AP EAMCET · PHYSICS · Nuclear Physics
The half-life of a radioactive element is \(10 \mathrm{~h}\). The fraction of initial radioactivity of the element that will remain after \(40 \mathrm{~h}\) is
- A \(\frac{1}{2}\)
- B \(\frac{1}{16}\)
- C \(\frac{1}{8}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{16}\)
Step-by-step Solution
Detailed explanation
After 4 half-lifes \(\frac{N_0}{2^4}=\frac{N_0}{16}\)
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