A sample of radon emitted initially `7 xx 10^(4) alpha`-particles per second. After some time, the emission rate became `2.1 xx 10^(4)`. If `t_((1)/(2))` for radon is 3.8 days, find the age of the sample

Answer

To find the age of the radon sample, we can follow these steps:

### Step 1: Understand the given data
We have:
- Initial emission rate of alpha particles, \( N_0 = 7 \times 10^4 \) particles/second
- Final emission rate of alpha particles, \( N = 2.1 \times 10^4 \) particles/second
- Half-life of radon, \( t_{1/2} = 3.8 \) days

### Step 2: Calculate the decay constant (\( \lambda \))
The decay constant (\( \lambda \)) can be calculated using the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
Substituting the half-life:
\[ \lambda = \frac{0.693}{3.8} \approx 0.182 \text{ days}^{-1} \]

### Step 3: Use the decay formula
The relationship between the initial and final quantities in radioactive decay is given by:
\[ N = N_0 e^{-\lambda t} \]
We can rearrange this to solve for time (\( t \)):
\[ t = -\frac{1}{\lambda} \ln\left(\frac{N}{N_0}\right) \]

### Step 4: Substitute the values into the equation
First, we need to calculate the ratio \( \frac{N}{N_0} \):
\[ \frac{N}{N_0} = \frac{2.1 \times 10^4}{7 \times 10^4} = \frac{2.1}{7} \approx 0.3 \]

Now, substituting into the time formula:
\[ t = -\frac{1}{0.182} \ln(0.3) \]

### Step 5: Calculate \( \ln(0.3) \)
Using a calculator:
\[ \ln(0.3) \approx -1.20397 \]

### Step 6: Calculate time (\( t \))
Substituting \( \ln(0.3) \) back into the equation:
\[ t = -\frac{1}{0.182} \times (-1.20397) \approx \frac{1.20397}{0.182} \approx 6.61 \text{ days} \]

### Step 7: Final calculation
To find the age of the sample, we can use the relationship:
\[ t = \frac{2.303}{\lambda} \log\left(\frac{N_0}{N}\right) \]
Substituting the values:
\[ t = \frac{2.303}{0.182} \log\left(\frac{7}{2.1}\right) \]
Calculating \( \log\left(\frac{7}{2.1}\right) \):
\[ \frac{7}{2.1} \approx 3.33 \quad \text{and} \quad \log(3.33) \approx 0.51 \]
Now substituting:
\[ t \approx \frac{2.303}{0.182} \times 0.51 \approx 6.61 \text{ days} \]

### Conclusion
The age of the radon sample is approximately **6.61 days**.