May 27th, 2019, 04:31 PM  #1 
Member Joined: Mar 2019 From: TTF area Posts: 88 Thanks: 1  Graph halflife
A radioisotope of Krypton has a halflife 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 halflives 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. 
May 27th, 2019, 05:27 PM  #2 
Member Joined: Oct 2018 From: USA Posts: 78 Thanks: 54 Math Focus: Algebraic Geometry 
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 
May 27th, 2019, 05:41 PM  #3 
Member Joined: Mar 2019 From: TTF area Posts: 88 Thanks: 1 
On demos have you put the number of undecayed atoms on the Y axis and the time on the X axis?

May 27th, 2019, 05:46 PM  #4 
Member Joined: Oct 2018 From: USA Posts: 78 Thanks: 54 Math Focus: Algebraic Geometry 
Yes

May 27th, 2019, 06:03 PM  #5  
Member Joined: Mar 2019 From: TTF area Posts: 88 Thanks: 1  Quote:
Last edited by helpmeddddd; May 27th, 2019 at 06:05 PM.  
May 27th, 2019, 06:18 PM  #6 
Member Joined: Oct 2018 From: USA Posts: 78 Thanks: 54 Math Focus: Algebraic Geometry 
Just change $f_{0}$ to $1024 \times 10^{24}$, so $\displaystyle f(t) =\left(1024 \times 10^{24} \right) e^{ \left( \frac{ \ln(0.5)t }{3} \right) }$ 
May 28th, 2019, 05:10 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,759 Thanks: 2138  
May 28th, 2019, 12:38 PM  #8 
Member Joined: Mar 2019 From: TTF area Posts: 88 Thanks: 1 
Show me this log graph?

May 28th, 2019, 03:59 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,759 Thanks: 2138 
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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