Fitting Curve by Exponential Growth Function - Rajini

[Rajini]

Hello All,
I am trying to fit a curve by exponential growth. See the attached photo for curve. I want to fit that curve with expo growth function.
We know that growth decay function is $N_{t}=N_{0}\times e^{\lambda t}$. Exponential part is positive as this is a growth function.
Is the function $N_{t}=b+[N_{0}\times e^{\lambda (t-t_0)}]$ is correct growth function for my curve shown in the photo? I am confused with $(t-t_0)$ part.
$b$=baseline, $N_0$=peak height or see photo, $\lambda$=1/meanlife time, $t_0$=see photo, from $t_0$ decays to the left side.
PS: I think the function $N_{t}=b+[N_{0}\times e^{\lambda (t-t_0)}]$ accounts for the curve just before/left side of $t_0$. But I need to use the full curve to fit. So kindly inform me how can I use the straight line at $t_0$ in the growth function.
Thanks for reply.
Cheers, Rajini.

 

[scottdave]

I think you will need two functions: one for before t0, and one for after t0. They should equal each other at t0 and have the same slope (looks like zero slope mabye?) at t0, as well, to be continuous, and look smooth.

 

[Rajini]

Hello Scottdave:
Yes two functions! It should be zero slop. But experimentally a slight slope is there...any idea of the function?
Thanks

 

[Khashishi]

We can't guess what the function is without knowing what the experiment is. Do you know why there is a slight slope there? Is it due to limited time resolution of your data? There could be some slew rate issue in your amplifiers, or maybe something in the physics you are modeling.

 

[Rajini]

Hello Khashishi,
That data is from coincidence count experiment obtained form Moessbauer spectrometer. The source/radioactive 57Co is used in such experiments. Kindly see the decay scheme of 57Co source more details. In short the 57Co decays to the 136 keV level and then to 14.4 keV level and then finally to ground state. In this experiment detection of the 122 keV level is used as the start signal (i.e. signal for the beginning of 14.4 keV gamma) and the detection of 14.4 keV is the stop signal (i.e. end of the decay of 14.4 keV energy level). 14.4 keV level's meanlife time is 142 ns, so if you fit that exponential growth function the meanlife time should be 142 ns, but I get 132 ns?. I get a exponential growth function, instead of exponential decay function...I don't why..may some wrong connections?? That curve is the result of decay of the 14.4 keV energy level from 57Co radioactive source.
Thanks,
rajini.

 

[Khashishi]

I don't really know much about this. It sounds like you have a three level system, initially in the 136 keV energy state. It decays to the 14.4 keV energy state, and then it decays to the ground state. So the population of the 14.4 keV state initially goes up since the rate of decay from the 136 keV state is greater than the rate of decay of the 14.4 keV state to the ground state. But then it hits a peak as the 136 keV population goes down, and the rate of decay to the ground state starts to exceed the rate of decay from the 136 keV state. Well, in that case, you don't have perfect exponential curves. I don't know anything about Moessbauer spectroscopy.

 

[Rajini]

Hi Khashishi,
An exponential decay is still possible (considered as 2 level system). The decay from 136 keV level to 14.4 keV level emit 122 keV gamma. The detection of 122 keV acts as a start signal for 14.4 keV level. Now the 14.4 keV level decays to ground state by emitting 14.4 keV gamma. The detection of 14.4 keV acts as a stop signal. Details in doi: 10.1038/nature13018 or Phys. Rev. A, 80, 063805, 2009. You can also see download for free : http://www.ortec-online.com/-/media/ametekortec/third%20edition%20experiments/nuclear-lifetimes-and-the-coincidence-method.pdf?la=en
I just need to know how to fit the exponential curve.
Cheers,
Rajini.