The decay constant for a radioactive nuclide is \(1.5 × 10^{−5}s^{−1}\). Atomic weight of the substance is 60 g mole−1. (\(N_A = 6×10^{23}\)). The activity of 1.0 µg of the substance is _____\(×10^{10}\) Bq.
The decay constant for a radioactive nuclide is \(1.5 × 10^{−5}s^{−1}\). Atomic weight of the substance is 60 g mole−1. (\(N_A = 6×10^{23}\)). The activity of 1.0 µg of the substance is _____\(×10^{10}\) Bq.
Show Hint
Activity (A) is the number of decays per unit time. It’s calculated as the product of the decay constant (λ) and the number of radioactive atoms (N). Remember to convert units consistently
Correct Answer: 15
Solution and Explanation
Step 1: Calculate the Number of Moles
The number of moles is given by:
\[ \text{No. of moles} = \frac{\text{Mass of sample}}{\text{Molar mass}} \]
Substitute the given values:
\[ \text{No. of moles} = \frac{1 \times 10^{-6}}{60} = \frac{10^{-7}}{6} \, \text{moles} \]
Step 2: Calculate the Number of Atoms
Using Avogadro’s number (\( N_A = 6 \times 10^{23} \)):
\[ \text{No. of atoms} = n \cdot N_A = \frac{10^{-7}}{6} \cdot 6 \times 10^{23} \]
Simplify:
\[ \text{No. of atoms} = 10^{16} \]
Step 3: Calculate the Initial Activity (\( A_0 \))
The activity at \( t = 0 \) is given by:
\[ A_0 = N_0 \lambda \]
Substitute \( N_0 = 10^{16} \) and \( \lambda = 1.5 \times 10^{-5} \):
\[ A_0 = 10^{16} \cdot 1.5 \times 10^{-5} \]
Simplify:
\[ A_0 = 15 \times 10^{10} \, \text{Bq} \]
Final Answer:
The initial activity is \( A_0 = 15 \times 10^{10} \, \text{Bq}. \)
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