AP EAMCET · PHYSICS · Nuclear Physics
A radioactive substance of half life 138.6 days is placed in a box. After \(n\) days only \(20 \%\) of the substance is present then the value of \(n\) is \([\ln (5)=1.61]\)
- A 693
- B 138.6
- C 277.2
- D 322
Answer & Solution
Correct Answer
(D) 322
Step-by-step Solution
Detailed explanation
Half life of a radioactive substance, \(n_{1 / 2}=138.6\) days If \(N_0\) be the initial amount of radioactive substance, then remaining amount after \(n\) days is given by \(N=20 \% \text { of } N_0=\frac{20}{100} \times N_0=\frac{N_0}{5}\) By radioative decay's law,…
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