A parent radioactive nucleus A (decay constant `lamda_(A))` converts into a radio-active nucleus B of decay constant `lamda_(b)`, initially, number of atoms of B is zero At any time `N_(a),N_(b)` are number of atoms of nuclei A and B respectively then maximum velue of `N_(b)`.

Options

A. `(lamda_(a)N_(a))/(lamda_(b))`
B. `(lamda_(b)N_(a))/(lamda_(a))`
C. `((lamda_(a)+lamda-(b))N_(a))/(lamda_(b))`
D. `(lamda_(b))/((lamda_(a)+lamda_(b)))N_(a)`

Explanation

To find the maximum value of the number of atoms of nucleus B (\(N_b\)), we need to analyze the relationship between the decay of nucleus A and the formation and decay of nucleus B.

### Step-by-Step Solution:

1. **Understanding the Decay Process**:
- Nucleus A decays into nucleus B. The decay constant for A is \(\lambda_A\) and for B is \(\lambda_B\).
- Initially, the number of atoms of B (\(N_b\)) is zero.

2. **Rate of Formation of B**:
- The rate at which nucleus B is formed from A is proportional to the number of atoms of A that are present. This can be expressed as:
\[ \text{Rate of formation of } B = \lambda_A N_a \]
- Here, \(N_a\) is the number of atoms of A at any time.

3. **Rate of Disintegration of B**:
- The rate at which nucleus B disintegrates is proportional to the number of atoms of B present:
\[ \text{Rate of disintegration of } B = \lambda_B N_b \]

4. **Condition for Maximum \(N_b\)**:
- At maximum \(N_b\), the rate of formation of B equals the rate of disintegration of B:
\[ \lambda_A N_a = \lambda_B N_b \]

5. **Expressing \(N_b\)**:
- Rearranging the above equation gives:
\[ N_b = \frac{\lambda_A}{\lambda_B} N_a \]

6. **Finding Maximum \(N_b\)**:
- To find the maximum value of \(N_b\), we need to express \(N_a\) in terms of time. The number of atoms of A at any time \(t\) can be expressed as:
\[ N_a = N_{a0} e^{-\lambda_A t} \]
- Where \(N_{a0}\) is the initial number of atoms of A.

7. **Substituting \(N_a\) into the Equation for \(N_b\)**:
- Substituting \(N_a\) into the equation for \(N_b\):
\[ N_b = \frac{\lambda_A}{\lambda_B} N_{a0} e^{-\lambda_A t} \]

8. **Finding the Maximum**:
- To find the maximum value of \(N_b\), we can differentiate \(N_b\) with respect to time \(t\) and set the derivative to zero. However, we can also observe that \(N_b\) will reach its maximum when \(N_a\) is at a certain value before it decays significantly.
- The maximum value of \(N_b\) occurs when the decay rates balance out, which we already established as:
\[ N_b = \frac{\lambda_A}{\lambda_B} N_a \]

### Final Answer:
The maximum value of \(N_b\) is given by:
\[ N_b = \frac{\lambda_A}{\lambda_B} N_a \]