I am struggling to understand the $Q$-factor for the $\beta^-$– decay process. My understanding of the $Q$-factor is that it's the difference in binding energies before and after the chemical reaction. So far, I have seen the $Q$-factor for $\beta^-$ – decay derived as follows (neglecting the binding energy of the electrons):
$$m(^A_ZX)=Zm_H + (A-Z)m_n + Zm_e \Rightarrow Zm_H + (A-Z)m_n = m(^A_ZX)-Zm_e$$
$$\Rightarrow Q=[m(^A_ZX)-Zm_e]c² – [m(^A_{Z+1}Y)-(Z+1)m_e + m_e]c²=[m(^A_ZX)-m(^A_{Z+1}Y)]c²$$
i.e. the Q-factor for $\beta^-$– decay is the difference in atomic mass of the parent and daughter atom converted into energy. I do not understand how this is correct as the binding energy of the nucleus before decay is given as (1) and after the reaction is given as (2). Substituting (2) from (1) gives a different result to the Q-factor above.
$$B_i=[Zm_H+(A-Z)m_n-m(^A_ZX)]c²\tag{1}$$
$$B_f=[(Z+1)m_H+(A-Z-1)m_n-m(^A_{Z+1}Y)+m_e]c²\tag{2}$$
$$\Rightarrow B_i – B_f=[-m_H + m_n – m(^A_ZX) + m(^A_{Z+1}Y) – m_e]c^2$$
Could someone please provide a clear definition of the $Q$-factor for radioactive decays and which of the above calculations is correct?
Best Answer
This is wrong. The Q-value is the difference between the rest masses of the initial state and the final state. The binding energy subtracts out the particle masses. Because the neutron mass is different from the proton mass, if your data source gives you just binding energies, you will be led astray.
(Beware that the “Q-factor” of a damped resonator is different from the “Q-value” of a nuclear decay.)
To see which calculation is correct, get some data and try to match it. Two systems where you might make progress are the decay of the free neutron (where the initial and final binding energies are both zero) and the decay of tritium to helium-3.
As I frequently advise people, I prefer the “mass excess” over the raw mass, the binding energy, or other equivalent data. The mass excess subtracts off the constant 1 a.m.u. per nucleon, and is usually already tabulated in energy units, so you don’t have to carry around four or five significant figures to get two or three significant figures in a mega-eV decay.