Question:
A radioactive substance has a half life of \(10^8\) years and an activity of \(10^4\ \text{Bq}\). The number of atoms of this substance present is:
A radioactive substance has a half life of \(10^8\) years and an activity of \(10^4\ \text{Bq}\). The number of atoms of this substance present is:
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The activity of a radioactive sample is directly proportional to the number of undecayed nuclei:
\[
A=\lambda N
\]
Always convert half-life into SI units before calculation.
Updated On: Jun 26, 2026
- \(9.1\times10^{19}\)
- \(6.7\times10^8\)
- \(4.5\times10^{19}\)
- \(5\times10^{20}\)
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The Correct Option is C
Solution and Explanation
Step 1: Use the relation between activity and number of atoms.
Activity is given by \[ A=\lambda N \] where \[ A=\text{activity} \] \[ \lambda=\text{decay constant} \] \[ N=\text{number of atoms} \] Thus, \[ N=\frac{A}{\lambda} \]
Step 2: Calculate the decay constant.
Decay constant is related to half-life by \[ \lambda=\frac{0.693}{T_{1/2}} \] Given, \[ T_{1/2}=10^8\ \text{years} \] Convert years into seconds: \[ 1\ \text{year}=365\times24\times3600 \] \[ 1\ \text{year}=3.15\times10^7\ \text{s} \] Therefore, \[ T_{1/2}=10^8\times3.15\times10^7 \] \[ T_{1/2}=3.15\times10^{15}\ \text{s} \] Hence, \[ \lambda=\frac{0.693}{3.15\times10^{15}} \] \[ \lambda\approx2.2\times10^{-16}\ \text{s}^{-1} \]
Step 3: Calculate the number of atoms.
Given activity: \[ A=10^4\ \text{Bq} \] Using \[ N=\frac{A}{\lambda} \] \[ N=\frac{10^4}{2.2\times10^{-16}} \] \[ N\approx4.5\times10^{19} \]
Step 4: Final conclusion.
Therefore, the number of atoms present is \[ \boxed{4.5\times10^{19}} \]
Activity is given by \[ A=\lambda N \] where \[ A=\text{activity} \] \[ \lambda=\text{decay constant} \] \[ N=\text{number of atoms} \] Thus, \[ N=\frac{A}{\lambda} \]
Step 2: Calculate the decay constant.
Decay constant is related to half-life by \[ \lambda=\frac{0.693}{T_{1/2}} \] Given, \[ T_{1/2}=10^8\ \text{years} \] Convert years into seconds: \[ 1\ \text{year}=365\times24\times3600 \] \[ 1\ \text{year}=3.15\times10^7\ \text{s} \] Therefore, \[ T_{1/2}=10^8\times3.15\times10^7 \] \[ T_{1/2}=3.15\times10^{15}\ \text{s} \] Hence, \[ \lambda=\frac{0.693}{3.15\times10^{15}} \] \[ \lambda\approx2.2\times10^{-16}\ \text{s}^{-1} \]
Step 3: Calculate the number of atoms.
Given activity: \[ A=10^4\ \text{Bq} \] Using \[ N=\frac{A}{\lambda} \] \[ N=\frac{10^4}{2.2\times10^{-16}} \] \[ N\approx4.5\times10^{19} \]
Step 4: Final conclusion.
Therefore, the number of atoms present is \[ \boxed{4.5\times10^{19}} \]
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