Saved Bookmarks
| 1. |
The radionuclide ""^(11)C decays according to ""_(6)^(11)C to ""_(5)^(11)B + e^(+) + upsilon, T_(1//2) = 20.3 min The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m(""_6^11C) = 11.011434 u and m(""_6^11B) = 11.009305u. Calculate Q and compare it with the maximum energy of the positron emitted. |
|
Answer» Solution :The disintegration ENERGY, `Q = (m_C - m_B - 2m_c)c^2 = (m_C - m_B -2m_c) (931.5 c^2)/(c^2) MEV` `2m_c = 2 xx 0.0005486 = 0.0010972` `:. Q = {:(11.011434-),(11.0104022):}/(0.0010318 xx 931.5)""{:(11.009305 + ),(.0010972):}/(11.0104022)` SINCE the maximum energy of positron emitted is 0.960 MeV, the neutrino carries only NEGLIGIBLE energy. |
|