After an interval of one day , `1//16th` initial amount of a radioactive material remains in a sample. Then, its half-life is .

Options

A. `6h`
B. `12 h`
C. `1.5 h`
D. `3h`

Explanation

To determine the half-life of a radioactive material given that after one day, \( \frac{1}{16} \) of the initial amount remains, we can follow these steps:

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

### Step 2: Set up the equation
From the problem, we know that after one day (let's denote this as \( t = 1 \) day), the remaining amount is \( \frac{1}{16} N_0 \). We can set up the equation:
\[ \frac{1}{16} N_0 = N_0 e^{-\lambda \cdot 1} \]

### Step 3: Simplify the equation
We can cancel \( N_0 \) from both sides (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{16} = e^{-\lambda} \]

### Step 4: Solve for \( \lambda \)
Taking the natural logarithm of both sides:
\[ -\lambda = \ln\left(\frac{1}{16}\right) \]
This can be rewritten as:
\[ \lambda = -\ln\left(\frac{1}{16}\right) = \ln(16) \]

### Step 5: Calculate the half-life
The half-life \( t_{1/2} \) is given by the formula:
\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]
Substituting \( \lambda = \ln(16) \):
\[ t_{1/2} = \frac{\ln(2)}{\ln(16)} \]

### Step 6: Simplify \( \ln(16) \)
Since \( 16 = 2^4 \):
\[ \ln(16) = \ln(2^4) = 4\ln(2) \]
Thus, we can substitute this back into the half-life formula:
\[ t_{1/2} = \frac{\ln(2)}{4\ln(2)} = \frac{1}{4} \text{ days} \]

### Step 7: Convert days to hours
Since \( \frac{1}{4} \) days is equivalent to:
\[ \frac{1}{4} \times 24 \text{ hours} = 6 \text{ hours} \]

### Conclusion
The half-life of the radioactive material is **6 hours**.

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