A radioactive nuclei with decay constant 0.5/s is being produced at a constant rate of 100 nuclei/s. If at t = 0 there were no nuclei, the time when there are 50 nuclei is:

We are given a radioactive nucleus with a decay constant λ = 0.5 per second and a constant production rate R = 100 nuclei per second. At time t = 0, there are no nuclei. We need to find the time when the number of nuclei N = 50.

The rate of change of the number of nuclei is given by the difference between the production rate and the decay rate. The decay rate is λN, so the differential equation is:

$$\frac{dN}{dt} = R - \lambda N$$

Substituting the given values R = 100 nuclei/s and λ = 0.5/s:

$$\frac{dN}{dt} = 100 - 0.5N$$

We can write this as:

$$\frac{dN}{dt} = 100 - \frac{1}{2}N$$

Rearranging the terms:

$$\frac{dN}{dt} + \frac{1}{2}N = 100$$

This is a linear first-order differential equation. To solve it, we find the integrating factor. The integrating factor is e raised to the integral of the coefficient of N, which is 1/2:

$$\text{Integrating factor} = e^{\int \frac{1}{2} dt} = e^{\frac{1}{2}t}$$

Multiply both sides of the equation by the integrating factor:

$$e^{\frac{1}{2}t} \frac{dN}{dt} + e^{\frac{1}{2}t} \cdot \frac{1}{2}N = 100 e^{\frac{1}{2}t}$$

The left side is the derivative of the product of N and the integrating factor:

$$\frac{d}{dt} \left( N e^{\frac{1}{2}t} \right) = 100 e^{\frac{1}{2}t}$$

Integrate both sides with respect to t:

$$\int \frac{d}{dt} \left( N e^{\frac{1}{2}t} \right) dt = \int 100 e^{\frac{1}{2}t} dt$$

$$N e^{\frac{1}{2}t} = 100 \int e^{\frac{1}{2}t} dt$$

The integral of e^{at} is (1/a)e^{at}, so with a = 1/2:

$$\int e^{\frac{1}{2}t} dt = \frac{1}{\frac{1}{2}} e^{\frac{1}{2}t} = 2 e^{\frac{1}{2}t}$$

Thus:

$$N e^{\frac{1}{2}t} = 100 \cdot 2 e^{\frac{1}{2}t} + C$$

$$N e^{\frac{1}{2}t} = 200 e^{\frac{1}{2}t} + C$$

Solve for N by dividing both sides by e^{$$\frac{1}{2}$$t}:

$$N = 200 + C e^{-\frac{1}{2}t}$$

Apply the initial condition: at t = 0, N = 0.

$$0 = 200 + C e^{0}$$

$$0 = 200 + C \cdot 1$$

$$C = -200$$

Substitute C back into the equation:

$$N = 200 - 200 e^{-\frac{1}{2}t}$$

Factor out 200:

$$N = 200 \left(1 - e^{-\frac{1}{2}t}\right)$$

Set N = 50 to find the time t:

$$50 = 200 \left(1 - e^{-\frac{1}{2}t}\right)$$

Divide both sides by 200:

$$\frac{50}{200} = 1 - e^{-\frac{1}{2}t}$$

$$\frac{1}{4} = 1 - e^{-\frac{1}{2}t}$$

Rearrange to solve for the exponential term:

$$e^{-\frac{1}{2}t} = 1 - \frac{1}{4}$$

$$e^{-\frac{1}{2}t} = \frac{3}{4}$$

Take the natural logarithm of both sides:

$$\ln \left( e^{-\frac{1}{2}t} \right) = \ln \left( \frac{3}{4} \right)$$

$$-\frac{1}{2}t = \ln \left( \frac{3}{4} \right)$$

Multiply both sides by -2:

$$t = -2 \ln \left( \frac{3}{4} \right)$$

Using the property of logarithms that -ln(a) = ln(1/a):

$$t = 2 \ln \left( \frac{4}{3} \right)$$

Thus, the time when there are 50 nuclei is $$ 2 \ln \left( \frac{4}{3} \right) $$ seconds.

Comparing with the options:

A. 1 s

B. $$ 2\ln\left(\frac{4}{3}\right) $$ s

C. $$ \ln 2 $$ s

D. $$ \ln\left(\frac{4}{3}\right) $$ s

Hence, the correct answer is Option B.