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A ""^(118)Cd radionuclide goes through the transformation chain. ""^(118)Cd{:(rarr),(30min):}""^(118)In{:(rarr),(45min):}""^(118)Sn (stable) The half lives are written below the respective arrows. At time t= 0 only Cd was present . Find the fraction of nuclei transformed into stable 5^(th) over 60 minutes. |
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Answer» Solution :`N_(1)=N_(0)e^(-lamda_(1)t)and N_(2)=(N_(0)lamda_1)/(lamda_(2)-lamda_1)(e^(-lamda_(1)t)-e^(-lamda_(2)))` `:.N_(3)=N_(0)-N_(1)-N_2` `N_(0)[1-e^(-lamda_(1)t)-lamda_(1)/(lamda_(2)-lamda_1)(e^(-lamda_(1)t)-e^(-lamda_(2)t))]` `:.N_3/N_0=1-e^(lamda_(1)^(t))-lamda_1/(lamda_(2)-lamda_(1))(e^(-lamda_(1)t)-e^(-lamda_(2)t))` `lamda=(0.693)/30=0.0231"min"^(-1)` `lamda_(2)=(0.693)/45 = 0.0154 "min"^(-1) and t = 60` minutes . On substituting the GIVEN values `N_3/N_0 =0.31 `. |
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