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InterviewSolution
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The half life of radioactive Radon is `3.8 days` . The time at the end of which `(1)/(20) th` of the radon sample will remain undecayed is `(given log e = 0.4343 ) `A. `3.8 days`B. `16.5 days`C. `33 days`D. `76 days` |
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Answer» Correct Answer - b `T_(1//2)=3.8 day` `:. lambda =(0.693)/(t_(1//2))=(0.693)/(3.8)=0.182` If the initial number of atoms is a =`A_(0)`, then after time t the number of atoms is `a//20`=A. We have to find `t`. ` t=(2.303)/(lambda) log.(A_(0))/(A)=(2.303)/(0.182) log.(a)/(a//20)=(2.303)/(0.182) log.20` `=16.46 day`. |
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