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The radio nuclide .^(11)C decays according to ._(6)^(11)C rarr _(5)^(11) B+e^(+)+v. The maximum energy of the emitted positron is 0.960 MeV. Given, the rest mass of c-11=11.011434 uand rest mass of B-11=11.009304 u, calculate 'Q' and compare it with the maximum energy of the positron emitted. |
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Answer» Solution :`""_(6)^(11) C to ""_(5)^(11)B + e^(+) + v + Q` `Q = [m_N (""_(6)^(11)C) - m ""_(5)^(11)B - m_e] c^2` where the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add `6m_e` in case of `""_(6)^(11)C` and `5m_e` in case of `_""_(5)^(11)B`. HENCE `Q = [m (""_(6)^(11)C) - m ""_(5)^(11)B - 2m ] c^2 ` (Note `m_e` has been doubled) Using given masses, `Q = 0.961 MeV` `Q = E_d+ E_e+ E_n` The DAUGHTER nucleus is too heavy compared to `e^+` and v, so it CARRIES negligible ENERGY `(Ed ~~ 0)`. If the kinetic energy `(E_v)` carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q, hence maximum `E_e ~~ Q`). |
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