The activity of a radioactive sample is measures as `N_0` counts per minute at `t = 0` and `N_0//e` counts per minute at `t = 5 min`. The time (in minute) at which the activity reduces to half its value is.

Options

A. `5log_(e)2`
B. `log_(e)"(2)/(5)`
C. `(5)/(log_(e)2)`
D. `5log_(10)2`

Explanation

To find the time at which the activity of a radioactive sample reduces to half its initial value, we can follow these steps:

### Step 1: Understand the decay formula
The activity \( N(t) \) of a radioactive sample at time \( t \) can be expressed using the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where:
- \( N_0 \) is the initial activity,
- \( \lambda \) is the decay constant,
- \( t \) is the time.

### Step 2: Set up the equation for the given conditions
At \( t = 0 \), the activity is \( N_0 \). At \( t = 5 \) minutes, the activity is given as \( \frac{N_0}{e} \). We can set up the equation:
\[ N(5) = N_0 e^{-\lambda \cdot 5} = \frac{N_0}{e} \]

### Step 3: Simplify the equation
Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ e^{-\lambda \cdot 5} = \frac{1}{e} \]

### Step 4: Take the natural logarithm of both sides
Taking the natural logarithm of both sides gives:
\[ -\lambda \cdot 5 = -1 \]

### Step 5: Solve for the decay constant \( \lambda \)
From the equation above, we can solve for \( \lambda \):
\[ \lambda = \frac{1}{5} \]

### Step 6: Find the half-life
The half-life \( t_{1/2} \) is the time at which the activity reduces to half its initial value. The relationship for half-life is given by:
\[ N_0/2 = N_0 e^{-\lambda t_{1/2}} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{2} = e^{-\lambda t_{1/2}} \]

### Step 7: Take the natural logarithm again
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -\lambda t_{1/2} \]

### Step 8: Substitute \( \lambda \) and solve for \( t_{1/2} \)
Substituting \( \lambda = \frac{1}{5} \):
\[ \ln\left(\frac{1}{2}\right) = -\frac{1}{5} t_{1/2} \]
Thus,
\[ t_{1/2} = -5 \ln\left(\frac{1}{2}\right) = 5 \ln(2) \]

### Step 9: Calculate \( t_{1/2} \)
Using the approximate value \( \ln(2) \approx 0.693 \):
\[ t_{1/2} \approx 5 \times 0.693 \approx 3.465 \text{ minutes} \]

### Step 10: Conclusion
The time at which the activity reduces to half its value is approximately \( 3.465 \) minutes.

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