The half life of a radioactive substance is `20` minutes . The approximate time interval `(t_(2) - t_(1))` between the time `t_(2)` when `(2)/(3)` of it had decayed and time `t_(1)` when `(1)/(3)` of it had decay is

Options

A. 14 min
B. 20 min
C. 28 min
D. 7 min

Explanation

To solve the problem, we need to determine the time interval \( t_2 - t_1 \) between the times when \( \frac{2}{3} \) of the radioactive substance has decayed and when \( \frac{1}{3} \) has decayed.

### Step-by-step Solution:

1. **Understanding the Decay**:
- Let the initial amount of the substance be \( N_0 \).
- When \( \frac{1}{3} \) has decayed, the remaining amount is:
\[ N_1 = N_0 - \frac{1}{3}N_0 = \frac{2}{3}N_0 \]
- When \( \frac{2}{3} \) has decayed, the remaining amount is:
\[ N_2 = N_0 - \frac{2}{3}N_0 = \frac{1}{3}N_0 \]

2. **Using the Half-Life Formula**:
- The relationship between the remaining quantity and the number of half-lives can be expressed as:
\[ N = N_0 \left( \frac{1}{2} \right)^n \]
- For \( N_1 \):
\[ \frac{2}{3}N_0 = N_0 \left( \frac{1}{2} \right)^{n_1} \]
Dividing both sides by \( N_0 \):
\[ \frac{2}{3} = \left( \frac{1}{2} \right)^{n_1} \]
- Taking logarithm on both sides:
\[ n_1 = \frac{\log(2/3)}{\log(1/2)} \]

3. **For \( N_2 \)**:
- For \( N_2 \):
\[ \frac{1}{3}N_0 = N_0 \left( \frac{1}{2} \right)^{n_2} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{3} = \left( \frac{1}{2} \right)^{n_2} \]
- Taking logarithm on both sides:
\[ n_2 = \frac{\log(1/3)}{\log(1/2)} \]

4. **Finding the Time Interval**:
- The time interval \( t_2 - t_1 \) is given by the difference in the number of half-lives:
\[ t_2 - t_1 = n_2 \cdot T_{1/2} - n_1 \cdot T_{1/2} \]
- Where \( T_{1/2} = 20 \) minutes (the half-life).
- Thus:
\[ t_2 - t_1 = (n_2 - n_1) \cdot 20 \]

5. **Calculating \( n_1 \) and \( n_2 \)**:
- Since \( n_1 \) and \( n_2 \) can be calculated, we can find \( n_2 - n_1 \):
\[ n_2 - n_1 = 1 \quad \text{(as derived from the logarithmic equations)} \]

6. **Final Calculation**:
- Therefore, substituting back:
\[ t_2 - t_1 = 1 \cdot 20 = 20 \text{ minutes} \]

### Final Answer:
The approximate time interval \( t_2 - t_1 \) is **20 minutes**.