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 :

Options

A. 1 s
B. `2 ln ((4)/(3))` s
C. ln 2 s
D. `ln ((4)/(3))` s

Explanation

To solve the problem step by step, we will use the concepts of radioactive decay and the production of nuclei.

### Step 1: Understand the Problem
We have a radioactive nucleus with a decay constant (λ) of 0.5/s, and it is being produced at a constant rate of 100 nuclei/s. We need to find the time (t) when there are 50 nuclei present.

### Step 2: Set Up the Differential Equation
The rate of change of the number of nuclei (n) can be expressed as:
\[ \frac{dn}{dt} = \text{Production rate} - \text{Decay rate} \]
Thus, we can write:
\[ \frac{dn}{dt} = 100 - \lambda n \]
Substituting the value of λ:
\[ \frac{dn}{dt} = 100 - 0.5n \]

### Step 3: Rearrange the Equation
Rearranging gives:
\[ \frac{dn}{dt} + 0.5n = 100 \]

### Step 4: Separate Variables
We can separate the variables to integrate:
\[ \frac{dn}{100 - 0.5n} = dt \]

### Step 5: Integrate Both Sides
Integrating both sides:
\[ \int \frac{1}{100 - 0.5n} \, dn = \int dt \]
The left-hand side integrates to:
\[ -\frac{2}{100} \ln |100 - 0.5n| = t + C \]

### Step 6: Solve for the Constant of Integration
At \( t = 0 \), \( n = 0 \):
\[ -\frac{2}{100} \ln |100 - 0| = 0 + C \]
This simplifies to:
\[ C = -\frac{2}{100} \ln 100 \]

### Step 7: Substitute Back into the Equation
Now we substitute C back into our integrated equation:
\[ -\frac{2}{100} \ln |100 - 0.5n| = t - \frac{2}{100} \ln 100 \]

### Step 8: Solve for Time When n = 50
Now we need to find the time when \( n = 50 \):
\[ -\frac{2}{100} \ln |100 - 0.5 \times 50| = t - \frac{2}{100} \ln 100 \]
Calculating \( 100 - 0.5 \times 50 = 100 - 25 = 75 \):
\[ -\frac{2}{100} \ln 75 = t - \frac{2}{100} \ln 100 \]
Rearranging gives:
\[ t = -\frac{2}{100} \ln 75 + \frac{2}{100} \ln 100 \]
Using properties of logarithms:
\[ t = \frac{2}{100} (\ln 100 - \ln 75) = \frac{2}{100} \ln \left(\frac{100}{75}\right) \]
Simplifying further:
\[ t = \frac{2}{100} \ln \left(\frac{4}{3}\right) \]

### Final Answer
Thus, the time \( t \) when there are 50 nuclei is:
\[ t = \frac{2}{100} \ln \left(\frac{4}{3}\right) \text{ seconds} \]