[demonat0r]

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.

[Mitch]

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.

[demonat0r]

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

[Arkcon]

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?

[Mitch]

The decay constant is...

[demonat0r]

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.

[Mitch]

The logic seems correct.

[Astrokel]

The value is correct.