`N_(1)` atoms of a radioactive element emit `N_(2)` beta partilces per second. The decay cosntant of the element is (in `s^(-1)`)

Options

A. `(N_(1))/(N_(2))`
B. `(N_(2))/(N_(1))`
C. `N_(1) ln (2)`
D. `N_(2) ln (2)`

Answer

To solve the problem, we need to find the decay constant (λ) of a radioactive element based on the given information about the number of atoms (N₁) and the number of beta particles emitted per second (N₂).

### Step-by-Step Solution:

1. **Understand the relationship between activity, decay constant, and number of atoms**:
The activity (A) of a radioactive substance is given by the formula:
\[ A = \lambda N \]
where:
- \( A \) is the activity (in particles per second),
- \( \lambda \) is the decay constant (in s⁻¹),
- \( N \) is the number of radioactive atoms.

2. **Identify the variables from the problem**:
From the problem, we know:
- The activity \( A \) is equal to \( N_2 \) (the number of beta particles emitted per second).
- The number of atoms \( N \) is equal to \( N_1 \).

3. **Substitute the known values into the activity formula**:
We can substitute \( A \) and \( N \) into the activity formula:
\[ N_2 = \lambda N_1 \]

4. **Rearrange the equation to solve for the decay constant (λ)**:
To find the decay constant \( \lambda \), we rearrange the equation:
\[ \lambda = \frac{N_2}{N_1} \]

5. **Conclusion**:
The decay constant \( \lambda \) is given by:
\[ \lambda = \frac{N_2}{N_1} \]

### Final Answer:
The decay constant of the element is \( \lambda = \frac{N_2}{N_1} \) (in s⁻¹).

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