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{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}
Thanku
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