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Specific Activity

by tehfrr
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tehfrr
#1
Dec12-06, 10:26 AM
P: 53
Does anyone know how to calculate the specific activity [Ci/g] for radionuclides from halflife times? (using to calculate activity, sources are encapsulated and cannot be opened)

Or, does anyone know of somewhere on the net the actual numerical values for each nuclide are? Neither my textbooks nor google has turned up anything useful for me. (In specific, Im looking for Pu-238, Pu-239, Pu-240, Pu-241, Pu-242, Am-241)
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Morbius
#2
Dec12-06, 10:39 AM
Sci Advisor
P: 1,160
Quote Quote by tehfrr View Post
Does anyone know how to calculate the specific activity [Ci/g] for radionuclides from halflife times? (using to calculate activity, sources are encapsulated and cannot be opened)

Or, does anyone know of somewhere on the net the actual numerical values for each nuclide are? Neither my textbooks nor google has turned up anything useful for me. (In specific, Im looking for Pu-238, Pu-239, Pu-240, Pu-241, Pu-242, Am-241)
Quite simple. The rate at which a nuclide decays is given by the formula:

[tex]{ dN \over dt} = -\lambda N[/tex]

where N is the number of radioactive nuclei, and lambda is the decay constant, which
is related to the half-life by:

[tex]\lambda = { ln (2) \over \tau_{1/2}}[/tex]

If you know the half-life you can calculate lambda. To get the number of radioactive
nuclei, N; if you know the mass; calculate the number of moles by dividing the mass
by the atomic weight of the nuclide. Multiply by Avogadro's Number to get N.

Now you can calculate the decay rate in terms of decays per second.

To convert to Curies, note that a Curie is defined as 3.7e+10 decays per second:

http://en.wikipedia.org/wiki/Curie

Divide by the mass; and you have the specific activity. Note that if you leave the
mass out of the step above where you calculate the number of moles; then you don't
have to divide by it later. So you will have specific activity in terms of half-life, and
atomic weight.

Dr. Gregory Greenman
Physicist
tehfrr
#3
Dec12-06, 11:06 AM
P: 53
That should work since I have the mass. Thank you.

theCandyman
#4
Dec12-06, 05:05 PM
P: 395
Specific Activity

I am glad this came up. I was bored while preparing for my radiation physics final and I decided to find out how many kilograms of U-238 are needed to have the same activity as one kilogram of Ra-226. The answer I got was around 3 BILLION kilograms of U-238. I may have made a mistake, and even though it was just for fun (Fun? What has school done to me?), I wouldn't mind someone double checking. See what you come up with, terfrr.
tehfrr
#5
Dec12-06, 05:35 PM
P: 53
According to the software Im using it would take 2.97E+6 grams of U-238. When I go home Ill try it by hand using Morbius' method and see if I come up with the same number.

(software gives SA Ra-226 0.99885 [Ci/g] and SA U-238 3.361E-7 [Ci/g]), divided Ra/U for grams U assuming 1g Ra
Morbius
#6
Dec13-06, 08:07 AM
Sci Advisor
P: 1,160
Quote Quote by theCandyman View Post
I am glad this came up. I was bored while preparing for my radiation physics final and I decided to find out how many kilograms of U-238 are needed to have the same activity as one kilogram of Ra-226. The answer I got was around 3 BILLION kilograms of U-238. I may have made a mistake, and even though it was just for fun (Fun? What has school done to me?), I wouldn't mind someone double checking. See what you come up with, terfrr.
Candyman,

I think you slipped a few decimal points; it's about 3 MILLION.

The calculation as I outlined above; if you take the mass out, so that you are calculating
specific activity; depends on the half-lifes and atomic weights. Thereforre, the ratio
of the specific activity will be given by the product of the proper ratios of the half-lifes
and atomic weights.

The half-life of U-238 is 4.468 Billion years; the half-life of Ra-226 is 1600 years.
The atomic weights are approximately 238 and 226, of course.

If the half-life is longer; then you need more of the substance for a given activity.
Likewise, if the atomic weight is larger, you have fewer nuclei per unit mass.

Therefore the ratio of the specific activities of U-238 to Ra-226 should be given by:

Ratio = ( U-238 half-life )/( Ra-226 half-life ) * ( U-238 atomic wt ) / ( Ra-226 atomic wt )
= ( 4.468e9 / 1600 ) * ( 238 / 226 ) = 2.94e+06
[This is essentially what tehfrr got with his software.]

So it takes 3 Million kilograms of U-238 to have the same activity as 1 kilogram of Ra-226

Dr. Gregory Greenman
Physicist
theCandyman
#7
Dec13-06, 11:06 PM
P: 395
Quote Quote by Morbius View Post
I think you slipped a few decimal points; it's about 3 MILLION.
I see, I forgot to convert grams to kilograms after I found the moles of U-238 needed.

I had used this to calculate. I should have thought of comparing the specific activities.

[tex] A_{U-238} = \lambda N = A_{Ra-226}[/tex]

Still, I am astounded by these huge numbers! If I did not do the calculation myself, I would disbelieve such a claim. Has anyone else been suprised by how the numbers turn out?
Astronuc
#8
Dec14-06, 07:01 AM
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P: 21,885
Comparing 4.468 Billion years to 1600 years - that's 6 orders of magnitude.

The ratio of U-238 to Cs-137 or Sr-90 is even greater, 8 orders of magnitude.

Then there are radionuclides that have half-lives of months, weeks, days, hours and seconds. The short the half life, the greater the specific initial or equilibrium specific activity. However, the shorter the half-life, the faster the particular radionuclide decays. On the other hand, that may mean decaying into another radionuclide of longer half life, before decaying into an inert isotope.
Morbius
#9
Dec14-06, 08:13 AM
Sci Advisor
P: 1,160
Quote Quote by theCandyman View Post
Still, I am astounded by these huge numbers! If I did not do the calculation myself, I would disbelieve such a claim. Has anyone else been suprised by how the numbers turn out?
Candyman,

As Astronuc points out - the huge ratio in the specific activities of these radionuclides is
almost entirely due to the huge ratio in half-lives.

Basically, U-238 is ALMOST stable!! It has a half-life of about 4.5 BILLION years which
is roughly the age of the Earth. So only one U-238 half-life has elapsed since the Earth
was formed.

If U-238 was stable, the half-life would be INFINITE. Then your specific activity ratio
would be a lot bigger!!!

Dr. Gregory Greenman
Physicist
olayi
#10
May21-11, 03:44 PM
P: 1
How do i calculate the activity of Th-234 that i would expect to find in a 0.25g sample of Uranyl nitrate (UO2(NO3)2.6H2O) and what mass of Th-234 is this activity equivalent to?
minerva
#11
Jun27-11, 10:17 AM
P: 82
Specific activity data for radionuclides is given on Wolfram Alpha... just look up whatever nuclide you like.

But I guess understanding how you can calculate it from half-life is a good thing to understand, anyhow :)

Quote Quote by olayi View Post
How do i calculate the activity of Th-234 that i would expect to find in a 0.25g sample of Uranyl nitrate (UO2(NO3)2.6H2O) and what mass of Th-234 is this activity equivalent to?
We need to either get some more data here or take some assumptions.

What's the uranyl nitrate isotopic composition? Natural? Depleted?

Can we assume that the uranium daughter products are all in secular equilibrium?


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