The half life of radioactive Radon is `3.8 days` . The time at the end of which `(1)/(20) th` of the radon sample will remain undecayed is `(given log e = 0.4343 ) `

Options

A. (a) 3.8 days
B. (b) 16.5 days
C. (c) 33 days
D. (d) 76 days

Explanation

To solve the problem, we need to find the time at which \( \frac{1}{20} \) of the Radon sample remains undecayed, given that the half-life of Radon is \( 3.8 \) days. We will use the decay constant and the exponential decay formula.

### Step-by-Step Solution:

1. **Identify the half-life and decay constant**:
The half-life \( T_{1/2} \) of Radon is given as \( 3.8 \) days. The decay constant \( \lambda \) can be calculated using the formula:
\[ \lambda = \frac{\ln 2}{T_{1/2}} \]
Substituting the values:
\[ \lambda = \frac{0.693}{3.8} \]

2. **Calculate the decay constant**:
Performing the calculation:
\[ \lambda \approx 0.1824 \text{ days}^{-1} \]

3. **Use the exponential decay formula**:
The number of undecayed nuclei at time \( t \) can be expressed as:
\[ N = N_0 e^{-\lambda t} \]
We want to find \( t \) when \( N = \frac{N_0}{20} \):
\[ \frac{N_0}{20} = N_0 e^{-\lambda t} \]

4. **Simplify the equation**:
Dividing both sides by \( N_0 \):
\[ \frac{1}{20} = e^{-\lambda t} \]

5. **Take the natural logarithm of both sides**:
\[ \ln \left(\frac{1}{20}\right) = -\lambda t \]

6. **Express \( \ln \left(\frac{1}{20}\right) \)**:
Using the property of logarithms:
\[ \ln \left(\frac{1}{20}\right) = -\ln(20) \]
Therefore, we have:
\[ -\ln(20) = -\lambda t \]

7. **Rearranging for \( t \)**:
\[ t = \frac{\ln(20)}{\lambda} \]

8. **Substituting \( \lambda \)**:
Now substituting the value of \( \lambda \):
\[ t = \frac{\ln(20)}{0.1824} \]

9. **Calculate \( \ln(20) \)**:
Using the given \( \ln(20) \approx 2.9957 \):
\[ t \approx \frac{2.9957}{0.1824} \approx 16.4 \text{ days} \]

10. **Final answer**:
The time at which \( \frac{1}{20} \) of the Radon sample remains undecayed is approximately \( 16.4 \) days.