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Old Feb 26th 2009, 03:12 PM   #1
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Arrow Radioactive decay problem

The chapter we are on is Radioactive Decay and the Nucleus. Here is the question:

There are 600 g of radioactive iodine-133. The half life of iodine-133 is 21 h.

a) How much of this radioactive substance would exist after 21 h. ?
b) After 42 h. ?
c) After 126 h. ?

P.S. If you could explain how to do this and maybe show me how to work this problem out, I would appreciate it!

- Confused
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Old Feb 26th 2009, 06:12 PM   #2
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For example, 126 = (6)(21) so the amount left after 126 hours is $\displaystyle \frac {600}{2^6} = \frac{600}{64} $.
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Old Jul 27th 2009, 09:49 PM   #3
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Decay equation:

$\displaystyle m = m_o(\frac{1}{2})^{\frac {\Delta T}{T_{half}}}$

$\displaystyle m_o$ is intial mass

$\displaystyle \Delta T$ is change in time

$\displaystyle T_{half}$ is the half-life time interval

$\displaystyle m$ is the mass remaining.

All you gotta to do now is substitute.
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Old Jul 28th 2009, 08:30 PM   #4
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Originally Posted by Deco View Post
Decay equation:

$\displaystyle m = m_o(\frac{1}{2})^{\frac {\Delta T}{T_{half}}}$
It's easier to read if you write it as follows

$\displaystyle m = m_o 2^{-\frac {\Delta T}{T_{half}}}$
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Old Jul 28th 2009, 08:32 PM   #5
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Whatever floats your boat.
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Old Jul 28th 2009, 08:33 PM   #6
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Originally Posted by Deco View Post
Whatever floats your boat.
Usually water.
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Old Jul 28th 2009, 08:37 PM   #7
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Hehe. .
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Old Jul 28th 2009, 08:45 PM   #8
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Originally Posted by Deco View Post
Hehe. .
I only mentioned it because the way Latex comes out it looks funny. It took me a few seconds for my eyes to adjust to what the expression was. Writing it in terms of exponentiating 2 makes it easier on my eyes and if its easier on my eyes there's a good chance its easier on others eyes too
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