Write nuclear reaction equation for α-decay of Pu94242Pu. - Physics

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Short Note

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

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Solution

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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Chapter 13: Nuclei - Exercise [Page 462]

APPEARS IN

NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 13.6 (ii) | Page 462
NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 6.2 | Page 462

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