Calculate the disintegration energy Q for the beta decay of 32P, as described by equation 32P→32S + e- + ν (\tau = 14.3 d). The needed atomic masses are 31.97391 u for 32P and 31.97207 u for 32S.
Answers
Because of the presence of the emitted electron, we must be careful to distinguish between nuclear and atomic masses. Let the boldface symbols mp and ms represent the nuclear masses 32P and 32S and let the italic symbols mp and ms represent their atomic masses. We take the disintegration energy Q to be Δmc2, where, from Eq. 32P→32S + e- + ν
Δm = mp – (ms + me),
in which me is the mass of the electron. If we add and subtract 15me on the right side of this equation, we obtain
Δm = (mp + 15me) – (ms + 16me)
The quantities in parentheses are the atomic masses in this way, the mass of the emitted electron is automatically taken into account. (This will not work for positron emission.)
The disintegration energy for the 32P decay is then
Q = Δmc2
= (31.97391 u – 31.97207 u) (931.5 MeV/u)
= 1.71 MeV
Therefore, from the above observation we conclude that, the disintegration energy Q for the beta decay of 32P would be 1.71 MeV.
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