- #1
Aakash Sunkari
- 13
- 1
Hey everyone,
I've learned about double beta decay and neutrinoless double beta decay recently. So we have two "conditions" for decay, 2v decay and 0v decay. Now, to the question I have:
There have been a lot of experiments measuring 2v decay rates, and there have been many experiments concerning a limit on Majorana mass for neutrinos in 0v decay. We have the following equation for 2v double beta decay rates:
(T 2v1/2)-1 = G2v | M2v | 2
Where G is the phase space factor and M is the nuclear matrix element.
So I've acquired this list of values that correspond to the G and M values for 2ν Double Beta Decay values, and I am calculating the rate for the 2v Decay of 48Ca:
M value (Nuclear Matrix Element): ~0.05
(G2v)-1 value: 9.7 X 1016 y x MeV-2
So my question is, in order to acquire the decay rate, do all I need to do is multiply these values? I know it seems silly to ask, but please bear with me. I am a high school student who is learning all of this physics for the first time (at a college), it just seems a little crazy that it could just be as simple as that
Thank you in advance!
-Aakash Sunkari
I've learned about double beta decay and neutrinoless double beta decay recently. So we have two "conditions" for decay, 2v decay and 0v decay. Now, to the question I have:
There have been a lot of experiments measuring 2v decay rates, and there have been many experiments concerning a limit on Majorana mass for neutrinos in 0v decay. We have the following equation for 2v double beta decay rates:
(T 2v1/2)-1 = G2v | M2v | 2
Where G is the phase space factor and M is the nuclear matrix element.
So I've acquired this list of values that correspond to the G and M values for 2ν Double Beta Decay values, and I am calculating the rate for the 2v Decay of 48Ca:
M value (Nuclear Matrix Element): ~0.05
(G2v)-1 value: 9.7 X 1016 y x MeV-2
So my question is, in order to acquire the decay rate, do all I need to do is multiply these values? I know it seems silly to ask, but please bear with me. I am a high school student who is learning all of this physics for the first time (at a college), it just seems a little crazy that it could just be as simple as that
Thank you in advance!
-Aakash Sunkari
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