What is the Composition of a Pu239 and Pu240 Mixture Based on Specific Activity?

[zorro]

Homework Statement

A mixture of Pu²³⁹ and Pu²⁴⁰ has a specific activity of 6 x 10⁹ dps. The half-lives of the isotopes are 2.44 x 10⁴ years and 6.58 x 10³ years respectively. Calculate the composition of the mixture.

The Attempt at a Solution

The last equation does not give permissible values.

Spoiler

39% and 61%

 

[Andrew Mason]

Using:

A = \lambda N

where A is the activity in disintegrations/sec, \lambda = ln2/T_{half} in seconds, and N is the number of atoms. The number of seconds in one year is 365.25 x 24 x 60 x 60 = 3.16 x 10^7 seconds

In one second the number of disintegrations from PU²³⁹ is:

A_{239} = \frac{\ln 2}{2.44 \times 10^4 \times 3.16 \times 10^7}N_{239}

Similarly for PU²⁴⁰

A_{240} = \frac{\ln 2}{6.58 \times 10^4 \times 3.16 \times 10^7}N_{240}

So the total activity is:

A_{239} + A_{240} = \frac{\ln 2}{2.44 \times 10^4 \times 3.16 \times 10^7}N_{239} + \frac{\ln 2}{6.58 \times 10^4 \times 3.16 \times 10^7}N_{240}

We know that total activity = 6 x 10^9 dps. This is for one gram. Work out N_239 in terms of N_240. This is a little tricky if you want to do it really accurately. But because they are so close in atomic mass, you can use an average molecular weight of 239 or 240 grams/mole and get sufficient accuracy.

AM

 

[zorro]

Hi AM,
I would be very grateful if you would explain me what is wrong with my solution.

 

[Andrew Mason]

Hi AM,
I would be very grateful if you would explain me what is wrong with my solution.

You are using just N. To understand why you can't do that, compare my equation with yours. Are they the same? Why are you using N? What does N represent?

Work out A239 and A240 for one gram of each element using the equations I have given you. Then determine the percentage of each element that you must have in order to have A239 + A240 = 6 x 10^9

AM

 

[zorro]

You are using just N. To understand why you can't do that, compare my equation with yours. Are they the same? Why are you using N? What does N represent?

I took the average disintegration constant for both the disintegrations by adding them (which is also done when a particular atom disintegrates in two parallel pathways). N is the total number of atoms of both the isotopes present at any instant.

 

[Andrew Mason]

I took the average disintegration constant for both the disintegrations by adding them (which is also done when a particular atom disintegrates in two parallel pathways). N is the total number of atoms of both the isotopes present at any instant.

What is your reasoning for stating:

\lambda = \frac{\ln 2}{3.15 \times 10^7 \times (2.44 \times 10^4 + 6.58 \times 10^3)}?

AM

 

[zorro]

It's

I took the total (average) disintegration constant assuming the total disintegration to occur only in a single process. N includes the sum of atoms of both isotopes.

 

[Andrew Mason]

It's

I took the total (average) disintegration constant assuming the total disintegration to occur only in a single process. N includes the sum of atoms of both isotopes.

By averaging the disintegration constants, are you not assuming that there are equal parts of each isotope?

Are you really averaging the constants here? Would an average not be:

\lambda_{avg} = \frac{\lambda_{239} + \lambda_{240}}{2} ?

AM

 

[zorro]

Actually I got this idea from a different problem. In that problem, a radioactive isotope disintegrates into two particles with disintegration constants λ1 and λ2. There the solution uses the formula saying (λ1+λ2) to be the average disintegration constant.
A = (λ1+λ2)N

But I think this is applicable only for one particular isotope. Here there are two different isotopes so that method can't be used (don't know why).

 

[Andrew Mason]

Actually I got this idea from a different problem. In that problem, a radioactive isotope disintegrates into two particles with disintegration constants λ1 and λ2. There the solution uses the formula saying (λ1+λ2) to be the average disintegration constant.
A = (λ1+λ2)N

But I think this is applicable only for one particular isotope. Here there are two different isotopes so that method can't be used (don't know why).

I don't know why you would take the average. I suggest you use the simple approach I have suggested.

AM

 

[SammyS]

I took the average disintegration constant for both the disintegrations by adding them (which is also done when a particular atom disintegrates in two parallel pathways). N is the total number of atoms of both the isotopes present at any instant.

In the case of one type of atom disintegrating in two different ways, of course there is only one N.

I this case, you have two different atoms. The whole idea is to find N₂₃₉/N₂₄₀ (or something similar) for the sample!

 

[zorro]

Thanks!