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Obtain the maximum kinetic energy of `beta`- particles, and the radiation frequencies of `gamma` decays in the decay scheme shown in Fig. You are given that `m(.^(198)Au)=197.968233 u` `m(.^(198)Hg)=197.966760 u`. |
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Answer» Energy corresponding to `r_(1)` `E_(1)=1.088-0=1.088 MeV` `= 1.088xx1.6xx10^(-13)` Joule Frequency `v_(1)=(E_(1))/(h)` `= (1.088xx1.6xx10^(-13))/(6.6xx10^(-34))` `= 2.63xx10^(20)Hz` similary `v_(2)=(E_(2))/(h)` `= (0.412xx1.6xx10^(-12))/(6.6xx10^(10))` `= 9.98xx10^(13)Hz` and `v_(3)=(E_(3))/(h)` `= ((1.088-0.412)xx1.6xx10^(-13))/(6.6xx10^(20)Hz)` Maximum K.E. of `beta_(1)` particle `K_("max")(beta_(1))=[m(._(79)Au^(198))-` mass of second excited state of `._(80)Hg^(198)]xx931 MeV` `= [m(._(79)Au^(198))-m(._(82)Hg^(198))-(1.088)/(931)]xx931 MeV` `= 931[197.968233-197.966760]-1.088 MeV` `= 1.371-1.088=0.283 MeV` similarly `k_("max")(beta_(2))=0.957 MeV` |
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