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Just saw this question in a school class 10 exam. Google search did not yield useful results. Can anyone please explain the answer here?

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  • $\begingroup$ Hi Gaurang. I've suggested a duplicate based on what I think you're asking. If the proposed duplicate doesn't answer your question perhaps you could edit your question to explain exactly what it is that you are asking. $\endgroup$ Commented Nov 27, 2015 at 11:03
  • $\begingroup$ @JohnRennie My query has been satisfied. Thank you! $\endgroup$ Commented Nov 27, 2015 at 17:16

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It depends what you mean by controlled.

Without going into the details of each decay process you mention ($\alpha, \beta, \gamma$), the decay of an unstable nucleus is inherently random. At a quantum mechanical level, we can think of the system changing from one eigen-state to another and that is fundamentally unpredictable (this is the bit that Einstein didn't like when he said something like, "God does not play dice").

Having said that, you can accelerate certain types of nuclear decay. In a nuclear reactor (or bomb), the decay of one nucleus is stimulated when it absorbs a neutron that is emitted from another decay. This is the famous chain-reaction that you often hear about.

Another way to do it is with a particle beam - a neutron beam into Thorium forces it to decay and produce heat.

So to answer your question, you can force certain types of nuclear reactions to proceed by bombarding the nuclei with neutrons (or maybe other particles). However, you cannot restrain a particle from decaying.

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  • $\begingroup$ I think the OP asked about the already emitted particles. I think the answer is certainly yes for alpha and beta due do their charge, via electric and magnetic fields. I am guessing this is not what the OP meant, however. $\endgroup$
    – Declan
    Commented Nov 27, 2015 at 12:56
  • $\begingroup$ @Declan I see what you mean... It's so vague that, on re-reading, he could mean anything. I think he's probably wondering if you can build an alpha-ray gun - something like that... $\endgroup$ Commented Nov 27, 2015 at 13:10

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