A) 0.124
B) 0.062
C) 0.093
D) 0.031
Correct Answer: D
Solution :
In time \[t=T,\] \[N=\frac{{{N}_{0}}}{2}\] In another half-life, (i.e., after 2 half-lives) \[N=\frac{1}{2}\frac{{{N}_{0}}}{2}=\frac{{{N}_{0}}}{4}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{2}}\] After yet another half-life (i.e., after 3 half-lives) \[N=\frac{1}{2}\left( \frac{{{N}_{0}}}{4} \right)=\frac{{{N}_{0}}}{8}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{3}}\] and so on. Hence, after n half-lives \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] \[={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/T}}\] where \[t=n\times T=\]total time of n half-lives. \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] \[={{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/T}}\] Here, \[n=\frac{t}{T}=\frac{19}{3.8}\] \[=5\] \[\therefore \] The fraction left \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}={{\left( \frac{1}{2} \right)}^{5}}=\frac{1}{32}\] \[=0.031\]
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