# Atoms

Dr. Brown has a radioisotope source containing 1.2 x 106 (also written as 1.2E+06) atoms of Plutonium-238 in 1985, how many atoms will remain in 352 years?

 Lightning55 Posts: 97, Reputation: 38 Junior Member #2 Dec 11, 2010, 01:30 PM
1.2 * 106 is NOT the same as 1.2E+06. I think you mean 1.2*(10^6). Try to solve this problem on your own.

Plutonium-238 has a half life of 87.7 years.
In a single half life, half of the radioactive atoms decay into stable atoms.

All you have to do is solve for the number of half lives, then reduce the original number of atoms by a half for each half life. Thus, you get (original number of atoms)*(1/2)^(number of half lives) for your answer.
gertrude88 (Dec 12, 2010 08:59 AM): It is helpful to a point if I understood the entire concept better. This is an online class so it makes it harder to learn these things for me.   Source:
 gertrude88 Posts: 5, Reputation: 10 Junior Member #3 Dec 12, 2010, 08:48 AM
Comment on Lightning55's post
Quote:
 Originally Posted by Lightning55 1.2 * 106 is NOT the same as 1.2E+06. I think you mean 1.2*(10^6). Try to solve this problem on your own. Plutonium-238 has a half life of 87.7 years. In a single half life, half of the radioactive atoms decay into stable atoms. All you have to do is solve for the number of half lives, then reduce the original number of atoms by a half for each half life. Thus, you get (original number of atoms)*(1/2)^(number of half lives) for your answer.
Still a little confused
 Lightning55 Posts: 97, Reputation: 38 Junior Member #4 Dec 12, 2010, 09:09 AM
Comment on Lightning55's post
Quote:
 Originally Posted by Lightning55 1.2 * 106 is NOT the same as 1.2E+06. I think you mean 1.2*(10^6). Try to solve this problem on your own. Plutonium-238 has a half life of 87.7 years. In a single half life, half of the radioactive atoms decay into stable atoms. All you have to do is solve for the number of half lives, then reduce the original number of atoms by a half for each half life. Thus, you get (original number of atoms)*(1/2)^(number of half lives) for your answer.
Well, a half-life is how long it takes for half of the atoms to turn into a stable form, which is NOT Plutonium-238. That's how many "halves" you would have (even partial halves). It is exponential regression.

Where exactly are you confused on?
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #5 Dec 12, 2010, 09:23 AM

If you however don't understand the answer, that doesn't mean that it's not helpful. The answer is simply in such a way that for you personally, you don't understand, even if the answer would be helpful to somebody else. In this case, simply ask for clarification, ask where you have a problem in understanding.

Now, coming to your question, I'm surprised that the question didn't give you the half life of the radioactive isotope... They should usually give them to you, or maybe it was mentioned somewhere in the give text or explanation given. (I don't know how e-learning works, sorry if that's not how it works)

One correction to be done in your formula, Lightning55 is this:

$N = N_o $$\frac12$$^{t/h}$

You missed h, the half life.

Here, N is the number of atoms you are looking for, that is the number after time t,
No is the initial number of atoms,
T is the time as I just said earlier (usually in years, but might change)
H is the half life (ie the time it takes for the number to become half the initial number of atoms)

Your question given as it is cannot be solved, unless you are expected to remember the half life of every radioactive isotope you'll come across with.
 Lightning55 Posts: 97, Reputation: 38 Junior Member #6 Dec 12, 2010, 09:59 AM
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Quote:
 Originally Posted by Unknown008 Gertrude88, I'll ask you to give thumbs down only if the answer given to you is misleading or not helpful at all. If you however don't understand the answer, that doesn't mean that it's not helpful. The answer is simply in such a way that for you personally, you don't understand, even if the answer would be helpful to somebody else. In this case, simply ask for clarification, ask where you have a problem in understanding. Now, coming to your question, I'm surprised that the question didn't give you the half life of the radioactive isotope... They should usually give them to you, or maybe it was mentioned somewhere in the give text or explanation given. (I don't know how e-learning works, sorry if that's not how it works) One correction to be done in your formula, Lightning55 is this: $N = N_o $$\frac12$$^{t/h}$ You missed h, the half life. Here, N is the number of atoms you are looking for, that is the number after time t, No is the initial number of atoms, T is the time as I just said earlier (usually in years, but might change) H is the half life (ie the time it takes for the number to become half the initial number of atoms) Your question given as it is cannot be solved, unless you are expected to remember the half life of every radioactive isotope you'll come across with.
Ah you're right. It was late last night and my formulas are going down the drain. Your formula makes it much clearer.
 msdyj38 Posts: 1, Reputation: 1 New Member #7 Mar 29, 2013, 02:44 PM
Dr. Brown has a radioisotope source containing 1.2 x 106 (also written as 1.2E+06) atoms of Plutonium-238 in 1985, how many atoms will remain in 352 years?

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