A) 11460 years
B) 17190 years
C) 22920 years
D) 45840 years
Correct Answer: C
Solution :
After n half-lives (i.e., at \[t=nT\]) the number of nuclides left undecayed, \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] \[Given\,\,=\,\,\frac{N}{{{N}_{0}}}=\frac{1}{16}\] \[\therefore \,\,\,\,\,\,\,\,\,\frac{1}{16}={{\left( \frac{1}{2} \right)}^{n}}\] \[{{\left( \frac{1}{2} \right)}^{4}}={{\left( \frac{1}{2} \right)}^{n}}\] Equating the powers, we obtain \[n=4\] That is, \[\frac{t}{T}=4\] or \[t=4T\] or \[t= 4 \times 5730 = 22920 years\] \[\left( \because ~\,\,\,\,T= 5730 years \right)\]You need to login to perform this action.
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