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Topic: Kinetics of Radioactive Decay Problem  (Read 13348 times)

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Offline demonat0r

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Kinetics of Radioactive Decay Problem
« on: January 18, 2009, 01:36:00 PM »
Americum-241 is widely used in smoke detectors.  The radiation released by this element ionizes particles that are then detected by a charged-particle collector.  The half-life of Americum-241 is 432.2 years, and it decays by emitting alpha particles.  How many alpha particles are emitted each second by a 5.00g sample of Americum-241?

I have no idea how to even start this problem. The examples in my textbook are nothing like this. Please help.

Offline Mitch

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Re: Kinetics of Radioactive Decay Problem
« Reply #1 on: January 18, 2009, 01:50:01 PM »


Activity equals A (alpha particles/s), lambda is the decay constant, and N is the number of atoms you have in 5.00 g sample of Americium-241.

Activity is what you will be solving for.
« Last Edit: January 18, 2009, 03:10:59 PM by Mitch »
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Offline demonat0r

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Re: Kinetics of Radioactive Decay Problem
« Reply #2 on: January 18, 2009, 02:03:44 PM »


Activity equals A (alpha particles/s), lambda is the decay constant, and N is the number of atoms you have in 5.00 g sample of Americium-241.

Activity is what you will be solving for.

what is the decay constant? or do you mean the rate constant, k?
i found the number of atoms in 5.00g = (5.00g / 241g/mol)(6.02e23) = 1.25e22 atoms of Am-141

Offline Arkcon

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Re: Kinetics of Radioactive Decay Problem
« Reply #3 on: January 18, 2009, 02:53:04 PM »

i found the number of atoms in 5.00g = (5.00g / 241g/mol)(6.02e23) = 1.25e22 atoms of Am-141

Great.  Now, what does half-life mean?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Mitch

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Re: Kinetics of Radioactive Decay Problem
« Reply #4 on: January 18, 2009, 03:08:44 PM »
The decay constant is...
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Offline demonat0r

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Re: Kinetics of Radioactive Decay Problem
« Reply #5 on: January 18, 2009, 04:20:33 PM »
So based on the equation A = (lambda) (number atoms)
A = (1.25e22 atoms)(0.0016037 years^-1) = 2.0047e19 alpha particles / year
= 6.356e11 alpha particles / second
could you please check if that's the correct answer? the answer to this problem is not in the back of the textbook.

Offline Mitch

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Re: Kinetics of Radioactive Decay Problem
« Reply #6 on: January 18, 2009, 05:00:47 PM »
The logic seems correct.
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Offline Astrokel

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Re: Kinetics of Radioactive Decay Problem
« Reply #7 on: January 18, 2009, 05:01:41 PM »
The value is correct.
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