A) 1000 yr
B) 2000 yr
C) 3000 yr
D) 4000 yr
Correct Answer: D
Solution :
\[{{N}_{0}}=\frac{3}{5}{{N}_{0}}{{e}^{-\lambda t}}\] \[e{{\lambda }^{t}}=\frac{5}{3}\] \[\therefore \] \[\log {{e}^{\lambda t}}={{\log }_{e}}\frac{5}{3}\] or \[\lambda t={{\log }_{e}}\frac{5}{3}\] or \[t=\frac{1}{\lambda }{{\log }_{e}}\frac{5}{3}\] \[=\frac{T}{0.693}\times 0.5\] \[\left( \because \,\,\,T=\frac{0.693}{\lambda } \right)\] \[=\frac{5570\times 0.5}{0.693}yr=4018.7yr\] \[=4000yr\]
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