Question
The radionuclide 11C decays according to `""_6^11C -> _5^11B + e^+ + v:"` `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) = 11.009305 u`
calculate Q and compare it with the maximum energy of the positron emitted
Solution
The given nuclear reaction is:
`""_6^11C -> _5^11B + e^+ v`
Half life of `_6^11C` nuclei `T_(1/2)` = 20.3 min
Atomic mass of `m(""_6^11C)`= 11.011434 u
Atomic mass of `m(""_6^11B) = 11.009305 u`
Maximum energy possessed by the emitted positron = 0.960 MeV
The change in the Q-value (ΔQ) of the nuclear masses of the `""_6^11C` nucleus is given as:
`triangleQ = [m'(""_6C^11) - [m'(""_5^11B) + m_e]]c^2` ...(1)
Where,
me = Mass of an electron or positron = 0.000548 u
c = Speed of light
m’ = Respective nuclear masses
If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case of `""^11C` and 5 me in the case of `""^11B`.
Hence, equation (1) reduces to:
`triangleQ = [m(""_6C^11) - m(""_5^11B) - 2m]c^2` are the atomic masses
Here `m(""_6C^11) and m(""_5^11B)` are the atomic masses
∴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 is almost comparable to the maximum energy of the emitted positron.
