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May 27th, 2019, 04:31 PM   #1
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Graph half-life

A radioisotope of Krypton has a half-life of 3.0s. A sample of krypton contains 1024x1024 undecayed atoms.

Calculate how many undecayed atoms will remain after half a minute

My solution → Firstly, I will change 3.0s to a minute 0.05 Then I figure out how many half-lives have been 0.5 (half a minute) / (divided by) 0.05=10 Then I, (1/2)^10=1/1024 change into a decimal 0.0009765625 Then lastly I will time (1024x1024) by (0.0009765625) to get 1x1024 undecayed atoms after half a minute.

So I think it would be beneficial to also display this in a graph to show the decrease (decay) in half a minute. Can somebody do this for me, or show me how to?

Last edited by greg1313; May 28th, 2019 at 12:52 AM.
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May 27th, 2019, 05:27 PM   #2
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Basic exponential form is:

$f(t) = f_{0}e^{kt}$

$t$ is time in seconds, $f_{0}$ is the starting value, and $k$ is a coefficient we'll need to find.

$f_{0} = 1024^{2}$

$f(3) = \frac{f_{0}}{2} = \left(\frac{1024^{2}}{2} \right)$

Now we solve for $k$

$\left(\frac{1024^{2}}{2} \right) = 1024^{2}e^{3k}$

$k = \left( \frac{ \ln(0.5) }{3} \right)$

So just graphing the equation

$\displaystyle f(t) =1024^{2}e^{ \left( \frac{ \ln(0.5)t }{3} \right) }$

Should do the trick, here it is in Desmos:

https://www.desmos.com/calculator/hj91obhfur
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May 27th, 2019, 05:41 PM   #3
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On demos have you put the number of undecayed atoms on the Y axis and the time on the X axis?
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May 27th, 2019, 05:46 PM   #4
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Yes
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May 27th, 2019, 06:03 PM   #5
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Quote:
Originally Posted by helpmeddddd View Post
A radioisotope of Krypton has a half-life of 3.0s. A sample of krypton contains 1024x10^24 undecayed atoms.

Calculate how many undecayed atoms will remain after half a minute

My solution → Firstly, I will change 3.0s to a minute 0.05 Then I figure out how many half-lives have been 0.5 (half a minute) / (divided by) 0.05=10 Then I, (1/2)^10=1/1024 change into a decimal 0.0009765625 Then lastly I will time (1024x10^24) by (0.0009765625) to get 1x10^24 undecayed atoms after half a minute.

So I think it would be beneficial to also display this in a graph to show the decrease (decay) in half a minute. Can somebody do this for me, or show me how to?
I forgot to place some ^ to the power of signs does this affect your graph?

Last edited by helpmeddddd; May 27th, 2019 at 06:05 PM.
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May 27th, 2019, 06:18 PM   #6
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Just change $f_{0}$ to $1024 \times 10^{24}$, so

$\displaystyle f(t) =\left(1024 \times 10^{24} \right) e^{ \left( \frac{ \ln(0.5)t }{3} \right) }$
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May 28th, 2019, 05:10 AM   #7
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Quote:
Originally Posted by helpmeddddd View Post
. . . (1/2)^10=1/1024 change . . .
It can be left as 1/1024. The answer is therefore 10^24.

Producing a graph isn't necessary, but why not opt for a log graph, which would be a straight line?
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May 28th, 2019, 12:38 PM   #8
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Show me this log graph?
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May 28th, 2019, 03:59 PM   #9
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Why? It's just a straight line segment (starting at (0, 27) approximately) with a gentle negative apparent slope. The apparent slope depends on the scaling used, but if you plot the base 10 logarithm of the function Greens gave, the gradient is about -0.1.
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