How Many Tubes Are Required to Achieve a Specific Isotope Concentration?

[L_McGrady]

Homework Statement

Want to separate two isotopes whose atomic masses differ by one neutron mass. We do this by filling a tube of height H with a gaseous mixture of the two isotopes

Derive a formula of the enrichment factor
[n1 (H)/ n2 (H)]/ [n1(0)/n2(0)]

If the gas from the top of the tube is fed into a second tube, and the processes is repeated N times, how many 10 m long tubes will be needed to make the final concentration .1 from the initial concentration of .001 at room temperature?

Homework Equations

μ= τ ln [nj/nq]
where nq= [mjτ/2∏hbar]^3/2

The Attempt at a Solution

Not sure where to even begin with this problem! Any help would be appreciated!

 

[anathema]

Thank you for your question! The enrichment factor, as defined in the post, is the ratio of the number of isotopes of the first type (n1) to the number of isotopes of the second type (n2) at a given height H, divided by the initial ratio of the number of isotopes at the bottom of the tube (n1(0) and n2(0)). This can be expressed as:

Enrichment factor = [n1(H)/n2(H)] / [n1(0)/n2(0)]

To derive a formula for this, we can use the equation for the chemical potential (μ) of a gas, which is given by:

μ= τ ln [nj/nq]

Where τ is the temperature, nj is the number of molecules of the jth isotope, and nq is the number of particles in a single quantum state. In this case, we can assume that the gas is in thermal equilibrium and that the two isotopes have the same temperature and volume. Therefore, we can simplify the equation to:

μ= τ ln [nj/nq] = τ ln [n1/nq] = τ ln [n2/nq]

Since the two isotopes have different masses, their number of particles in a single quantum state (nq) will be different. However, we can assume that the ratio of their number of particles in a single quantum state will remain constant throughout the tube. Therefore, we can rewrite the equation as:

μ= τ ln [n1/nq] = τ ln [n2/nq] = τ ln [n1(0)/nq] = τ ln [n2(0)/nq]

Now, we can substitute this into the equation for the enrichment factor:

Enrichment factor = [n1(H)/n2(H)] / [n1(0)/n2(0)] = [exp(μ1/τ)/exp(μ2/τ)] / [exp(μ1(0)/τ)/exp(μ2(0)/τ)] = exp[(μ1-μ2)/τ] / exp[(μ1(0)-μ2(0))/τ]

Using the formula for the chemical potential, we can rewrite this as:

Enrichment factor = exp[(τ ln [n1(0)/nq] - τ ln [n2(0)/nq