Question

# The radio nuclide 11C decays according to116C→115B+e++ν: T1/2=20.3 minThe maximum energy of the emitted positron is 0.960 MeV. Given the mass values:m(116C)=11.011434u and m(116B)=11.009305uCalculate Q and compare it with the maximum energy of the positron emitted.

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#### 116C→116B+e++ν+QQ=[mN(116C)−mN(116B)−me]c2Where, the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add 6me in case of 11C and 5me in case of 11B. Hence,Q=[m(116C)−m(116B)−2me]c2 (Note me has been doubled)Using given masses, Q = 0.961 MeV.Q = Ed+Ee+EνThe daughter nucleus is too heavy compared to e+ and ν, so it carries negligible energy (Ed ≈ 0). If the kinetic energy (Eν) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q, hence, maximum Ee ≈ Q.

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