A) \[0.783MeV\]
B) \[0.840MeV\]
C) \[0.589MeV\]
D) \[0.687MeV\]
Correct Answer: A
Solution :
| Electron's energy is maximum if no neutrino is emitted. Then, |
| \[{{K}_{\max }}=\left( {{m}_{n}}-{{m}_{p}}-{{m}_{e}} \right).{{c}^{2}}\] |
| \[=\{{{m}_{n}}-\left( {{m}_{p}}+{{m}_{e}} \right)\}.{{c}^{2}}\] |
| As, \[{{m}_{p}}+{{m}_{e}}=mass\,of\,H-atom\]. |
| We have, |
| \[{{K}_{\max }}=\left( {{m}_{n}}-{{m}_{H}} \right).{{c}^{2}}=\left( 840\mu u \right).{{c}^{2}}\] |
| \[=840\times {{10}^{-6}}u\times 932\frac{MeV}{u}=.0.783MeV\] |
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