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

Short Note

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

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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APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.6 (ii) | Page 462
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 6.2 | Page 462
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