A) 20 days
B) 15 days
C) 10 days
D) 5 days
Correct Answer: C
Solution :
Key Idea \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\] where n = number of half-life period \[{{N}_{0}}=\] initial concentration N = concentration left after n half-life period. \[T=n{{\times }_{1/2}}\] where, T = total time (-1/2 = half-life period Here, \[N=\frac{1}{32},\,\,\,T=50\] days \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\] \[\frac{1}{32}={{\left( \frac{1}{2} \right)}^{n}}\] \[{{\left( \frac{1}{2} \right)}^{5}}={{\left( \frac{1}{2} \right)}^{n}}\] \[n=5\] \[\because \] \[T=n\times {{t}_{1/2}}\] \[\therefore \] \[{{t}_{1/2}}=\frac{T}{n}=\frac{50}{5}=10\] day
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