13.13 The radionuclide C611 decays according to CB+e++v:511611T1/2 =20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( C)611 = 11.011434 u and m ( B511 ) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.

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7 months ago

13.13 The given values are

m ( C)611 = 11.011434 u and m ( B511 ) = 11.009305 u

The given nuclear reaction:

CB+e++v511611

Half life of C611 nuclei, T1/2 =20.3 min

The maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the C611

ΔQ = m'(C)611-{m'B511+me}c2(1)

where

me = Mass of an electron or positron = 0

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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