Question

Write nuclear reaction equation for α-decay of `""_94^242"Pu"`.

Answer 1

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"`