A) \[2\times {{10}^{9}}years\]
B) \[4\times {{10}^{9}}years\,\]
C) \[6\times {{10}^{9}}years\]
D) \[8\times {{10}^{9}}years\]
Correct Answer: D
Solution :
[d] \[{{N}_{x}}=\frac{{{N}_{0}}}{2}{{e}^{-{{\lambda }_{1}}t}}=0.2{{N}_{0}}\,\,\,{{N}_{y}}=\frac{{{N}_{0}}}{2}{{e}^{-{{\lambda }_{2}}t}}=0.8{{N}_{0}}\] \[{{e}^{({{\lambda }_{1}}-{{\lambda }_{2}})}}=4\Rightarrow t=8\times {{10}^{9}}\] years
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