Radioactive element decays to form a stable nuclide, then the rate of decay of reactant `((d N)/(d t))` will vary with time `(t)` as shown in figure.

Options

A.
B.
C.
D.

Answer

To solve the problem regarding the decay of a radioactive element to form a stable nuclide, we will analyze the rate of decay over time and derive the necessary expressions step by step.

### Step-by-Step Solution:

1. **Understanding the Decay Process**:
The decay of a radioactive element can be described by the equation:
\[ N(t) = N_0 e^{-\lambda t} \]
where:
- \( N(t) \) is the number of undecayed nuclei at time \( t \),
- \( N_0 \) is the initial number of nuclei,
- \( \lambda \) is the decay constant,
- \( t \) is the time.

2. **Differentiating the Decay Equation**:
To find the rate of decay, we differentiate \( N(t) \) with respect to time \( t \):
\[ \frac{dN}{dt} = \frac{d}{dt}(N_0 e^{-\lambda t}) = -\lambda N_0 e^{-\lambda t} \]
This indicates that the rate of decay is proportional to the number of undecayed nuclei present at that time.

3. **Analyzing the Rate of Decay at Specific Times**:
- At \( t = 0 \):
\[ \frac{dN}{dt} \bigg|_{t=0} = -\lambda N_0 e^{0} = -\lambda N_0 \]
- As \( t \) approaches infinity:
\[ \frac{dN}{dt} \bigg|_{t \to \infty} = -\lambda N_0 e^{-\lambda \cdot \infty} = 0 \]

4. **Conclusion**:
The rate of decay \( \frac{dN}{dt} \) starts at \( -\lambda N_0 \) when \( t = 0 \) and approaches \( 0 \) as \( t \) approaches infinity. This indicates that the decay rate decreases over time.

5. **Choosing the Correct Option**:
Based on the analysis, we can conclude that the rate of decay varies from \( -\lambda N_0 \) at \( t = 0 \) to \( 0 \) as \( t \) approaches infinity. Thus, the correct option is:
- **D**: The rate of decay starts at \( -\lambda N_0 \) and decreases to \( 0 \).