A parallel beam of Be(A=9) atoms is formed by evaporation...

A parallel beam of $\mathrm{Be}(\mathrm{A}=9)$ atoms is formed by evaporation from an oven heated to 1000 K through a small hole. (a) If the beam atoms are to traverse a 1 meter path length with less than $1 / e$ loss resulting from collisions with background gas atoms at room temperature ( 300 K ), what should be the pressure in the vacuum chamber? Assume a collision cross-section of $10^{-16} \mathrm{~cm}^2$, and ignore collisions between 2 beam atoms. (b) What is the mean time $(\bar{\tau})$ for the beam atoms to travel one meter? Show how the exact value for $\bar{\tau}$ is calculated from the appropriate velocity distribution. Do not evaluate integrals. Make a simple argument to get a numerical estimate for $\bar{\tau}$. (c) If the Be atoms stick to the far wall, estimate the pressure on the wall due to the beam where the beam strikes the wall. Assume the density of particles in the beam is $10^{10} / \mathrm{cm}^3$. Compare this result with the pressure from the background gas. FIGURE CAN'T COPY.

A parallel beam of $\mathrm{Be}(\mathrm{A}=9)$ atoms is formed by evaporation from an oven heated to 1000 K through a small hole.
(a) If the beam atoms are to traverse a 1 meter path length with less than $1 / e$ loss resulting from collisions with background gas atoms at room temperature ( 300 K ), what should be the pressure in the vacuum chamber? Assume a collision cross-section of $10^{-16} \mathrm{~cm}^2$, and ignore collisions between 2 beam atoms.
(b) What is the mean time $(\bar{\tau})$ for the beam atoms to travel one meter? Show how the exact value for $\bar{\tau}$ is calculated from the appropriate velocity distribution. Do not evaluate integrals. Make a simple argument to get a numerical estimate for $\bar{\tau}$.
(c) If the Be atoms stick to the far wall, estimate the pressure on the wall due to the beam where the beam strikes the wall. Assume the density of particles in the beam is $10^{10} / \mathrm{cm}^3$. Compare this result with the pressure from the background gas.
FIGURE CAN'T COPY.

Key Concepts

Momentum Transfer and Local Pressure

When particles in a beam impact a surface, they transfer momentum, leading to a localized pressure different from that of the background gas. This concept is essential in vacuum technology and surface science as it relates the density and velocity of incident particles to the resulting pressure on a surface. Understanding this momentum transfer is key in designing experiments and interpreting measurements where beam-induced pressures can affect system performance or measurement accuracy.

Beam Kinematics and Transit Time Calculations

Beam kinematics involves determining the motion parameters of particles within a beam, such as their average speed and transit time. Calculations often integrate the Maxwell-Boltzmann distribution to derive an exact expression for the mean time it takes for a particle to travel a specific distance. Simplified estimates are derived from typical or most probable speeds, providing a practical approach while highlighting the underlying physics of particle motion in a beam.

Maxwell-Boltzmann Velocity Distribution

The Maxwell-Boltzmann velocity distribution describes the statistical spread of speeds among particles in thermal equilibrium at a given temperature. This distribution is crucial when calculating average quantities such as the mean travel time of particles over a given distance, as well as for understanding energy distributions in gases. Its application is widespread in kinetic theory, where integrating over this distribution yields exact and average values of physical parameters like mean velocity, collision rates, and transport properties.

Collision Cross Section

The collision cross section quantifies the effective area that a moving particle presents for collisions with other particles. This parameter is central in estimating collision rates in gaseous systems because it directly relates to the probability of particles interacting as they traverse a medium. In many experimental setups, knowing the collision cross section helps in designing systems that reduce unwanted scattering or reactions, thereby preserving the properties of the beam or gas under study.

Vacuum Pressure and Mean Free Path

In vacuum physics, achieving sufficiently low pressure is crucial to ensure that particles, such as those in a molecular beam, traverse a desired distance without suffering significant collisions with background gas molecules. The mean free path, defined as the average distance a particle travels before a collision, is inversely related to the gas density (and thus the pressure) and the collision cross section. A proper understanding of these concepts allows one to set vacuum chamber pressures that minimize unwanted interactions, which is fundamental in fields like atomic physics and surface science.

Transcript

00:01 Okay, in this problem we have a liquid cesium sample that we're superheating to produce an atomic beam of some kind.

00:09 It's kind of interesting.

00:10 The temperature of our oven is about 400 degrees celsius, and it's about 55 cubic centimeters.

00:16 The vapor pressure of the cesium will be 17 millimeters mercury, and we're given the diameter of the cesium atoms, which is about 0 .33 nanometers.

00:26 In parts a, b, and c, we're asked to find respectively the root mean square velocity, which we can just, get from kinetic theory and then we're asked to find the frequency that an individual atom collides with other atoms so that we're going to call that f sub c and then we're asked in part c to find the total frequency of all the atoms colliding per second so a number of collisions per second total in our sample okay so to start us off the root mean square velocity is given from kinetic theory it's root 3k t over m where k is boltzman's constant we have the temperature of our sample and m is the mass of the individual atoms.

01:11 An equivalent expression is root 8 kt over pi m which is this is equation 18 -7b.

01:32 So just plugging everything in we have remember our bolstman constant is one always 1 .38 times 10 the negative 23 joules per kelvin the temperature in kelvin and sv in kelvin 670, kelvin, that's just 400 plus 273, which is how we do the conversion.

01:53 So we have pi times 133 times 1 .66 times 10 to negative 27 kilograms.

02:03 Okay? and crunching the numbers, the root mean square velocity will come out to be about 330 meters per second.

02:17 It's actually 327 .3, but it's good to approximate it up to something nice.

02:21 All right.

02:24 Now to get a collision frequency, we need to get the mean -free path, which is the distance a individual animal travel before it makes a collision.

02:36 We'll combine this with the ideal gas law.

02:41 So to start, the ideal gas law, pv equals big n kt, and it's a total number of particles in the gas.

02:48 K is both means constant as usual.

02:50 We need to get n over v, which is p over kt.

02:56 We're going to use this for later because the traditional.

02:58 Expression for mean -free path involves n over v and that actual expression is kind of gross to be honest 1 over 4 pi root 2 r squared where r is the radius of the individual particles times n over v which you can see we're going to replace n over v by something involving pressure and temperature instead and so once we make that substitution we get kt over 4 pi root 2 r squared times p all right at this point, we can plug in boltzman's constant, temperature, the radius of our particles, which we're given, and the pressure, which we're also given in millimeters mercury.

03:51 So it's actually, it probably would be useful to actually put these numbers in because there's a little bit of nuance in how we're doing these units.

03:57 It gets a little gross.

04:00 Boltzman's constant is as usual.

04:03 Temperature is as usual for pi root two.

04:11 Remember, the radius is half the diameter.

04:13 That's a common mistake, because we have to use half.

04:16 1 .65 times 10 to the negative 10 meters squared.

04:24 Our pressure 17 millimeters mercury needs to be converted into pascales, which is 133.

04:33 Sorry, i'm almost out of space here.

04:36 133 pascales per 1 millimeter mercury.

04:41 I'm not going to include that...