yes yes that's my question.never thought u were misunderstanding what i wanted ::)my poor English
There are several possible ways for radioactive nuclides to decay, most notably alpha decay, beta decay, electron capture and spontaneous fission: Alpha decay involved the emission of a stripped 4He nucleus from the atom. Beta decay involves the emission of an electron or a positron (along with neutrinos) from the nucleus. During B+ decay a proton is converted into a neutron and the reverse is true for B- decay. Electron capture is the process by which an orbital electron is absorbed in the nucleus, destroys a nuclear proton and creates a neutron. Spontaneous fission involves the splitting of a nucleus into two or more smaller nuclei.
In order to simplify the 238
U problem, let’s ignore the spontaneous fission decay mode and focus only on why 238
U decays via alpha emission instead of beta decay or electron capture.
The Q value of a reaction determines whether or not a reaction is thermodynamically stable or not. In the case of nuclear reactions, the Q value is simply the mass of the reactants minus the mass of the products (the reverse of most of chemistry). A positive Q value implies that the reaction is exothermic and that over time the reaction will tend towards the products. The reverse is true for a negative Q value. In this case, the reaction is endothermic and over time will favor the reactants.
The values below are the Q values for each reaction:
Alpha decay: 4.27 MeV
B- decay: -0.147
B+ decay: -4.483
Electron Capture: -3.461
As you can see, the only decay mode that will release energy is alpha decay. As the other modes all require energy, 238
U will not decay by them.
On a fascinating side note, the Table of Radioactive Isotopes lists 238
U as having a bb decay, which I assume means double beta decay.