ExamBro
ExamBro
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 \([\ln (5)=1.61]\)

  1. A 693
  2. B 138.6
  3. C 277.2
  4. D 322
Verified Solution

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,…
Same subject
Explore more questions on app