JEE Advanced · Physics · 28. Nuclear Physics
The activity of a freshly prepared radioactive sample is \(10^{10}\) disintegrations per second, whose mean life is \(10^{-9} \mathrm{~s}\). The mass of an atom of this radioisotope is \(10^{-25} \mathrm{~kg}\). The mass (in \(\mathrm{mg}\) ) of the radioactive sample is
- A 5
- B 8
- C 9
- D 1
Answer & Solution
Correct Answer
(D) 1
Step-by-step Solution
Detailed explanation
Activity \(\left(-\frac{d N}{d t}\right)=\lambda N=\left(\frac{1}{t_{\text {mean }}}\right) \times N\)
\(
\begin{aligned}
& \therefore \quad N=\left(-\frac{d N}{d t}\right) \times t_{\text {mean }} \\
& \quad=\text { Total number of atoms } \\
& \text { Mass of one atom is } 10^{-25} \mathrm{~kg}=m \text { (say) } \\
& \therefore \quad \text { Total mass of radioactive substance } \\
& =(\text { number of atoms) } \\
& \quad \times(\text { mass of one atom) } \\
& =\left(-\frac{d N}{d t}\right)\left(t_{\text {mean }}\right)(m)
\end{aligned}
\)
Substituting the values, we get
Total mass of radioactive substance
\(
=1 \mathrm{mg}
\)
\(\therefore\) Answer is 1 .
Analysis of Question
Question is very simple. Again in my opinion one can solve it.
\(
\begin{aligned}
& \therefore \quad N=\left(-\frac{d N}{d t}\right) \times t_{\text {mean }} \\
& \quad=\text { Total number of atoms } \\
& \text { Mass of one atom is } 10^{-25} \mathrm{~kg}=m \text { (say) } \\
& \therefore \quad \text { Total mass of radioactive substance } \\
& =(\text { number of atoms) } \\
& \quad \times(\text { mass of one atom) } \\
& =\left(-\frac{d N}{d t}\right)\left(t_{\text {mean }}\right)(m)
\end{aligned}
\)
Substituting the values, we get
Total mass of radioactive substance
\(
=1 \mathrm{mg}
\)
\(\therefore\) Answer is 1 .
Analysis of Question
Question is very simple. Again in my opinion one can solve it.
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