A nucleus undergoes β- decay and becomes
, Calculate the maximum kinetic energy of electrons emitted assuming that the daughter nucleus and anti- neutrino carry negligible kinetic energy.
Asked by Topperlearning User | 9th Jul, 2014, 12:48: PM
β- decay of is given as:
Where,
Q = kinetic energy of the daughter nucleus
Ignoring the rest mass of the anti-neutrino and the mass of the electron
,
the mass defect involved in the nuclear reaction is given as:
Δm = m -m
= 22.994466 - 22.989770
= 0.004696 u = 0.004696 x 931.5 MeV/c2 = 4.374 MeV/c2
Q = Δ mc2 = 4.374 MeV
Hence, when the energy carried by is zero, the maximum kinetic energy of β particle is 4.374 MeV.
Answered by | 9th Jul, 2014, 02:48: PM
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