What Is the Effective Target Area per Argon Atom?

[mishima]

Homework Statement

The intensity of a collimated, parallel beam of K atoms is reduced 3.0% by a layer of Ar gas 1.0 mm thick at a pressure of 6.0 x 10^-4 mmHg. Calculate the effective target area per argon atom. (this is B-4 from "introductory exercises in physics")

Homework Equations

I've been trying transmission T = I/I naught = e^(-x/l)
Where x is 1mm, T is .97, and l is the mean free path which includes the cross section.

The Attempt at a Solution

Im assuming the cross section of argon is the same thing as the effective target area per argon atom they are asking for, so I solve for that. I use l = {(2^.5) A n}^-1. A is the area I am looking for and n is the number of molecules per volume. To get n I convert the pressure given into molecules/volume. Not sure why but I can't seem to get anything close to the stated answer of 1.4 x 10^-14 cm^2.

 

[mishima]

The problem with the above is that I was incorrectly converting pressure to g/cm^3. I thought that, given the definition of mmHg as "The millimeter of mercury (symbol: mmHg) is defined as the pressure exerted at the base of a column of fluid exactly 1 mm high, when the density of the fluid is exactly 13.5951 g/cm3, at a place where the acceleration of gravity is exactly 9.80665 m/s2" I could just do a ratio like 1/13.5951 = 6x10^-4/x where x is the density. Not sure why this doesn't work, but every other step in the solution is pretty trivial so...

I suppose my real problem is not being able to find molecules/volume with only the information supplied by the text. Is there any way to get density with neither temperature nor volume?

 

[mishima]

Yep, I am sure this is the right path to a solution. I just don't know how to find n in the following formula from what is given. Any ideas?

 

[mishima]

Well, after all this, the solution was pretty trivial. Big thing I overlooked for a long time was just to assume the temperature of the gas to be 0 C. I thought since the density I was calculating was different the temperature had to be as well, but I guess that's not the case. Using the ideal gas law to solve for n, the number density, gives

P = pRT/M

Where P is pressure, p is density, R is gas constant, M is molar mass. But p is also

p = nM/N

Where n is number density, and N is avogadro's number. So, solving for n

n = P / kT

Using .08 Pa for P and 273 K for T gives an n which then works in the formula

Probability of hit = ns dx

Where s is the cross section, dx is .1 cm, Probability is the given 3%. All this gives the right answer.

I guess I couldve also just solved it algebraically like

s = ProbkT/Pdx

Well, I learned a lot on the side from this anyways...