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Neon-23 decays in the following way, ....

Neon-23 decays in the following way,
`._10^23Nerarr_11^23Na+ _(-1)^0e+barv`
Find the minimum and maximum kinetic energy that the beta particle`(._-1^0e)`can have. The atomic masses of `.^23Ne` and `.^23 Na` are `22.9945u` and `22.9898u`, respectively.

Text Solution

Verified by Experts

Here, atomic masses are given (not the nuclear masses), but still we can use them for calculating the mass defect because mass of electrons get cancelled both sides.
Thus, Mass defect
`Deltam`=(22.9945-22.9898)=0.0047 u
`therefore` Q=(0.0047u)(931.5MeV/u)=4.4 MeV
Hence, the energy of beta particles can range from 0 to 4.4 MeV.
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