A) 500 yr
B) 600 yr
C) 800 yr
D) 950 yr
Correct Answer: A
Solution :
Key Idea\[\left( \frac{N}{{{N}_{0}}} \right)={{\left( \frac{1}{2} \right)}^{n}}\] where, n is the number of half-life periods \[{{N}_{0}}\]is the initial concentration (or amount) N is amount of substance left after n half-life periods and,\[T=n\times {{t}_{1/2}}\] where, \[{{t}_{1/2}}=\]half-life period T = total time Given, that \[{{N}_{0}}=0.2\,g,\,N=0.025\,\,g,\,\,T=1500\,\,yr\] By using, \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\] \[\frac{0.025}{0.2}={{\left( \frac{1}{2} \right)}^{n}}\] \[\frac{1}{8}={{\left( \frac{1}{2} \right)}^{n}}\] \[{{\left( \frac{1}{2} \right)}^{3}}={{\left( \frac{1}{2} \right)}^{n}}\] n = 3 Now by using, \[{{t}_{1/2}}=\frac{T}{n}=\frac{1500}{3}=500\,yr\] Therefore, the half-life of the object is 500 yr.
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