Charge Is Quantized — Question 9

Question 9

An odd electron total limits exact equal sharing between small islands

Two identical isolated islands share exactly five excess electrons. Each island must have an integer electron count. Suppose the lowest-energy configurations are those with the smallest possible difference between counts, and symmetry assigns equal probabilities to equivalent configurations. No electron escapes.

1. List the allowed lowest-energy count pairs and charges. Explain why assigning charge $-2.5e$ to each island is not an individual outcome.

2. Find the mean charge and variance of either island over repeated preparations. Compute the sum of the two charges in every preparation.

Question 9 - Solution

An odd electron total limits exact equal sharing between small islands

(a) Method and calculation

$$N_A+N_B=5,\quad N_A,N_B\in\mathbb Z_{\ge0},\quad\min|N_A-N_B|=1.$$$$(N_A,N_B)=(2,3)\text{ or }(3,2),\quad(Q_A,Q_B)=(-2e,-3e)\text{ or }(-3e,-2e).$$

(b) Interpretation and further calculation

$$\langle Q_A\rangle=-\tfrac12(2e+3e)=-2.5e,\quad\operatorname{Var}(Q_A)=\tfrac12[(.5e)^2+(-.5e)^2]=.25e^2.$$$$Q_A+Q_B=-5e\quad\text{for every outcome}.$$The continuum equal-sharing approximation becomes an ensemble statement when only a few excess electrons are present.

Physical check

The island charges are correlated: whenever one has two excess electrons, the other has three.