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. 3.8 days
B. 16.5 days
C. 33 days
D. 76 days

Answer

To solve the problem, we need to determine the time at which \( \frac{1}{20} \) of the original radon sample remains undecayed. We will use the half-life and the decay constant in our calculations.

### 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:
\[ \lambda = \frac{0.693}{3.8 \text{ days}} \approx 0.1824 \text{ days}^{-1} \]

**Hint:** Remember that \( \ln(2) \) is approximately 0.693.

2. **Set Up the Decay Equation:**
The amount of undecayed sample \( N_t \) after time \( t \) can be expressed as:
\[ N_t = N_0 e^{-\lambda t} \]
We want to find \( t \) when \( N_t = \frac{N_0}{20} \).

3. **Substitute into the Decay Equation:**
Setting up the equation:
\[ \frac{N_0}{20} = N_0 e^{-\lambda t} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{20} = e^{-\lambda t} \]

**Hint:** Dividing by \( N_0 \) simplifies the equation.

4. **Take the Natural Logarithm of Both Sides:**
Taking the natural logarithm:
\[ \ln\left(\frac{1}{20}\right) = -\lambda t \]
This can be rewritten as:
\[ -\ln(20) = -\lambda t \quad \Rightarrow \quad \ln(20) = \lambda t \]

5. **Calculate \( \ln(20) \):**
We can express \( \ln(20) \) using logarithmic properties:
\[ \ln(20) = \ln(2 \times 10) = \ln(2) + \ln(10) \]
Using \( \ln(2) \approx 0.693 \) and \( \ln(10) \approx 2.303 \):
\[ \ln(20) \approx 0.693 + 2.303 = 2.996 \]

**Hint:** Use known values for \( \ln(2) \) and \( \ln(10) \) to find \( \ln(20) \).

6. **Solve for Time \( t \):**
Now substituting back into the equation:
\[ t = \frac{\ln(20)}{\lambda} \]
Substituting \( \lambda \):
\[ t = \frac{2.996}{0.1824} \approx 16.43 \text{ days} \]

**Hint:** Ensure you perform the division carefully to avoid calculation errors.

7. **Final Answer:**
Rounding to one decimal place, the time at which \( \frac{1}{20} \) of the radon sample remains undecayed is approximately:
\[ t \approx 16.4 \text{ days} \]

Therefore, the nearest answer is **16.5 days**.