# Help understanding nuclear property

by Sean Reed
Tags: nuclear, property
 Sci Advisor HW Helper P: 4,738 You distribution is gaussian (since you have collected much data), FWHM = $\Delta E = 2.35\sigma$ 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^(½) since $\sigma = \sqrt{\bar{n}}$. Now take the ratio FWHM/E, we see that the ratio (the resolution if you like) is poportional to E^(-½). So the width is also a measurment of the gamma-ray energy.
 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: $$\sigma^{2}=\frac{1}{(E-E_{0})^{2}+\Gamma^{2}/4}$$ ie. the probability of decaying from |E> to |E0>, where $$\Gamma$$ 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 $$\Gamma$$ (prob falls to half height when $$\(E=E_{0}+or-\Gamma/2$$