A) \[\frac{1}{\sqrt{2}}\]
B) \[\left( 1-\frac{1}{\sqrt{2}} \right){{N}_{0}}\]
C) \[20%\]
D) \[\frac{1}{8}\]
Correct Answer: B
Solution :
\[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\]and\[n=\frac{T}{{{t}_{1/2}}}\] So, \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{T/{{t}_{1}}/2}}\] where,\[{{N}_{0}}=\]initial amount of radioactive substance \[N=\] amount of radioactive substance left after time\[T\]. \[T=\]time\[=2\,\,days\] \[{{t}_{1/2}}=\]half-life\[=4\,\,days\] \[\therefore \] \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{2/4}}\] \[\therefore \] \[\frac{N}{{{N}_{0}}}=\frac{1}{\sqrt{2}}\] \[\therefore \] \[N={{N}_{0}}\times \frac{1}{\sqrt{2}}\] \[\therefore \]Amount of substance decayed in two days will be \[{{N}_{0}}-{{N}_{0}}\times \frac{1}{\sqrt{2}}=\left( 1-\frac{1}{\sqrt{2}} \right){{N}_{0}}\]
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