If 'f' denotes the ratio of the number of nuclei decayed `(N_(d))` to the number of nuclei at `t = 0 (N_(0))` then for a collection of radioactive nuclei, the rate of change of 'f' with respect to time is given as :
[`lambda` is the radioactive decay constant]

Options

A. `-lambda(1-e^(-lambda t))`
B. `lambda (1-e^(-lambda t))`
C. `lambda e^(-lambda t)`
D. `-lambda e^(-lambda t)`

Answer

To solve the problem, we need to find the rate of change of the ratio of the number of decayed nuclei \( f \) with respect to time \( t \). Let's break it down step by step.

### Step-by-Step Solution:

1. **Define the Variables**:
- Let \( N_0 \) be the initial number of nuclei at \( t = 0 \).
- Let \( N_d \) be the number of nuclei that have decayed by time \( t \).
- The ratio \( f \) is defined as:
\[ f = \frac{N_d}{N_0} \]

2. **Relate \( N_d \) and \( N \)**:
- The number of nuclei remaining at time \( t \) can be expressed using the radioactive decay formula:
\[ N = N_0 e^{-\lambda t} \]
- The number of decayed nuclei \( N_d \) can be expressed as:
\[ N_d = N_0 - N = N_0 - N_0 e^{-\lambda t} = N_0(1 - e^{-\lambda t}) \]

3. **Substitute \( N_d \) into the equation for \( f \)**:
- Now substituting \( N_d \) into the equation for \( f \):
\[ f = \frac{N_d}{N_0} = \frac{N_0(1 - e^{-\lambda t})}{N_0} = 1 - e^{-\lambda t} \]

4. **Differentiate \( f \) with respect to \( t \)**:
- To find the rate of change of \( f \) with respect to time, we differentiate \( f \):
\[ \frac{df}{dt} = \frac{d}{dt}(1 - e^{-\lambda t}) \]
- The derivative of a constant (1) is 0, and using the chain rule for the exponential function:
\[ \frac{df}{dt} = 0 - (-\lambda e^{-\lambda t}) = \lambda e^{-\lambda t} \]

5. **Final Result**:
- Thus, the rate of change of \( f \) with respect to time \( t \) is:
\[ \frac{df}{dt} = \lambda e^{-\lambda t} \]

### Conclusion:
The correct answer is:
\[ \frac{df}{dt} = \lambda e^{-\lambda t} \]