Question -
Answer -
The given nuclear reactionis:
Atomic mass of 
= 11.011434 u
Atomicmass of 
Maximum energy possessed by theemitted positron = 0.960 MeV
Thechange in the Q-value (ΔQ) of the nuclear masses ofthe 
nucleus is given as:Where,
me =Mass of an electron or positron = 0.000548 u
c = Speedof light
m’ =Respective nuclear masses
If atomic masses are used insteadof nuclear masses, then we have to add 6 me in thecase of
and 5 me inthe case of 
.
Hence, equation (1) reduces to:
∴ΔQ =[11.011434 − 11.009305 − 2 × 0.000548] c2
= (0.001033 c2)u
But 1 u = 931.5 Mev/c2
∴ΔQ =0.001033 × 931.5 ≈ 0.962 MeV
The value of Q isalmost comparable to the maximum energy of the emitted positron.