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Offline johnnieslingshot

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Decay Question
« on: Sep 08, 2009, 06:28 »
This is a question on a homework assignment for a Radioactive Material Handling class I am taking.  Is there anyone here who can give me some insight on how to answer it?  This is the only question that I have not been able to answer on the assignment.  The question is:  "A material will decay to 75% of its original activity in 12.5 years.  What is its half-life?"  Any help will be highly appreciated.

thenuttyneutron

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Re: Decay Question
« Reply #1 on: Sep 08, 2009, 06:53 »
Easy Way above.

Hard way below.

.75*X=X*e^(r*t)

You know X, and t.  You must figure out what r, decay constant, is by taking the natural log of both sides.

ln(.75*X)=r*t*ln(X)

Once you know what r is, put it in the equation.  Find what halflife T is.

.5=X*e^(r*T)

ln(.5)=r*T*ln(X)

T=r*ln(X)/(ln(.5))
« Last Edit: Sep 08, 2009, 06:55 by The Nutty Neutron »

Offline johnnieslingshot

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Re: Decay Question
« Reply #2 on: Sep 08, 2009, 09:07 »
Hey Henry, how's your headache?  :)    I think I have it figured out now.  By using your logic I decided to plug in the original activity of 100Ci for 100% and the end activity of 75Ci for 75% into the decay formula.  So now I have:

75Ci = 100Ci e^[(-.693/t1/2)(12.5)]
.75 = e^[(-.693/t1/2)912.5)]
Ln .75 = Ln e^[(-.693/t1/2)(12.5)]
-.288... = (-.693/t1/2)(12.5)
-.023... = (-.693/t1/2)
t1/2(-.023...) = -.693
t1/2 = -.693/-.023
t1/2 = 30.13

Sounds like it might be a 137Cs source

Thank you for all of the input Henry and Nutty.  If anybody disagrees with this logic please let me know.  I am open to obtaining further knowledge.

Slingshot

Offline RDTroja

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Re: Decay Question
« Reply #3 on: Sep 08, 2009, 09:34 »
The way I read this the material has decayed by 25% in 12.5 years. So if you multiply 12.5 times 2 then you would have the answer for the half life. 25 years. I hope I am reading  the post right. If not I will try again. Take care.

That is the easy way but it is wrong. The answer is 30.11 years. Here's how:

...

Damn... Slingshot beat me to it (I got interrupted in the middle of the reply and it took me too long to get back.) The difference (30.13 v. 30.11) is rounding error.

Nutty Neutron made his too hard to follow by including the X, which you don't need as long as you know the .75 part.
« Last Edit: Sep 08, 2009, 09:36 by RDTroja »
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thenuttyneutron

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Re: Decay Question
« Reply #4 on: Sep 08, 2009, 12:17 »
Hey Henry, how's your headache?  :)    I think I have it figured out now.  By using your logic I decided to plug in the original activity of 100Ci for 100% and the end activity of 75Ci for 75% into the decay formula.  So now I have:

75Ci = 100Ci e^[(-.693/t1/2)(12.5)]
.75 = e^[(-.693/t1/2)912.5)]
Ln .75 = Ln e^[(-.693/t1/2)(12.5)]
-.288... = (-.693/t1/2)(12.5)
-.023... = (-.693/t1/2)
t1/2(-.023...) = -.693
t1/2 = -.693/-.023
t1/2 = 30.13

Sounds like it might be a 137Cs source

Thank you for all of the input Henry and Nutty.  If anybody disagrees with this logic please let me know.  I am open to obtaining further knowledge.

Slingshot

This is better than mine.  I still have the engineering mind that I have to put variable everywhere so I can take the formula and use it on many problems as needed :)  The key to this is finding the decay constant and then it is plug and chug.  I can see how my "X" was getting confusing.

I did find one error on my previous post (sorry I had just gotten off a long midnight and I pulled it out of thin air), the last line should have 1/T.  If you see that mistake, the way I put down will work.  I am glad you found that other material to help because it is the same thing only better written.

30.12 years is what I get with my method.
« Last Edit: Sep 08, 2009, 12:26 by The Nutty Neutron »

moondoggie

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Re: Decay Question
« Reply #5 on: Sep 08, 2009, 05:13 »
I usually get to it this way:

T1/2 = -(ln2)(T)/ln(A/Ao)

T1/2 = -(.693 * 12.5)/ln(.75)

T1/2 = -8.6643/-.2877

T1/2 = 30.117 y

Depends how far you carry out the decimal points for accuracy.

Offline johnnieslingshot

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Re: Decay Question
« Reply #6 on: Sep 09, 2009, 06:44 »
This class that I am taking deals with the proper handling of accountable sources and also shipping of RAM.  So the depth that the course goes through is understandable to me.  Heck, we are even doing equilibrium equations. 

Once the light finally came on and I saw the the 75% was actually the A/Ao part of the equation I realized that I was trying to make it harder than it actually was.  I will find out Thursday evening what the actual answer is and the way the instructor wanted it done.  I am sure that the answer will be 30.1..... depending on rounding.  I really appreciate all of the input from you folks.

Slingshot

JsonD13

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Re: Decay Question
« Reply #7 on: Sep 09, 2009, 02:55 »
The way I read this the material has decayed by 25% in 12.5 years. So if you multiply 12.5 times 2 then you would have the answer for the half life. 25 years. I hope I am reading  the post right. If not I will try again. Take care.

Yeah, can't really use this due to decay being exponential in nature.

Jason

Offline johnnieslingshot

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Re: Decay Question
« Reply #8 on: Sep 14, 2009, 07:36 »
Hey ya'll,  the answer was 30.1...yrs.  The instructor worked it out like this:

A/Ao= e^[(-.693/t1/2)912.5)]
.75 = e^[(-.693/t1/2)912.5)]
Ln .75 = Ln e^[(-.693/t1/2)(12.5)]
-.288... = (-.693/t1/2)(12.5)
-.023... = (-.693/t1/2)
t1/2(-.023...) = -.693
t1/2 = -.693/-.023
t1/2 = 30.1....yrs

Sorry it took me so long to post.  I had a real busy weekend and did not log on.  Thanks again for all of the input.

 


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