What is the activity of a 20 ng sample of `""^(92)Kr`, which has a half life of 1.84 s?

Answer

To find the activity of a 20 ng sample of \( ^{92}\text{Kr} \) with a half-life of 1.84 seconds, we can follow these steps:

### Step 1: Understand the formula for activity
The activity \( R \) of a radioactive sample can be calculated using the formula:
\[ R = \lambda \cdot N \]
where \( \lambda \) is the decay constant and \( N \) is the number of radioactive nuclei present in the sample.

### Step 2: Calculate the decay constant \( \lambda \)
The decay constant \( \lambda \) can be calculated using the half-life \( t_{1/2} \) with the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Given that the half-life \( t_{1/2} = 1.84 \) seconds, we can substitute this value into the equation:
\[ \lambda = \frac{\ln(2)}{1.84} \]

### Step 3: Calculate the number of nuclei \( N \)
To find the number of nuclei \( N \), we can use the formula:
\[ N = \frac{m}{M} \cdot N_A \]
where:
- \( m \) is the mass of the sample (20 ng = \( 20 \times 10^{-9} \) g),
- \( M \) is the molar mass of \( ^{92}\text{Kr} \) (which is 92 g/mol),
- \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \) nuclei/mol).

Substituting the values:
\[ N = \frac{20 \times 10^{-9}}{92} \cdot 6.022 \times 10^{23} \]

### Step 4: Substitute \( \lambda \) and \( N \) into the activity formula
Now we can substitute the values of \( \lambda \) and \( N \) into the activity formula:
\[ R = \lambda \cdot N \]

### Step 5: Calculate the final activity \( R \)
After calculating \( \lambda \) and \( N \), we can find the activity \( R \).

### Detailed Calculations
1. Calculate \( \lambda \):
\[ \lambda = \frac{\ln(2)}{1.84} \approx \frac{0.693}{1.84} \approx 0.376 \, \text{s}^{-1} \]

2. Calculate \( N \):
\[ N = \frac{20 \times 10^{-9}}{92} \cdot 6.022 \times 10^{23} \approx \frac{20 \times 10^{-9} \cdot 6.022 \times 10^{23}}{92} \approx 1.31 \times 10^{13} \, \text{nuclei} \]

3. Calculate \( R \):
\[ R = 0.376 \cdot 1.31 \times 10^{13} \approx 4.93 \times 10^{12} \, \text{Bq} \]

### Final Answer
The activity of the 20 ng sample of \( ^{92}\text{Kr} \) is approximately \( 4.93 \times 10^{12} \, \text{Bq} \).

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