Question:
A radioactive element of mass \(1\ \text{kg}\) after \(N\) years is left with only \(125\ \text{g}\). If the half-life of the element is \(12.5\) years, then the value of \(N\) is
A radioactive element of mass \(1\ \text{kg}\) after \(N\) years is left with only \(125\ \text{g}\). If the half-life of the element is \(12.5\) years, then the value of \(N\) is
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After \(n\) half-lives, the remaining mass is
\[
m=m_0\left(\frac{1}{2}\right)^n.
\]
If the remaining fraction is \(\frac{1}{8}\), then
\[
\frac{1}{8}=\left(\frac{1}{2}\right)^3,
\]
so \(3\) half-lives have passed.
Updated On: Jun 26, 2026
- \(37.5\ \text{years}\)
- \(25.0\ \text{years}\)
- \(50.0\ \text{years}\)
- \(75.0\ \text{years}\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the initial and final masses.
Initial mass is \[ m_0=1\ \text{kg}=1000\ \text{g}. \] Final mass is \[ m=125\ \text{g}. \]
Step 2: Find the fraction remaining.
\[ \frac{m}{m_0} = \frac{125}{1000}. \] \[ = \frac{1}{8}. \] Now, \[ \frac{1}{8}=\left(\frac{1}{2}\right)^3. \] So, the substance has passed through \[ 3 \] half-lives.
Step 3: Use the half-life value.
Given half-life is \[ T_{1/2}=12.5\ \text{years}. \] Therefore, total time is \[ N=3T_{1/2}. \] \[ N=3\times12.5. \] \[ N=37.5\ \text{years}. \]
Step 4: Final conclusion.
Hence, the value of \(N\) is \[ \boxed{37.5\ \text{years}} \] Therefore, the correct option is \[ \boxed{(1)} \]
Initial mass is \[ m_0=1\ \text{kg}=1000\ \text{g}. \] Final mass is \[ m=125\ \text{g}. \]
Step 2: Find the fraction remaining.
\[ \frac{m}{m_0} = \frac{125}{1000}. \] \[ = \frac{1}{8}. \] Now, \[ \frac{1}{8}=\left(\frac{1}{2}\right)^3. \] So, the substance has passed through \[ 3 \] half-lives.
Step 3: Use the half-life value.
Given half-life is \[ T_{1/2}=12.5\ \text{years}. \] Therefore, total time is \[ N=3T_{1/2}. \] \[ N=3\times12.5. \] \[ N=37.5\ \text{years}. \]
Step 4: Final conclusion.
Hence, the value of \(N\) is \[ \boxed{37.5\ \text{years}} \] Therefore, the correct option is \[ \boxed{(1)} \]
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