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Help understanding nuclear property

by Sean Reed
Tags: nuclear, property
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Sean Reed
#1
Apr13-08, 04:01 AM
P: 4
Using a small germanium gamma ray detector I have been collecting data on some newly acquired gamma decay sources. The program I have been using gives me a plot of counts vs energy (well channel, but the channels are proportional to the energy) Thus after collecting data I have a bell shaped curve with the peak centered near or on the energy of the decay I am looking at. What I am confused on is what importance in this does the FWHM have with regards to the source its self? I know some of the cause of the curve is noise in the equipment, but what meaning does it have with the source.

I feel like there is some meaning the FWHM has about the source because I used a large detector to measure the decays per second of the source, then I had to switch to a small detector and when I did I only looked at the FWHM of the curves.

Any help is greatly appreciated
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malawi_glenn
#2
Apr13-08, 05:14 AM
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You distribution is gaussian (since you have collected much data),

FWHM = [itex] \Delta E = 2.35\sigma [/itex]
http://mathworld.wolfram.com/GaussianFunction.html

sigma is related to the number of electrons causing the signal (particle-hole paris in the Ge-detector), hence the higher energy the more particle-hole paris. So the FWHM is proportional to E^() since [itex] \sigma = \sqrt{\bar{n}} [/itex].

Now take the ratio FWHM/E, we see that the ratio (the resolution if you like) is poportional to E^(-).

So the width is also a measurment of the gamma-ray energy.
Sean Reed
#3
Apr13-08, 04:30 PM
P: 4
So am I correct that the importance of the FWHM is that it is another way to measure the energy of the decay?

malawi_glenn
#4
Apr13-08, 04:31 PM
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Help understanding nuclear property

sort of yes
vertices
#5
Apr13-08, 08:55 PM
P: 230
"So am I correct that the importance of the FWHM is that it is another way to measure the energy of the decay?"

Not quite.

I assume you are talking about the photon energies?

A relavent formula here is the Briet Wigner Cross section:

[tex]\sigma^{2}=\frac{1}{(E-E_{0})^{2}+\Gamma^{2}/4}[/tex]

ie. the probability of decaying from |E> to |E0>, where [tex]\Gamma[/tex] is the decay rate (cos of heisenberg uncertainty relation), expressed in units of energy. Ofcourse, the count rate is proportional to the probability.

By inspection, you can see that the FWHM is equal to [tex]\Gamma[/tex] (prob falls to half height when [tex]\(E=E_{0}+or-\Gamma/2[/tex]
Sean Reed
#6
Apr14-08, 03:32 PM
P: 4
So when I am using the detector for the alpha particles of Am-241 samples, why will they have different FWHM values? Specifically I have two samples of Am-241 and the stronger sample (more Ci) always gives me a smaller FWHM.

thanks
malawi_glenn
#7
Apr14-08, 03:39 PM
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vertices: that is the intrinsic width of resonances.

Now we where discussing the detector resolution.

Sean_Reed:

Yes, since you will also get a higher peak (more registered alphas) due to higher intensity of alphas, the FWHM will go down. There are many parameters that governs the FWHM. What text book do you use for this laboration?


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