The rate of disintegration of a radioactive substance falls from `(40)/(3)` dps to `(5)/(3)` dps in 6 hours. The half - life of the radioactive substance is

Options

A. `(6)/(7)` hours
B. 2 hours
C. 3 hours
D. 1 hours

Answer

To solve the problem, we need to determine the half-life of a radioactive substance given that its rate of disintegration falls from \( \frac{40}{3} \) disintegrations per second (dps) to \( \frac{5}{3} \) dps over a period of 6 hours.

### Step-by-Step Solution:

1. **Understanding the Disintegration Rates**:
- Initial disintegration rate, \( A_1 = \frac{40}{3} \) dps.
- Final disintegration rate, \( A_2 = \frac{5}{3} \) dps.

2. **Finding the Ratio of Disintegration Rates**:
- The ratio of the initial and final disintegration rates is:
\[ \frac{A_1}{A_2} = \frac{\frac{40}{3}}{\frac{5}{3}} = \frac{40}{5} = 8 \]
- This implies that the final number of nuclei is \( \frac{1}{8} \) of the initial number of nuclei.

3. **Using the Half-Life Formula**:
- The number of nuclei after \( n \) half-lives is given by:
\[ N = N_0 \left(\frac{1}{2}\right)^n \]
- Here, \( N_0 \) is the initial number of nuclei, and \( N \) is the final number of nuclei.
- Since we found that \( N = \frac{1}{8} N_0 \), we can set up the equation:
\[ \frac{1}{8} N_0 = N_0 \left(\frac{1}{2}\right)^n \]
- Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{8} = \left(\frac{1}{2}\right)^n \]

4. **Solving for \( n \)**:
- We know that \( \frac{1}{8} = \left(\frac{1}{2}\right)^3 \), thus:
\[ n = 3 \]
- This means it takes 3 half-lives to reduce the number of nuclei to \( \frac{1}{8} \) of its original amount.

5. **Calculating the Half-Life**:
- We are given that this reduction occurs over a period of 6 hours. Therefore, we can find the half-life \( t_{1/2} \) as follows:
\[ 3 \cdot t_{1/2} = 6 \text{ hours} \]
- Solving for \( t_{1/2} \):
\[ t_{1/2} = \frac{6 \text{ hours}}{3} = 2 \text{ hours} \]

### Conclusion:
The half-life of the radioactive substance is **2 hours**.

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