InterviewSolution
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InterviewSolution
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A `Bi^(210)` radionuclide decays via the chain `Bi^(210)(beta^(-)-"decay")/(lambda_(1))Po^(210)(alpha-decay)/(lambda_(2)) Pb^(206)` (stable), where the decay constants are `lambda_(1) = 1.6 xx 10^(-6)s^(-1), T_(1//2) ~~ 5 days , lambda_(2) = = 5.8 xx 10^(-8)s^(-1), T_(1//2) ~~ 4.6` months. `alpha & beta` activites of the `1.00 mg` a month after its manufacture is `(x)/(5) xx 10^(11)` Find `x.2^((1)/(4.6)) = 0.86` |
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Answer» Correct Answer - `7` ` Bi^(250)underset(lambda_(1))rarrPo^(210)underset(lambda_(2))rarrPb^(208)` Let `N_(A), N_(B)` and `N_(C )` be the no of nuclei of `Bi^(210), Po^(210)` and `Pb^(208)` respectively. `N_(0)` be the no of nuclei of `Bi^(210)` at `t=0` `N_(0)=(1xx10^(-3))/(210)xx6.02xx10^(23) " "~~2.9xx10^(18)` `(dN_(A))/(dt)= -lambda_(1)N_(A)"_____"`I `(dN_(B))/(dt)=-lambda_(2)N_(B)+lambda_(1)N_(A)"______"`(II) `(dN_(c ))/(dt)= lambda_(2)N_(B)"_____"`III Solving equ. I , we get `N_(A)= "No" e^(-)lambda_(1)t` From equation II, `(dN_(B))/(dt)=-lambda_(2)N_(B)+lambda_(1)N_(o)e^(-lambda_(1)t)` This is first order line differential equation. On solving we get `N_(B)=e^(-lambda_(2)t)[(lambda_(1)N_(0))/(lambda_(2)-lambda_(1))e^((lambda_(2)-lambda_(2))t)+C]` where `C` is a constant The value of `C` can be obtained by taking `t-0, N_(B)=0` `rArrC=(lambda_(1)N_(0))/(lambda_(1)-lambda_(2))` `:. N_(B)=(lambda_(1)N_(o))/(lambda_(1)-lambda_(2))[e^(-lambda_(2)t)-e^(-lambda_(1)t)] :. beta`-activity `A_(B)=|(dN_(A))/(dt)|=N_(0)lambda_(1)e^(-lambda_(1)t)=2.9xx10^(18)xx1.6xx10^(-6)xx(1)/(2^(6))S^(-1)=0.72xx10^(11)S^(-1)` `alpha`-activity `A_(alpha)=lambda_(2)N_(B)=(lambda_(1)lambda_(2)N_(0))/(lambda_(1)-lambda_(2))(e^(-lambda_(2)t)-e^(-lambda_(1)t))~~(lambda_(1)lambda_(2)N_(0))/(lambda_(1))e^(-lambda_(2)t)=(lambda_(2)N_(0))/((1)/(2^(4xx6)))=1.4xx10^(11)s^(-1)` |
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