Question:
Half-life period of a radioactive substance is \(10\,\text{min}\). Then amount of substance decayed in \(40\,\text{min}\) will be:
Half-life period of a radioactive substance is \(10\,\text{min}\). Then amount of substance decayed in \(40\,\text{min}\) will be:
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After \(n\) half-lives: remaining \(=\left(\frac{1}{2}\right)^n\), decayed \(=1-\left(\frac{1}{2}\right)^n\).
Updated On: Apr 16, 2026
- \(25\%\)
- \(50\%\)
- \(75\%\)
- None of these
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The Correct Option is D
Solution and Explanation
Concept: After \( n \) half-lives,
\[
\text{Remaining fraction} = \left(\frac{1}{2}\right)^n
\]
Step 1: \[ n = \frac{40}{10} = 4 \]
Step 2: \[ \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]
Step 3: \[ 1 - \frac{1}{16} = \frac{15}{16} = 93.75\% \]
Step 1: \[ n = \frac{40}{10} = 4 \]
Step 2: \[ \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]
Step 3: \[ 1 - \frac{1}{16} = \frac{15}{16} = 93.75\% \]
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