A radioactive substance with decay constant of `0.5 s^(–1)` is being produced at a constant rate of 50 nuclei per second. If there are no nuclei present initially, the time (in second) after which 25 nuclei will be present is

Options

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

Answer

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

### Step 1: Define the variables and equations
We know that:
- The decay constant \( \lambda = 0.5 \, \text{s}^{-1} \)
- The production rate of nuclei \( R = 50 \, \text{nuclei/s} \)
- The number of nuclei present at time \( t \) is \( n(t) \).

The rate of change of the number of nuclei can be expressed as:
\[ \frac{dn}{dt} = R - \lambda n \]
Substituting the values:
\[ \frac{dn}{dt} = 50 - 0.5n \]

### Step 2: Rearranging the equation
Rearranging the equation gives:
\[ \frac{dn}{50 - 0.5n} = dt \]

### Step 3: Integrate both sides
Now we will integrate both sides. The left side will be integrated with respect to \( n \) and the right side with respect to \( t \):
\[ \int \frac{dn}{50 - 0.5n} = \int dt \]

### Step 4: Solve the left side integral
The left side can be solved using a substitution. Let \( u = 50 - 0.5n \), then \( du = -0.5dn \) or \( dn = -2du \):
\[ \int \frac{-2du}{u} = -2 \ln |u| + C = -2 \ln |50 - 0.5n| + C \]

### Step 5: Solve the right side integral
The right side integrates to:
\[ t + C' \]

### Step 6: Combine the results
Setting the constants \( C \) and \( C' \) to a single constant \( C \):
\[ -2 \ln |50 - 0.5n| = t + C \]

### Step 7: Solve for \( n \)
Exponentiating both sides gives:
\[ 50 - 0.5n = e^{-\frac{t + C}{2}} \]
Rearranging gives:
\[ 0.5n = 50 - e^{-\frac{t + C}{2}} \]
\[ n = 100 - 2e^{-\frac{t + C}{2}} \]

### Step 8: Initial condition
At \( t = 0 \), \( n(0) = 0 \):
\[ 0 = 100 - 2e^{-\frac{C}{2}} \implies 2e^{-\frac{C}{2}} = 100 \implies e^{-\frac{C}{2}} = 50 \implies C = -2 \ln(50) \]

### Step 9: Substitute back to find \( n(t) \)
Substituting back, we get:
\[ n = 100 - 2e^{-\frac{t - 2 \ln(50)}{2}} = 100 - 2 \cdot 50 e^{-\frac{t}{2}} = 100 - 100 e^{-\frac{t}{2}} \]

### Step 10: Set \( n(t) = 25 \) and solve for \( t \)
Setting \( n(t) = 25 \):
\[ 25 = 100 - 100 e^{-\frac{t}{2}} \]
\[ 100 e^{-\frac{t}{2}} = 75 \implies e^{-\frac{t}{2}} = 0.75 \]
Taking the natural logarithm:
\[ -\frac{t}{2} = \ln(0.75) \implies t = -2 \ln(0.75) \]

### Step 11: Calculate \( t \)
Calculating \( t \):
\[ t \approx -2 \times (-0.2877) \approx 0.5754 \, \text{s} \]

Thus, the time after which 25 nuclei will be present is approximately **0.575 seconds**.