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प्रश्न
Write nuclear reaction equation for α-decay of `""_94^242"Pu"`.
उत्तर
α is a nucleus of helium `(""_2^4"He")` and β is an electron (e− for β− and e+ for β+). In every α-decay, there is a loss of 2 protons and 4 neutrons. In every β+-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β−-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.
For the given case, the various nuclear reaction can be written as:
`""_94^242"Pu" -> ""_92^238"U" + ""_2^4"He"`
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