InterviewSolution
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InterviewSolution
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A radioactive nucleus X decays to a nucleus Y with a decay constant `lambda_X=0.1s^-1`, Y further decays to a stable nucleus Z with a decay constant `lambda_Y=1//30s^-1`. Initially, there are only X nuclei and their number is `N _0=10^20`. Set up the rate equations for the populations of X, Y and Z. The population of Y nucleus as a function of time is given by `N_Y(t)={N _0lambda_X//(lambda_X-lambda_Y)}[exp(-lambda_Yt)-exp(-lambda_Xt)]`. Find the time at which `N_Y` is maximum and determine the population X and Z at that instant. |
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Answer» Correct Answer - A::D (a) Let at time `t=t`, number of nuclei of Y and Z are `N_Y` and `N_Z`. Then, Rate equations of the populations of X, Y and Z are `((dN_X)/(dt))=-lambda_XN_X` …(i) `((dN_Y)/(dt))=lambda_XN_X-lambda_YN_Y`…(ii) and `((dN_Z)/(dt))=lambda_YN_Y`...(iii) (b) Given, `N_Y(t)=(N_0lambda_X)/(lambda_X-lambda_Y)[e^(-lambda_Yt)-e^(-lambda_Xt)]` For `N_Y` to be maximum `(dN_Y(t))/(dt)=0` i.e `lambda_XN_X=lambda_YN_Y` ...(iv) [from Eq. (ii)] or `lambda_X(N_0e^(-lambda_Xt))=lambda_Y(N_0lambda_X)/(lambda_X-lambda_Y)[e^(-lambda_Yt)-e^(-lambda_Xt)]` or `(lambda_X-lambda_Y)/(lambda_Y)=(e^(-lambda_Yt))/(e^(-lambda_Xt))-1` `lambda_X/lambda_Y=e^((lambda_X-lambda_Y)t)` or `(lambda_X-lambda_Y)t1n(e)=1n(lambda_X/lambda_Y)` or `t=(1)/(lambda_X-lambda_Y)1n(lambda_X/lambda_Y)` Substituting the values of `lambda_X` and `lambda_Y`, we have `t=(1)/((0.1-1//30))1n((0.1)/(1//30))=151n(3)` or `t=16.48s` (c) The population of X at this moment, `N_X=N_0e^(-lambda_Xt)=(10^20)e^(-(0.1)(16.48))` `N_X=1.92xx10^19` `N_Y=(N_Xlambda_X)/(lambda_Y)` [From Eq. (iv)] `=(1.92 xx 10^19)((0.1)/(1//30))` `=5.76xx10^19` `N_Z=N_0-N_X-N_Y` `=10^20-1.92xx10^19-5.76xx10^19` `N_Z=2.32xx10^19` |
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