A) 330 years
B) 460 years
C) 660 years
D) 920 years
Correct Answer: C
Solution :
[c] \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] so, for 1% decay \[\frac{1}{100}{{N}_{0}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\Rightarrow \frac{1}{100}={{\left( \frac{1}{2} \right)}^{n}}\] \[\Rightarrow \,\,\,\,\,\,\,\,n=\frac{2}{\log 2}\Rightarrow \frac{t}{T}=\frac{2}{\log 2}\Rightarrow t=6.6T\] \[=6.6\times 100=660\] years If \[\alpha \] and B are emitted simultaneously.
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