JEE Advanced 2013Paper 1
- Exam
- JEE Advanced
- Year
- 2013
- Paper
- Paper 1
- Subject
- Physics
- Chapter
- Nuclear Physics
JEE Advanced 2013 · Paper 1 · Physics · Nuclear Physics
A freshly prepared sample of a radioisotope of half-life 1386 s has...
A freshly prepared sample of a radioisotope of half-life 1386 s has activity disintegrations per second. Given that In 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is
Correct answer
0
Step-by-step solution
\text{f} = \left(1 - {\text{e}}^{- λ \text{t}}\right) = 1 - {\text{e}}^{- λ \text{t}} ≈ 1 - \left(1 - λ \text{t}\right) = λ \text{t}
f = 0.04
Hence \text{%} decay ≈ 4 %
f = 0.04
Hence \text{%} decay ≈ 4 %