A) \[2\times {{10}^{8}}\]
B) \[4\times {{10}^{8}}\]
C) \[1\times {{10}^{8}}\]
D) \[7\times {{10}^{8}}\]
Correct Answer: D
Solution :
: For radioactive decay, \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{t/T}}\] where\[T=\]half life \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/T}}=(8\times {{10}^{8}})\times {{\left( \frac{1}{2} \right)}^{60/20}}=\frac{8\times {{10}^{8}}\times 1}{8}\] \[=1\times {{10}^{8}}nuclei\]. \[\therefore \] Number of nuclei left undecayed\[=1\times {{10}^{8}}\] \[\therefore \] Number of nuclei that have decayed \[=(8-1)\times {{10}^{8}}=7\times {{10}^{8}}\].
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