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.

Answer

A, D

Explanation

(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`