In the beta decay of Na24, the combined electron neutrino momentum has a magnitude equal to 4 Mevc. What is the recoil energy of the daughter nucleus, given that its mass is 23.99u?
Take [c=3×108m/s,1u=1.66×10−27kg]
A
2.78×10−4 MeV
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B
3.56×10−4 MeV
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C
0.72×10−4 MeV
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D
1.25×10−4 MeV
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Solution
The correct option is B3.56×10−4 MeV Here, momentum of the electron and neutrino pair is p=4MeVc=4×1.6×10−133×108 p=2.13×10−21kgms
Also mass of the daughter nucleus is m=23.99u=23.99×1.66×10−27kg
As parent nucleus decays at rest, so according to conservation of linear momentum, the momentum of daughter nucleus and momentum of electron neutrino pair must be same
If P′⇒ momentum of daughter nucleus then, P′=p
Now recoil energy Er is Er = P′22m=p22m ⇒(2.13×10−21)22×23.99×1.66×10−27J ⇒2.13×2.13×10−152×23.99×1.66××1.66×10−13MeV