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### Topic: half life problem  (Read 3946 times)

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#### clockworks204

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##### half life problem
« on: July 22, 2010, 03:58:00 PM »
Actinium (227Ac) has a half life of 22 years.  One gram of actinium contains 2.65x1021 atoms.  What is the activity of one gram of actinium? What is the activity is curies? (one curie is 3.7x1010 disintegrations per second)

I know that the formula for unstable atoms left over after t time is N=Noe-0.693t/t1/2
I know that t1/2= 22 yrs, No= 2.65x1021 atoms.  I'm not sure where to go from here.

#### MrTeo

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##### Re: half life problem
« Reply #1 on: July 22, 2010, 04:21:19 PM »
The formula you've written comes from the solution of this differential equation:

${ \frac{dN}{dt}=-kN }$

Here as you can see the first term is the activity of the amount of compound given (disintegrations per second) so you only need to find out the decay constant (if you're not sure about that continue reading) and then you'll easily find the activity in Curie as the problem asks you.

Bonus:
Here is the known formula and how you can get it from the differential equation I gave you:

${ \frac{dN}{dt}=-kN \\
\frac{dN}{N}=-kdt \\
\int{\frac{dN}{N}}=-k\int{dt} \\
\ln{N}=-kt+c \\
N=e^c \cdot e^{-kt}=c'e^{-kt} }$

Knowing that at t=0 we have N0 atoms gives us the final formula:

${ N\left(0\right)=N_0 \\
N\left(0\right)=c'e^{-k\cdot 0}=c'=N_0 \\
N=N_0e^{-kt} }$

Finally the relation with half-life:

${ \frac{N_0}{2}=N_0e^{-kt_{1/2}} \\
-\ln{0.5}=kt_{1/2} \\
k \approx \frac{0.693}{t_{1/2}} }$
The way of the superior man may be compared to what takes place in traveling, when to go to a distance we must first traverse the space that is near, and in ascending a height, when we must begin from the lower ground. (Confucius)

#### clockworks204

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##### Re: half life problem
« Reply #2 on: July 22, 2010, 06:38:49 PM »
Thank you for your reply.  I've been studying what you've posted for a bit, and have tried applying it.
k=.693/22 years = .0315 for the decay constant, and use your first equation to multiply .0315 by 2.65x1021 to get 8.3x1019.

According to the textbook, the answer for this should be 2.6x1012.  The activity in curies is supposed to be 71 curies.

I've tried this problem for a while now, and still can't come up with the answers from the textbook

#### MrTeo

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##### Re: half life problem
« Reply #3 on: July 23, 2010, 02:44:16 AM »
Just convert 22yr to seconds
The way of the superior man may be compared to what takes place in traveling, when to go to a distance we must first traverse the space that is near, and in ascending a height, when we must begin from the lower ground. (Confucius)

#### clockworks204

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##### Re: half life problem
« Reply #4 on: July 23, 2010, 10:58:08 AM »
Got it!  You've been a great *delete me*