Nuclei of a radioactive element `X` are being produced at a constant rate `K` and this element decays to a stable nucleus `Y` with a decay constant `lambda` and half-life `T_(1//3)`. At the time `t=0`, there are `N_(0) `nuclei of the element X.
The number `N_(Y)` of nuclei of `Y` at time `t` is .

Options

A. `Kt-(K-lambda N_(0))/(lambda)e^(-lambda t)+(K-lambda(N_(0)))/(lambda)`
B. `Kt+(K-lambda N_(0))/(lambda)e^(-lambda t)-(K-lambda(N_(0)))/(lambda)`
C. `Kt+(K-lambda N_(0))/(lambda)e^(-lambda t)`
D. `Kt+(K-lambda N_(0))/(lambda)e^(-lambda t)`

Explanation

To solve the problem, we need to determine the number of nuclei of the stable element Y at time t, given that the radioactive element X is produced at a constant rate K and decays to Y with a decay constant λ. Here's the step-by-step solution:

### Step 1: Understand the initial conditions
At time \( t = 0 \), there are \( N_0 \) nuclei of element X, and no nuclei of Y, so \( N_Y(0) = 0 \).

### Step 2: Write the differential equation for the number of Y nuclei
The rate of change of the number of Y nuclei is given by the rate at which X decays to Y. This can be expressed as:
\[ \frac{dN_Y}{dt} = \lambda N_X \]
where \( N_X \) is the number of X nuclei at time t.

### Step 3: Determine the expression for \( N_X(t) \)
The number of X nuclei at time t can be expressed as:
\[ N_X(t) = N_0 e^{-\lambda t} + \frac{K}{\lambda} (1 - e^{-\lambda t}) \]
This equation accounts for the initial number of X nuclei decaying over time and the constant production of X at rate K.

### Step 4: Substitute \( N_X(t) \) into the differential equation
Substituting \( N_X(t) \) into the equation for \( \frac{dN_Y}{dt} \):
\[ \frac{dN_Y}{dt} = \lambda \left( N_0 e^{-\lambda t} + \frac{K}{\lambda} (1 - e^{-\lambda t}) \right) \]
This simplifies to:
\[ \frac{dN_Y}{dt} = \lambda N_0 e^{-\lambda t} + K (1 - e^{-\lambda t}) \]

### Step 5: Integrate to find \( N_Y(t) \)
Integrate both sides with respect to time t:
\[ N_Y(t) = \int \left( \lambda N_0 e^{-\lambda t} + K (1 - e^{-\lambda t}) \right) dt \]
This results in:
\[ N_Y(t) = -N_0 e^{-\lambda t} + Kt + C \]
where C is the constant of integration.

### Step 6: Apply initial conditions to find C
At \( t = 0 \), \( N_Y(0) = 0 \):
\[ 0 = -N_0 + 0 + C \implies C = N_0 \]
Thus, we have:
\[ N_Y(t) = -N_0 e^{-\lambda t} + Kt + N_0 \]

### Step 7: Simplify the expression for \( N_Y(t) \)
Rearranging gives:
\[ N_Y(t) = Kt + N_0(1 - e^{-\lambda t}) \]

### Final Expression
The number of nuclei of Y at time t is:
\[ N_Y(t) = Kt + N_0(1 - e^{-\lambda t}) \]