A) 106 yr
B) 1060 yr
C) 1.06 yr
D) 10.6 yr
Correct Answer: A
Solution :
Key Idea \[\frac{N}{{{N}_{o}}}={{\left( \frac{1}{2} \right)}^{n}}\] where n = number of half-life periods \[{{N}_{0}}=\] initial concentration of substance N = amount of concentration left after n half-life period. Total time \[(T)=n\times {{t}_{1/2}}\] Amount disintegrated \[=99.9%=\frac{999}{1000}\] Amount left \[=1-\frac{999}{1000}\] \[=\frac{1}{1000}\approx \frac{1}{1024}\] \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\] \[\frac{1}{1024}={{\left( \frac{1}{2} \right)}^{n}}\] \[{{\left( \frac{1}{2} \right)}^{10}}={{\left( \frac{1}{2} \right)}^{n}}\] n = 10 Total time \[=n{{\times }_{1/2}}\] Here, \[{{t}_{1/2}}=10.6\,\,yr\] Total time \[=10\times 10.6\] = 106 yr
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