Question

30: (a)The mean free pathis the average distance a molecule travels before colliding with another molecule. The mean free path, l, is given by l 5 kT/(12sP), where kis Boltzmann’s constant, Tis temperature (K), Pis pressure (Pa), and s is the collision cross section. For a molecule with a diameter d, the collision cross section is pd2. The collision cross section is the area swept out by the molecule within which it will strike any other molecule it encounters. The magnetic sector mass spectrometer is maintained at a pressure of |1025 Pa so that ions do not collide with (and deflect) one another as they travel through the mass

analyzer. What is the mean free path of a molecule with a diameter of 1 nm at 300 K in the mass analyzer?

(b)The vacuum in an orbital mass separator is 1028 Pa. Find the mean free path in the orbital for the same conditions as (a).

Short answer

Mean free path of 1Nm :

^{$$\begin{gathered} \lambda ~{10}^{-5}=93.2290m \\ \lambda ~{10}^{-8}=93.2290m \end{gathered}$$}

Step-by-step solution

Mean Free Path

Mean free path is the average distance over which a moving particle travels before substantially changing its direction or energy, typically as a result of one or more successive collisions with other particles.

Speed of the heaviest ion

Mean free path of 1 nm at 300 k~10^-5;

$$\begin{gathered} \lambda =\frac{\left(1.3806852\times {10}^{-23}{\displaystyle \frac{J}{K}}\right)\left(300K\right)}{\sqrt{2}\left[\pi \right(1\times {10}^{-9}m{)}^2\left]\right({10}^{-5}Pa)} \\ \lambda =93.2290m \end{gathered}$$

Mean free path of 1 nm at 300 k ~ 10^-8;

$$\begin{gathered} \lambda =\frac{\left(1.3806852\times {10}^{-23}{\displaystyle \frac{J}{K}}\right)\left(300K\right)}{\sqrt{2}\left[\pi (1\times {10}^{-9}m{)}^2\right]\left({10}^{-8}Pa\right)} \\ \lambda =93.2290m \end{gathered}$$

The final answer is:

$$\begin{gathered} \lambda ~{10}^{-5}=93.2290m \\ \lambda ~{10}^{-8}=93.2290m \end{gathered}$$