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`(._-1^0
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.
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here ,atomic masses are given (not the nuclear masses ),But still we can electrons get cancelled both sides thus ,
Mass defect , `Delta m-m(""_(10)Ne^(23))-M(""_(11)Na^(23))`
`=(22.995-22.9898)=0.0047u`
Kinetic energy is given by `Q=Delta m.c^(2)`
`therefore Q-(0.0047 u )(931.5 Me V//u)=4.4 MeV `
hence , the energy of beta particles can range form O to 4.4 MeV .
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