Chemical Forums
Specialty Chemistry Forums => Nuclear Chemistry and Radiochemistry Forum => Topic started by: khalid on July 27, 2005, 08:38:31 PM
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Hello, I'm new to the forum, nice to meet you all ^^
My question:
the Isotop Sr90 has a half life of 28.1 years. how much would remain from a 1g of Sr after 25 years.
I toke the topic of half-life in geology and general chem before, but i dont have my textbook at the moment to review and find the answer ><
By dividing 28.1/25, this would give us the number of half lifes in 28.1 years i think?
But i dont know how to take it from there, and how to find the mass after 28.1 years.
A 2nd part of the question asks about the percentage of the initial acivities remains after 25 years. i found the decay constant, k in 1/sec., but couldnt continou.
I would ask the instractor if i could, but am taking this course as an independent study. sorry if my question was not very clear, english is my 2nd language.
Thanks alot and i hope to hear from you soon.
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Hello, I'm new to the forum, nice to meet you all ^^
My question:
the Isotop Sr90 has a half life of 28.1 years. how much would remain from a 1g of Sr after 25 years.
I toke the topic of half-life in geology and general chem before, but i dont have my textbook at the moment to review and find the answer ><
By dividing 28.1/25, this would give us the number of half lifes in 28.1 years i think?
But i dont know how to take it from there, and how to find the mass after 28.1 years.
A 2nd part of the question asks about the percentage of the initial acivities remains after 25 years. i found the decay constant, k in 1/sec., but couldnt continou.
I would ask the instractor if i could, but am taking this course as an independent study. sorry if my question was not very clear, english is my 2nd language.
Thanks alot and i hope to hear from you soon.
the formula for radioactive decay is:
N(t) = N(0) * (1/2)^(t/tau) with N = numbers of cores and tau = half life time.
so 1g Sr = 0,0114129194 mol = 6,8706 x 10^21 cores.
so: N(25 years) = 6,8706 x 10^21 * (1/2)^25/28,1
--> N(25 years) = 3,7083 x 10^21 cores
--> that's 0,0061599 mol Sr = 0,540 g Sr.
Same for the activities of a radioactive isotope:
A(t) = A(0) * (1/2)^(t/tau)
we want the percentage of activity f the initial activities after 25 years, so that A(25)/A(0) * 100%
so the percentage is given in: (1/2)^(t/tau)
--> 1/2^(25/28,1) * 100% = 54 %
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The radioactive decay formula is actually.
N(t) = N(0) * e-0.693t/tau If you define tau as the half life.
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The radioactive decay formula is actually.
N(t) = N(0) * e-0.693t/tau If you define tau as the half life.
it's the same as you have it in ln-form and i have it simple in log-form.
here's a derivation
ln x = 2,303 log x
so we have N(t) = N(0) * (1/2)^(t/tau)
take the log on both sides:
log N(t) = log N(0) + (t/tau)log 0,5
convert to ln:
2,303 log N(t) = 2,303 log N(0) + 2,303(t/tau)log 0,5
thus:
ln N(t) = ln N(0) + (t/tau) ln 0,5 --> ln 0,5 = -0,693
now both sides e^:
N(t) = N(0) * e^-0,693(t/tau)
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Best to keep things in the natural log form. It flows nicely from the differential form of the decay law.
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Best to keep things in the natural log form. It flows nicely from the differential form of the decay law.
i agree, that form the differential equation dN/dt = -lambda * N we'll get a power of e, but when you don't know how to solve differential equations, it's easier to say that the decay is exponential with a constant factor: 1/2
--> from maths, the formula for a exponential decay ( dutch variables used)
H = b * g^t --> N = N(0) x (1/2)^tau
to let it count for every t, just remember that 3 months for example is 3/12 year and thus we get:
N = N(0) x (1/2)^(t/tau)
it's not wrong and easier for people that don't know differential equations. But again, i agree that solving the differential equation is better.
(here in holland we get radioactive decay in secondary school without ever seeing any differential equation. That's why i did it this way. The answer will still be the same :))
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Whatever floats your boat, just don't get the youth confused.
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Thanks alot for the help.
I was able to solve all 7 problems with the info you gave me. ^^;