The radionuclide ${}^{11}C$ decays according to ${}_{6}^{11}C \to {}_{5}^{11}B + e^{+} + v; T_{1 / 2} = 20.3 \min .$
The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:
$m ({}_{6}^{11}C) = 11.011434 u and m ({}_{5}^{11}B) = 11.009305 u .$
Calculate Q and compare it with the maximum energy of the positron emitted.

Solution

For the given reaction, mass defect is,
$∆ m = [m ({}_{6}^{11}C - 6 m_{e}) - [m ({}_{5}^{11}B) - 5 m_{e} + m_{e}] = m ({}_{6}^{11}C) - m ({}_{5}^{11}B) - 2 m_{e} = 11.011434 u - 11.009305 u - 2 \times 0.000548 u = 0.001033 u$

Now, Q-value is ,

$Q = 0.001033 \times 931.5 MeV = 0.962 MeV$

which, is the maximum energy of the positron.
We have,
$Q = E_{d} + E_{e} + E_{v}$
The daughter nucleus is too heavy compared to e⁺ and v. So, it carries negligible energy (E_d ≈ 0). If the kinetic energy (E_v) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q.
Hence, maximum E_e≈ Q.

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