How to Fit Half Lives of Polonium Isotopes to an Exponential Curve

[Galgenstrick]

Homework Statement

We are given a table of half lives of polonium isotopes and their corresponding alpha particle K.E. and asked to fit the data to the curve of :

half life=Ae^(B/sqrt(K.E.)) to find the constants A and B.

Homework Equations

half life=Ae^(B/sqrt(K.E.))

The Attempt at a Solution

I am graphing the half life vs. alpha particle K.E., and fitting the data using an exponential trend-line does not fit the data at all, it over 1000% off. Putting this data into logger pro and using this exact equation Ae^(B/sqrt(K.E.)) does not fit the data either its about 400% off. I think I am missing a detail. Please help!

 

[kuruman]

Take the logarithm of your equation on both sides. What do you get?

 

[Galgenstrick]

Take the logarithm of your equation on both sides. What do you get?

I am not sure I see how this would help. This equation is just used to fit the data, Y vs. X, in this case half life vs. K.E. of the alpha particles omitted.

It might help if I include the data. here it is (sorry for the weird formatting):

half life......KE (Mev)
138.4 days ......5.30
3E-7 seconds...8.78
1.64 E-4 seconds...7.68
0.15 seconds...6.78
3.05 minutes...6.00

 

[kuruman]

I am not sure I see how this would help.

I am. Just write down the algebraic equation, take the natural logarithm and if you don't see what is going on, ask again.

 

[Galgenstrick]

Taking the ln of both sides and simplifying, I get the following:

ln(half life)=B/sqrt(K.E.)+ln(A)

Is this correct? I did this by hand.

Now if I do this, I can take the ln of my half life values in my data instead of graphing it in log scale? Am I on the right track? Thank you for helping by the way.

 

[kuruman]

Taking the ln of both sides and simplifying, I get the following:

ln(half life)=B/sqrt(K.E.)+ln(A)
Is this correct? I did this by hand.

Very good, that is correct.

Now if I do this, I can take the ln of my half life values in my data instead of graphing it in log scale? Am I on the right track? Thank you for helping by the way.

You are on the right track. Look at the equation that you derived. Let y = ln(half-life) and x = 1/sqrt(K.E.) and substitute y and x in the equation. Do you recognize what the equation has become?

 

[Galgenstrick]

I get Y=B*x+ln(A)

I am embarrassed to say I do not recognize this, besides it being a line..

If I were to guess here at what you are getting at, if I graph these values in log scale, and do an exponential curve fit, I would get a line. I tried this, and maybe it is correct. The R^2 vale is only .95, and the % difference between the actual half life, and the calculated half life is almost 400% for one of the points.

This should be the same as taking the ln of the half life values, graphing, then doing a linear fit.

 

[kuruman]

I get Y=B*x+ln(A)

I am embarrassed to say I do not recognize this, besides it being a line..

More precisely it represents a straight line that obeys the equation

Y = slope*x + intercept

So what is "slope" equal to and what is "intercept" equal to?
What if you plotted y vs. x and did a linear regression to fit the data? How would you interpret the slope and intercept of the fitted line?

 

[Galgenstrick]

The intercept would be the ln(A), so to find A I would just take e^ln(A). and B would be the slope. Would this be correct?

 

[kuruman]

That would be correct.

 

[Galgenstrick]

This is the original way I did this problem, but the large % difference between the actual half life, and the regression line's calculated half life was large enough to make me think twice (about 400%). Maybe I am thinking about this too hard, thank you for the reassurance.

 

[kuruman]

You are thinking about it too hard. Logarithmic scales can be tricky. Don't forget that your dependent variable varies by 13 orders of magnitude, so don't think linearly. What is R² for your straight line fit? That should be a better indicator of how good the fit is than your linear 400% estimate.

 

[Galgenstrick]

You are thinking about it too hard. Logarithmic scales can be tricky. Don't forget that your dependent variable varies by 13 orders of magnitude, so don't think linearly. What is R² for your straight line fit? That should be a better indicator of how good the fit is than your linear 400% estimate.

The R^2 is 0.9599.

 

[kuruman]

I got 0.991. Did you use consistent units for the half-lives? I converted all to minutes. If you did, we need to compare constants A and B.

 

[Galgenstrick]

I got 0.991. Did you use consistent units for the half-lives? I converted all to minutes. If you did, we need to compare constants A and B.

You are right, yes, it is 0.991, I forgot to change my K.E. to 1/sqrt(K.E.) before graphing. Thank you so much for your help!