Half-life of a radioactive substance is 20 min. 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 decayed is

Options

A. 14 min
B. 20 min
C. 25 min
D. 7 min

Answer

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

### Step-by-Step Solution:

1. **Understand the Half-Life Concept**:
The half-life of a radioactive substance is the time required for half of the substance to decay. In this case, the half-life is given as 20 minutes.

2. **Determine the Initial Amount**:
Let's assume the initial amount of the substance is \( N_0 \).

3. **Calculate the Amount Remaining at \( t_1 \)**:
At time \( t_1 \), when \( \frac{1}{3} \) of the substance has decayed, the remaining amount is:
\[ N(t_1) = N_0 - \frac{1}{3}N_0 = \frac{2}{3}N_0 \]

4. **Calculate the Amount Remaining at \( t_2 \)**:
At time \( t_2 \), when \( \frac{2}{3} \) of the substance has decayed, the remaining amount is:
\[ N(t_2) = N_0 - \frac{2}{3}N_0 = \frac{1}{3}N_0 \]

5. **Relate the Decay to Half-Life**:
The decay process is exponential. We can use the concept of half-life to relate the times \( t_1 \) and \( t_2 \). The relationship between the remaining amount and the half-life can be expressed as:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
where \( T_{1/2} \) is the half-life.

6. **Set Up the Equations**:
For \( t_1 \):
\[ \frac{2}{3}N_0 = N_0 \left( \frac{1}{2} \right)^{\frac{t_1}{20}} \]
Simplifying gives:
\[ \frac{2}{3} = \left( \frac{1}{2} \right)^{\frac{t_1}{20}} \]

For \( t_2 \):
\[ \frac{1}{3}N_0 = N_0 \left( \frac{1}{2} \right)^{\frac{t_2}{20}} \]
Simplifying gives:
\[ \frac{1}{3} = \left( \frac{1}{2} \right)^{\frac{t_2}{20}} \]

7. **Solve for \( t_1 \) and \( t_2 \)**:
Taking logarithms:
\[ \log\left(\frac{2}{3}\right) = \frac{t_1}{20} \log\left(\frac{1}{2}\right) \implies t_1 = -20 \frac{\log\left(\frac{2}{3}\right)}{\log\left(\frac{1}{2}\right)} \]
\[ \log\left(\frac{1}{3}\right) = \frac{t_2}{20} \log\left(\frac{1}{2}\right) \implies t_2 = -20 \frac{\log\left(\frac{1}{3}\right)}{\log\left(\frac{1}{2}\right)} \]

8. **Calculate the Time Interval**:
The time interval \( t_2 - t_1 \) can be approximated by the difference of the two times:
\[ t_2 - t_1 \approx 20 \text{ minutes} \]

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