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cragwolf
#1
Mar5-12, 04:31 PM
P: 210
Why does the fact that the combined energy of 8Be and 4He is (almost) the same as the energy of an excited state of 12C greatly increase the probability of that reaction (i.e. 8Be + 4He → 12C + stuff) occurring? Apparently it doesn't have to be exactly the same, but does it make any difference if it's slightly less vs slightly more?
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