A) \[1.8\times {{10}^{9}}\] year
B) \[2.3\times {{10}^{10}}\]year
C) \[5.1\times {{10}^{7}}\] year
D) \[6.2\times {{10}^{6}}\]year
Correct Answer: A
Solution :
[a] Suppose x is the number of \[P{{b}^{206}}\]nulei. The number of \[{{U}^{238}}\] nuclei will be 3x, Thus \[3x+x={{N}_{0}}\] We know that \[N={{N}_{0}}{{e}^{-\lambda t}}\] or \[3x=4x{{e}^{-\lambda t}}\] \[\therefore \,\,\,\,\,\,\,\,{{e}^{\lambda t}}=\frac{4}{3}\] or \[t=\frac{In\,4/3}{\lambda }=\frac{\operatorname{Im}\,\,4/3}{(0.693/{{t}_{1/2}})}\] \[=\frac{In\,\,4/3}{(0.693/4.5\times {{10}^{9}})}=1.868\times {{10}^{9}}\] years.
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