The radionuclide `._6C^(11)` decays according to `._6C^(11)to ._5B^(11) +e^(+) +v`: half life =20.3min. The maximum energy of the emitted positron is `0.960 MeV`. Given the mass values
`m(._6C^(11))=11.011434u, m(._6B^(11))=11.009305u`
Calculate Q and compare it with maximum energy of positron emitted.

Explanation

The given nuclear reaction is:
`""_(6)^(11)Cto""_(5)^(11)B+e^(+)+v`
Half life of `""_(6)^(11)C` nuclei, `T_(1//2)` = 20.3 min
Atomic mass of `m(""_(6)^(11)C)` = 11.011434 u
Atomic mass of `m(""_(6)^(11)B)` = 11.009305 u
Maximum energy possessed by the emitted position = 0.960 MeV
The change in the Q-value `(DeltaQ)` of the nuclear masses of the `""_(6)^(11)C` nucleus is given as:
`DeltaQ=[m'(""_(6)C^(11))-[m'(""_(5)^(11)B)+m_(c)]]c^(2)" "...(1)`
Where,
`m_(e)` = Mass of an election or position = 0.000548 u
c = Speed of light
m' = Respective nuclear masses
If atomic masses are used instead of nuclea masses, then we have to add 6 `m_(e)` in the case of `""^(11)C` and 5 `m_(e)` in the case of `""^(11)B`. ltbr. Hence, equation (1) reduces to:
`DeltaQ=[m(""_(6)C^(11))-m(""_(5)^(11)B)-2m_(e)]c^(2)`
Here, `m(""_(6)^(11)C)and m(""_(5)^(11)B)` are the atomic masses.
`thereforeDeltaQ=[11.011434-11.009305-2xx0.000548]c^(2)`
`=(0.001033c^(2))` u
But 1 u = 931.5 `MeV//c^(2)`
`thereforeDeltaQ=0.001033xx931.5~~0.962` MeV
The value of Q is almost comparable to the maximum energy of the emitted position.