GATE-2018 (24) :Wyllie – Time average equation

GATE_2019

85) The abundance of ²³⁴ U in secular equilibrium with its parent ²³⁸U will be _____________* 10^-3 %( given that half-life of the ²³⁸U and ²³⁴U are 4.467* 10⁹ y and 2.44 * 10⁵ y respectively and the abundance of ²³⁸U is 99.28%).

(Thanks to Chandra Sekar, ANU)

Solution:

The Radioactive decay equilibrium can be defined as the 

 

$ \frac{N_{1}}{N_{2}}= \frac{\lambda_{2} - \lambda_{1}}{\lambda_{1}}$--------(1)

Where 

Abundance of ²³⁸U is given $(N_{1})$=99.28% and

Half life are given for  ²³⁸U and ²³⁴U using these values we can calculate the decay constants.

$\lambda_{1}$ = $\frac{0.693}{t_{1 \frac{1}{2}}}$

half-life of ²³⁸U =4.467* 10⁹ y

   =$\frac{0.693}{4.467*10^9}$ 

Decay constant $\lambda_{1}= 0.1551*10^{-9}$ 

$\lambda_{2}$ = $\frac{0.693}{t_{2 \frac{1}{2}}}$

half-life of ²³⁴U=2.44 * 10⁵ y

    =$\frac{0.693}{2.44*10^5}$

  $= 0.2840*10^{-5}$  


Decay constant $\lambda_{2}  = 2840 * 10^{-9}$ 


Substitute the  above values in eq (1)

  

 $ \frac{N_{1}}{N_{2}} = \frac{\lambda_{2} - \lambda_{1}}{\lambda_{1}}$


 $ \frac{N_{1}}{N_{2}} = \frac{2840 * 10^{-9} -0.1551*10^{-9}}{0.1551*10^{-9}}$


 $ \frac{N_{1}}{N_{2}}= \frac{2839.8449*10^{-9}}{0.1551*10^{-9}}$

 
$ \frac{N_{1}}{N_{2}}= 18309.7672$


 $ N_{2}=\frac{N_{1}}{18309.7672}$


$ N_{2} = \frac{99.28}{18309.7672}$

$N_{2} = 0.005422$ 

  $ N_{2} = 5.42*10^{-3}$%

Reference:http://oregonstate.edu/instruct/ch374/ch418518/Chapter%203.pdf