A) 6.38 h
B) 12 h
C) 4.18 day
D) 1.2day
Correct Answer: A
Solution :
The activity or decay rate R of a radioactive substance is the number of decays per second. \[\therefore \] \[R=\lambda N\] or \[R=\lambda {{N}_{0}}{{\left( \frac{1}{2} \right)}^{t/{{T}_{1/2}}}}\] or \[R={{R}_{0}}{{\left( \frac{1}{2} \right)}^{t/{{T}_{1/2}}}}\] where\[{{R}_{0}}=\lambda {{N}_{0}}\]is the activity of the radioactive substance at time\[t=0\]. According to question \[\frac{R}{{{R}_{0}}}=1-\frac{75}{100}=25%\] \[\therefore \] \[\frac{25}{100}={{\left( \frac{1}{2} \right)}^{t/{{T}_{1/2}}}}\] or \[{{\left( \frac{1}{2} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{t/{{T}_{1/2}}}}\] or \[\frac{t}{{{T}_{1/2}}}=2\] \[\therefore \] \[t=2{{T}_{1/2}}=2\times 3.20=6.40\,\,h\] or \[t\approx 6.38\,\,h\]
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