A radioactive substance `X` decys into another radioactive substance `Y`. Initially, only `X` was present . `lambda_(x)` and `lambda_(y)` are the disintegration constant of `X` and Y. `N_(y)` will be maximum when.

Options

A. `(N_(y))/(N_(x)-N_(y))=lambda_(y)/(lambda_(x)-lambda_(y))`
B. `(N_(x))/(N_(x)-N_(y))=lambda_(x)/(lambda_(x)-lambda_(y))`
C. `lambda_(y)N_(y)=lambda_(x)lambda_(x)`
D. `lambda_(y)N_(x)=lambda_(y)lambda_(x)`

Answer

To solve the problem of when the number of substance Y, denoted as \( N_y \), will be maximum, we can follow these steps:

### Step 1: Understand the decay process
Initially, we have only substance X present. As time progresses, substance X decays into substance Y. The decay of substance X is characterized by its disintegration constant \( \lambda_x \), and the decay of substance Y is characterized by its disintegration constant \( \lambda_y \).

### Step 2: Write the rate of change of substances
The rate of formation of substance Y (\( N_y \)) can be expressed as the difference between the rate at which substance X decays and the rate at which substance Y decays. Mathematically, this can be represented as:
\[ \frac{dN_y}{dt} = \lambda_x N_x - \lambda_y N_y \]
where \( N_x \) is the number of nuclei of substance X at time \( t \) and \( N_y \) is the number of nuclei of substance Y at the same time.

### Step 3: Set the condition for maximum \( N_y \)
To find the maximum number of substance Y, we need to set the rate of change of \( N_y \) to zero:
\[ \frac{dN_y}{dt} = 0 \]
This leads to the equation:
\[ \lambda_x N_x - \lambda_y N_y = 0 \]

### Step 4: Rearranging the equation
From the equation above, we can rearrange it to express the relationship between \( N_x \) and \( N_y \):
\[ \lambda_x N_x = \lambda_y N_y \]
This implies that the maximum number of substance Y occurs when the decay rate of X equals the decay rate of Y.

### Step 5: Express \( N_x \) in terms of initial conditions
Initially, when only substance X is present, we can express \( N_x \) in terms of the initial number of nuclei \( N_{x0} \) and the time \( t \):
\[ N_x = N_{x0} e^{-\lambda_x t} \]
Substituting this back into the equation gives:
\[ \lambda_x (N_{x0} e^{-\lambda_x t}) = \lambda_y N_y \]
From this, we can solve for \( N_y \):
\[ N_y = \frac{\lambda_x}{\lambda_y} N_{x0} e^{-\lambda_x t} \]

### Conclusion
The maximum number of substance Y, \( N_y \), will occur when the decay rates of X and Y are equal, which can be mathematically expressed as:
\[ \lambda_x N_x = \lambda_y N_y \]