If the amount of a radioactive substance is increased three times, the number of atoms disintegrated per unit time would
Be doulbe b.Be triple
c. Remian one-third d. Not change

Options

A. a. Be double
B. b.Be triple
C. c. remain one-third
D. d. Not change

Explanation

To solve the question, we need to understand the relationship between the amount of a radioactive substance and its activity, which is defined as the number of atoms disintegrated per unit time.

### Step-by-Step Solution:

1. **Understanding Activity**:
- The activity (A) of a radioactive substance is defined as the number of disintegrations (or decays) per unit time. It is given by the formula:
\[ A = \lambda \times N \]
where \( \lambda \) is the decay constant and \( N \) is the number of radioactive atoms present.

2. **Initial Condition**:
- Let's denote the initial amount of the radioactive substance as \( N \). The initial activity can be expressed as:
\[ A_{\text{initial}} = \lambda \times N \]

3. **Increasing the Amount**:
- If the amount of the radioactive substance is increased three times, the new amount of the substance becomes:
\[ N_{\text{new}} = 3N \]

4. **New Activity Calculation**:
- Using the new amount in the activity formula, the new activity becomes:
\[ A_{\text{new}} = \lambda \times N_{\text{new}} = \lambda \times (3N) = 3(\lambda \times N) = 3A_{\text{initial}} \]

5. **Conclusion**:
- Since the new activity is three times the initial activity, we can conclude that the number of atoms disintegrated per unit time would be triple the initial amount. Therefore, the answer to the question is:
\[ \text{b. Be triple} \]