A) \[1.96\times {{10}^{9}}\,\operatorname{yr}\]
B) \[3.92\times {{10}^{9}}\,\operatorname{yr}\]
C) \[4.2\times {{10}^{9}}\,\operatorname{yr}\]
D) \[8.2\times {{10}^{9}}\,\operatorname{yr}\]
Correct Answer: C
Solution :
\[\frac{X}{Y}=\frac{1}{7}\] In such question always add numerator and denominator to get initial number of atoms. \[\operatorname{X}+Y=8={{N}_{o}}\,\,\,before\,\,declay\]. \[\frac{N}{{{N}_{o}}}={{\left( \frac{1}{2} \right)}^{n}}\] \[\frac{1}{8}={{\left( \frac{1}{2} \right)}^{n}}\] \[{{\left( \frac{1}{2} \right)}^{3}}={{\left( \frac{1}{2} \right)}^{n}}\] \[\operatorname{n}=3\] \[\operatorname{n}=\frac{T}{{{t}_{1/2}}}\Rightarrow T=n\times {{t}_{1/2}}\] \[\operatorname{T}=3\times 3\times 1.4\times 1{{0}^{9}}\] \[\operatorname{T}=4.2\times 1{{0}^{9}}\,year\]
You need to login to perform this action.
You will be redirected in
3 sec
