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Obtain the maximum kinetic energy of beta- particle and the radiation frequencies corresponding to gamma- decays in the figure given below. (rest mass of au -198.968233 u and r.m. of Hg=198=197.966760 u) |
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Answer» Solution :Energies of `gamma`- PHOTONS `gamma_(1), gamma_(2), gamma_(3)` are `E_(1)=(1.088-0) MeV=1.088 MeV=1.088xx1.6xx10^(-13)=1.7408xx10^(-13)J` `E_(2)=(0.412-0) MeV=0.412 MeV=0.412 MeV=0.412xx1.6xx10^(-13)=0.6592xx10^(-13)J` `E_(3)=(1.088-0.452) MeV=0.676 MeV=0.672xx1.6xx10^(-13)=1.0816xx10^(-13)J` but frequency of RADIATION `v=(E)/(h)`  `therefore""v_(1)=(1.7408xx10^(-13))/(6.625xx10^(-34))=0.2628xx10^(21) Hz=2.628xx10^(24)Hz` `v_(2)=(0.6592xx10^(-13))/(6.625xx10^(-34))=0.0955xx10^(21)Hz=9.55xx10^(19)Hz` `v_(3)=(1.0816xx10^(-13))/(6.625xx10^(-34))=0.1633xx10^(21)Hz=1.633xx^(20)Hz` `beta_(1)^(-)` decay, `_(79)^(198) Au rarr _(80)^(198)Hg+_(-1)^(0)e+Q_(2)( bar beta_(1))+Q_(1)(gamma_(1))` The maximum KINETIC energy of `beta _(1)^(-)=[197.968233-197.966760]xx931.5 MeV=0.284 MeV` `beta_(2)^(-)` decay, `_(79)^(198) Au rarr _(80)^(198)Hg+_(-1)^(0)e+Q_(2)( bar beta_(2))+Q_(2)(gamma_(2))` The maximum kinetic energy of `beta _(2)^(-)=[197.96822233-197.966760]xx931.5 MeV-0.412 MeV=0.96 MeV` |
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