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प्रश्न
Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:
12N → 12C* + e+ + v
12C* → 12C + γ (4.43MeV).
The atomic mass of 12N is 12.018613 u.
(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)
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उत्तर
Given:-
Atomic mass of 12N, m(12N) = 12.018613 u
12N → 12C* + e+ + v
12C* → 12C + γ (4.43 MeV)
Net reaction is given by
12N → 12C + e+ + v + γ (4.43 MeV)
Qvalue of the `β^+` decay will be
Qvalue = [m(`""^12N`) - (m(12C*) + 2me)]c2
`= [12.018613 xx 931 "MeV" - (12 xx 931 + 4.43) "MeV" - (2 xx 511) "keV"]`
= [11189.3287 - 11176.43 - 1.022] MeV
`= 11.8767 "MeV" = 11.88 "MeV"`
The maximum kinetic energy of beta particle will be 11.88 MeV, assuming that neutrinos have zero energy.
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