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$\alpha$ radiation consist of positive charged helium nuclei, $\beta$ radiation of negative charged electrons. So why don't the $\alpha$ particles take those electrons to get neutral?

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you mean $\beta^-$ decay, there is also $\beta^+$ decay consisting of positrons –  Tobias Kienzler Mar 24 '11 at 13:34
    
Yes, I mean $\beta^-$ decay. –  martin Mar 24 '11 at 14:46
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I agree with jwenting but in some sense, I feel that he is not answering the question: why there's no "combined $\alpha$ plus $\beta$ decay in which a nucleus emits e.g. a helium atom?

Well, let me start with the $\beta$-decay. Nuclei randomly - after some typical time, but unpredictably - may emit an electron because a neutron inside the nuclei may decay via $$ n\to p+e^- +\bar\nu$$ which may be reduced to a more microscopic decay of a down-quark, $$ d\to u + e^- +\bar\nu.$$ This interaction, mediated by a virtual W-boson, is why a nucleus - with neutrons - may sometimes randomly emit an electron. So the $\beta$-decay is due to the weak nuclear force. On the other hand, the $\alpha$-decay is due to the strong nuclear force: the nucleus literally breaks into pieces, with a very stable combination of 2 protons and 2 neutrons appearing as one of the pieces (helium nucleus).

The two processes above are independent, and each of them can kind of be reduced to a single elementary interaction whose origin is different. This independence and different origin is why the "combined" decay, with an emission of both electron (or two electrons) and a Helium nucleus, is extremely unlikely. Such an emission of a whole atom (which is electrically neutral but it is surely not "nothing"!) could only occur if several of the elementary decay interactions would occur at almost the same time which is extremely unlikely.

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when they meet, it might happen. But calculate the chances of 2 particles meeting that have the right energy levels to actually combine. You'll find those chances are miniscule.

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Do you have a reference for this calculation? –  martin Mar 24 '11 at 14:48
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