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GATE_2019_85

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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

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