y, p_z)$ must obey Heisenberg’s uncertainty relations $\Delta q_\alpha \Delta p_\alpha \geq \frac{\hbar}{2}$ where $\hbar$ is the Planck constant divided by $2\pi$ and $\alpha = x, y, z$.

We shall assume the electron bubble to be isotropic and its interface with liquid helium is a sharp spherical surface. The liquid is kept at a constant temperature very close to $0$ K with its surface tension $\sigma$ given by $3.75 \times 10^{-4} \text{ N m}^{-1}$ and its electrostatic responses to the electron bubble may be neglected.

Consider an electron bubble in liquid helium with an equilibrium radius $R$. The electron, of mass $m$, moves freely inside the bubble with kinetic energy $E_k$ and exerts pressure $P_e$ on the inner side of the bubble-liquid interface. The pressure exerted by liquid helium on the outer side of the interface is $P_\text{He}$.

Find a relation between $E_k$ and $P_e$

I'm having a bit of trouble with this problem, and with the official solutions.

By assuming that 'the electron moves freely' and 'its interface ... is a sharp spherical surface'; I believe that the potential energy is $0$ inside the sphere and $\infty$ outside. Hence this is the case of an infinite spherical well (which I do NOT know how to solve, and I believe is out of the scope of this question).

Of course, the wave function must vanish at the surface of the sphere. Hence the electron is not uniformly distributed throughout the sphere. As a result, the usual method to find the pressure as in the kinetic theory of gases (like the derivation given in Blundell and Blundell section 6.1) would be irrelevant.

It seems puzzling then that the result comes to be identical, ie $PV = \frac{2}{3} E_k$.

Of course, if I knew the wavefunction, or more specifically $E_k$; I could then employ classically $4\pi R^2 P = \frac{\partial E_k}{\partial R}$ (or since $E_k$ should be quantized, going through the standard calculations through $Z$, the partition function).

I have looked at the solutions, and I feel they treat the problem superficially. How would I solve this problem rigorously?

Answer 1

Model extensions include a term covering the initial shock-driven acceleration of the bubble wall, an automated method determining shock front position and pressure decay and a complete energy balance for the partitioning of absorbed energy into vaporization, bubble and shock wave energy and dissipation through viscosity and condensation. Pressure is thus directly proportional to temperature density per geometrical spacetime unit, as only indirectly hinted at by Gay-Lussac’s law. Here :

1. There is NOT a very large number N of electrons , all identical and each having mass m.

2. The electrons do not obey Newton’s laws and are in discontinuous motion, which is chaotic rather than random AND anisotropic, that is, not the same in all directions.

3. The electrons are much smaller than the average distance between them, so their total volume is much less than that of their spherical bubble container (which has volume V).

The root mean square velocity is the square root of the average of the square of the velocity. The reason we use the rms velocity instead of the average is that for a typical electron chamber , the net velocity is zero since the electrons are moving in all directions.