Question:
A radioactive element \(A\) converts into another stable element \(B\). Half-life of \(A\) is \(1.5\,\text{hrs}\). After time \(t\), the ratio of atoms of \(A\) and \(B\) is found to be \(1:8\), then \(t\) in hours is
A radioactive element \(A\) converts into another stable element \(B\). Half-life of \(A\) is \(1.5\,\text{hrs}\). After time \(t\), the ratio of atoms of \(A\) and \(B\) is found to be \(1:8\), then \(t\) in hours is
Show Hint
If a radioactive element \(A\) decays into stable \(B\), then
\[
B=N_0-N.
\]
Use the given ratio \(A:B\) to first find the remaining fraction of \(A\), then compare it with powers of \(\frac12\).
Updated On: Jun 18, 2026
- \(6\)
- \(8\)
- Between \(3\) to \(4.5\)
- Between \(4.5\) to \(6\)
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The Correct Option is D
Solution and Explanation
Step 1: Let the initial number of atoms of \(A\) be \(N_0\).
After time \(t\), let the number of atoms of \(A\) remaining be \[ N \] Since \(A\) converts into stable \(B\), the number of atoms of \(B\) formed is \[ N_0-N \]
Step 2: Use the given ratio.
Given, \[ A:B=1:8 \] Therefore, \[ \frac{N}{N_0-N}=\frac{1}{8} \] \[ 8N=N_0-N \] \[ 9N=N_0 \] \[ N=\frac{N_0}{9} \]
Step 3: Use radioactive decay formula.
For radioactive decay, \[ N=N_0\left(\frac12\right)^{t/T_{1/2}} \] Given, \[ T_{1/2}=1.5\,\text{hrs} \] So, \[ \frac{N_0}{9}=N_0\left(\frac12\right)^{t/1.5} \] Cancelling \(N_0\), \[ \frac{1}{9}=\left(\frac12\right)^{t/1.5} \]
Step 4: Locate the time interval.
After \(3\) half-lives, \[ N=\frac{N_0}{2^3}=\frac{N_0}{8} \] After \(4\) half-lives, \[ N=\frac{N_0}{2^4}=\frac{N_0}{16} \] But here, \[ N=\frac{N_0}{9} \] Since \[ \frac{1}{16}<\frac{1}{9}<\frac{1}{8}, \] the required time is between \(3\) and \(4\) half-lives.
Step 5: Convert half-lives into hours.
One half-life is \[ 1.5\,\text{hrs} \] Therefore, \(3\) half-lives correspond to \[ 3\times 1.5=4.5\,\text{hrs} \] and \(4\) half-lives correspond to \[ 4\times 1.5=6\,\text{hrs} \] Thus, \[ t \] lies between \[ 4.5\,\text{hrs} \] and \[ 6\,\text{hrs}. \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\text{Between }4.5\text{ to }6\text{ hours}} \]
After time \(t\), let the number of atoms of \(A\) remaining be \[ N \] Since \(A\) converts into stable \(B\), the number of atoms of \(B\) formed is \[ N_0-N \]
Step 2: Use the given ratio.
Given, \[ A:B=1:8 \] Therefore, \[ \frac{N}{N_0-N}=\frac{1}{8} \] \[ 8N=N_0-N \] \[ 9N=N_0 \] \[ N=\frac{N_0}{9} \]
Step 3: Use radioactive decay formula.
For radioactive decay, \[ N=N_0\left(\frac12\right)^{t/T_{1/2}} \] Given, \[ T_{1/2}=1.5\,\text{hrs} \] So, \[ \frac{N_0}{9}=N_0\left(\frac12\right)^{t/1.5} \] Cancelling \(N_0\), \[ \frac{1}{9}=\left(\frac12\right)^{t/1.5} \]
Step 4: Locate the time interval.
After \(3\) half-lives, \[ N=\frac{N_0}{2^3}=\frac{N_0}{8} \] After \(4\) half-lives, \[ N=\frac{N_0}{2^4}=\frac{N_0}{16} \] But here, \[ N=\frac{N_0}{9} \] Since \[ \frac{1}{16}<\frac{1}{9}<\frac{1}{8}, \] the required time is between \(3\) and \(4\) half-lives.
Step 5: Convert half-lives into hours.
One half-life is \[ 1.5\,\text{hrs} \] Therefore, \(3\) half-lives correspond to \[ 3\times 1.5=4.5\,\text{hrs} \] and \(4\) half-lives correspond to \[ 4\times 1.5=6\,\text{hrs} \] Thus, \[ t \] lies between \[ 4.5\,\text{hrs} \] and \[ 6\,\text{hrs}. \]
Step 6: Final conclusion.
Therefore, \[ \boxed{\text{Between }4.5\text{ to }6\text{ hours}} \]
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