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NCERT Class XII Chapter
Nuclei
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Question : 13 of 31
Marks: +1, -0
The radionuclide 611C\mathrm{^{11}_{6}C} decays according to
611C\mathrm{^{11}_{6}C} → 511B+e+\mathrm{^{11}_{5}B}+e^{+} + v ; T1/2T_{1/2} = 20.3 min
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
m (611C)(\mathrm{^{11}_{6}C}) = 11.011434 u and m (511B)(\mathrm{^{11}_{5}B}) = 11.009305 u
Calculate Q and compare it with the maximum energy of the positron emitted
Solution:  
The given equation,
611C\mathrm{^{11}_{6}C} → 511B+e+\mathrm{^{11}_{5}B}+e^{+} + v + Q
The Q value is, Q =
[m(611C)−6me−m(511B)+5me−me]c2\left[ m(\mathrm{^{11}_{6}C}) - 6m_e - m(\mathrm{^{11}_{5}B}) + 5m_e - m_e \right] c^2
Q = [11.011434 – 11.009305 – 2 × 0.000548] c2c^2
931.5 MeV ≈ 0.96 MeV
As we know that different positrons comes out with different possible energies shared between daughter nucleus and positron.
So, the Q value of reaction is almost same as the maximum energy of positron emitted.
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