Question

Equation $20.3$is the mean free path of a particle through a gas of identical particles of equal radius. An electron can be thought of as a point particle with zero radius.

a. Find an expression for the mean free path of an electron through a gas.

b. Electrons travel $3\mathrm{km}$through the Stanford Linear Accelerator. In order for scattering losses to be negligible, the pressure inside the accelerator tube must be reduced to the point where the mean free path is at least $50\mathrm{km}$. What is the maximum possible pressure inside the accelerator tube, assuming $T={20}^{°}C$$?$Give your answer in both $\mathrm{Pa}$and $\mathrm{atm}$.

Short answer

a.Expression for free path of electron through a gas ${\lambda }_{-e}$is$\frac{1}{\pi \sqrt{2}N/V{r}^2}$.

b.Maximum pressure inside the accelerator tube in localid="1648440566181" $pa$is localid="1648440574419" $1.8\times {10}^{-6}pa$and in localid="1648440586059" $atm$is localid="1648440599622" $1.79\times {10}^{-10}atm$.

Step-by-step solution

Expression for free path of electron through a gas  (part a)

(a) The difference is only that instead of the moving particle having a radius or $r$, our electron will have zero radius. As a result, the radius of the cylinder in which we think the particle moves will be half as large, implying a factor of one fourth when squared, which in the result will cancel the $4$factor in the denumerator. That is, our expression will be

${\lambda }_{e^{-}}=\frac{1}{\pi \sqrt{2}N/V{r}^2}$

Calculation for pressure (part b)

(b)The mean free path must be expressed in terms of temperature rather than numerical density. Using the ideal gas law as a guide, we can have

$pV=N{k}_{B}T\Rightarrow \frac{N}{V}=\frac{p}{k_{B}T}$

${\lambda }_{e^{-}}=\frac{k_{B}T}{\pi \sqrt{2}p{r}^2}$

$p=\frac{k_{B}T}{\pi \sqrt{2}{\lambda }_{e^{-}}{r}^2}$

$p=\frac{1.38·{10}^{-23}·293}{\pi \sqrt{2}·5·{10}^4}·{\left(1·{10}^{-10}\right)}^2=1.82·{10}^{-6}\mathrm{Pa}$

$p=1.79\times {10}^{-10}atm$