A) 12.5
B) 8
C) 0
D) 25
Correct Answer: D
Solution :
Key Idea \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\] where n is the number of half-life period \[{{n}_{0}}\] is the initial concentration N is the concentration left after n half-life period. \[T=n\times {{t}_{1/2}}\] where, T = total time \[{{t}_{1/2}}=\]half- life period Here, \[{{t}_{1/2}}=10\] yr T = 20 yr \[n=\frac{T}{{{t}_{1/2}}}=\frac{20}{10}=2\] \[\therefore \] \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{4}\] \[N=\frac{{{N}_{0}}}{4}=\frac{1}{4}=0.25\] or 25% N = 25%
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