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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 | 09 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 | 09 Jul, 2014, 02:48: PM
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