A parent nucleus X decays into a daughter nuc | vedclass

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(A) The decay process is given by $X \xrightarrow{T_{1/2, X} = 40000 \, yr} Y \xrightarrow{T_{1/2, Y} = 20 \, yr} Z$.Since the number of $Y$ nuclei re

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$2 	imes 10^{17}$

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(A) The decay process is given by $X \xrightarrow{T_{1/2, X} = 40000 \, yr} Y \xrightarrow{T_{1/2, Y} = 20 \, yr} Z$.Since the number of $Y$ nuclei remains constant over time,the rate of production of $Y$ must equal the rate of decay of $Y$.This condition is known as secular equilibrium.Therefore,$\lambda_X N_X = \lambda_Y N_Y$.Using the relation $\lambda = \frac{\ln 2}{T_{1/2}}$,we get $\frac{\ln 2}{T_X} N_X = \frac{\ln 2}{T_Y} N_Y$.This simplifies to $\frac{N_X}{T_X} = \frac{N_Y}{T_Y}$.Rearranging for $N_Y$,we get $N_Y = N_X 	imes \frac{T_Y}{T_X}$.Substituting the given values: $N_Y = (4 	imes 10^{20}) 	imes \frac{20}{40000}$.$N_Y = (4 	imes 10^{20}) 	imes \frac{1}{2000} = 2 	imes 10^{17}$ nuclei.

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