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Physics
High Energy, Nuclear, Particle Physics
Q value for a radioactive decay
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[QUOTE="eneacasucci, post: 6899668, member: 733610"] Thank you so much, so we have shown that for the alpha decay Q>0 I've just one last question: From the conservation of energy, we see that the Q-value can be expressed also as the difference between the kinetic energy (K) of the final particles and the initial one ##Q=K_2+K_{\alpha}-K_1##, if we choose a reference frame where A1 is at rest we get ##K_1=0## and ##Q=K_2+K_{\alpha}>[COLOR=rgb(0, 0, 0)]0[/COLOR]## since the kinetic energy can't be negative. I suppose that if we change the reference system, for example considering the reference system for which the second nucleus A2 is at rest, the Q value should not change but it is simply less immediate to establish its sign, right? If the Q value is >0 for one reference system, can I say it is for any reference system I choose (It should not be a quantity that depends on the chosen reference system)? [/QUOTE]
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High Energy, Nuclear, Particle Physics
Q value for a radioactive decay
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