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.

Explanation

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.