Question

The half-life of $^{198}\text{Au}$ is 3 days. If atomic weight of $^{198}\text{Au}$ is 198 g/mol then the activity of 2 mg of $^{198}\text{Au}$ is [in disintegration/second] :

• A. $16.18 \times 10^{12}$
• B. $2.67 \times 10^{12}$
• C. $6.06 \times 10^{18}$
• D. $32.36 \times 10^{12}$

Hint:

Always convert the half-life into seconds when activity is asked in Bq (disintegrations per second).
Use the relation \(A = \frac{0.693 \cdot m \cdot N_A}{T_{1/2} \cdot M}\) to save time in such calculations.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Activity (\(A\)) of a radioactive sample is the number of disintegrations per unit time.
It is given by the relation \(A = \lambda N\), where \(\lambda\) is the decay constant and \(N\) is the number of radioactive nuclei present in the sample.
Step 2: Key Formula or Approach:
1. Decay constant: \[\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{T_{1/2}}\]
2. Number of nuclei: \[N = \frac{m}{M} \times N_A\]
3. Activity: \[A = \lambda \times N\]
Step 3: Detailed Explanation:
Given:
Half-life \(T_{1/2} = 3 \text{ days} = 3 \times 24 \times 3600 \text{ seconds} = 259,200 \text{ s}\).
Mass \(m = 2 \text{ mg} = 2 \times 10^{-3} \text{ g}\).
Atomic weight \(M = 198 \text{ g/mol}\).
Avogadro's number \(N_A = 6.022 \times 10^{23} \text{ mol}^{-1}\).

First, calculate the number of nuclei \(N\):
\[N = \frac{2 \times 10^{-3}}{198} \times 6.022 \times 10^{23} \approx 6.0828 \times 10^{18} \text{ nuclei}\]

Next, calculate the decay constant \(\lambda\):
\[\lambda = \frac{0.6931}{259200} \approx 2.674 \times 10^{-6} \text{ s}^{-1}\]

Now, calculate the activity \(A\):
\[A = (2.674 \times 10^{-6}) \times (6.0828 \times 10^{18})\]
\[A \approx 16.26 \times 10^{12} \text{ disintegrations/second}\]
Looking at the options, \(16.18 \times 10^{12}\) is the closest standard value.
Step 4: Final Answer:
The activity of the gold sample is approximately \(16.18 \times 10^{12}\) disintegrations per second.