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. `log_e2/5`
B. `5/(log_e2)`
C. `5log_10 2`
D. `5log_e 2`

Explanation

To solve the problem step by step, we will use the concepts of radioactive decay and the formula for activity.

### Step 1: Understand the given data
At time \( t = 0 \), the activity \( A_0 \) is \( N_0 \) counts per minute. At \( t = 5 \) minutes, the activity \( A \) is \( \frac{N_0}{e} \) counts per minute.

### Step 2: Use the radioactive decay formula
The activity of a radioactive sample at time \( t \) is given by the formula:
\[ A(t) = A_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

### Step 3: Set up the equation for \( t = 5 \) minutes
At \( t = 5 \) minutes, we can write:
\[ \frac{N_0}{e} = N_0 e^{-5\lambda} \]

### Step 4: Simplify the equation
Dividing both sides by \( N_0 \):
\[ \frac{1}{e} = e^{-5\lambda} \]
This implies:
\[ e^{-1} = e^{-5\lambda} \]

### Step 5: Equate the exponents
Since the bases are the same, we can equate the exponents:
\[ -1 = -5\lambda \]
From this, 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 is given by:
\[ \frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{2} = e^{-\lambda t_{1/2}} \]

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

### Step 8: Substitute the value of \( \lambda \)
Substituting \( \lambda = \frac{1}{5} \):
\[ t_{1/2} = 5 \ln(2) \]

### Final Answer
The time at which the activity reduces to half its value is \( 5 \ln(2) \) minutes.

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