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` u and rest mass of `B-11=11.009304` u, calculate 'Q' and compare it with the maximum energy of the positron emitted.

Explanation

The rest mass of the products:
`{:(" "e^(+)"is 0.96 MeV"),(ul(" v is zero ")),(" r.m. of products "=10250. 66 MeV):}`
`Deltam =(10251.64-10250.66) MeV=0.9850 MeV`
Fraction of energy of positron with respect to the total energy `=(0.960 MeV)/(0.985 MeV)=0.975`
position carries `97.5%` of the total energy released.