The Distribution of Molecular Speeds — Question 4

Question 4

A narrow speed window selects a probability rather than a density

A Maxwell gas has most probable speed $v_p=400$ m/s and density $f(v)=4v^2e^{-v^2/v_p^2}/(\sqrt\pi v_p^3)$. A detector samples $10^6$ independent molecules without speed bias. Its accepted window is 399–401 m/s.

1. Estimate the accepted fraction and expected count using the midpoint approximation.

2. For a window of total width $w$ centered at $v_p$, derive the leading relative correction to $wf(v_p)$ from $f^{\prime\prime}(v_p)/f(v_p)=-4/v_p^2$. Evaluate it at $w=2$ m/s.

Question 4 - Solution

A narrow speed window selects a probability rather than a density

(a) Method and calculation

$$f(v_p)=\frac4{e\sqrt\pi v_p}\approx2.07554\times10^{-3}\,\mathrm{s/m}.$$$$P\simeq(2\,\mathrm{m/s})f(v_p)\approx.00415107,\quad\langle N_{\mathrm{accepted}}\rangle\approx4151.07.$$

(b) Interpretation and further calculation

$$\int_{-w/2}^{w/2}\left[f(v_p)+\tfrac12f^{\prime\prime}(v_p)x^2+\cdots\right]dx=wf(v_p)+\frac{w^3}{24}f^{\prime\prime}(v_p)+\cdots.$$$$\frac{P-wf(v_p)}{wf(v_p)}\simeq-\frac{w^2}{6v_p^2}=-4.16667\times10^{-6}.$$The midpoint estimate is slightly high because the density is concave downward near its maximum.

Physical check

An expected count can be noninteger; the observed count is an integer with statistical variation.