A radioactive substance decays to `((1)/(16))^(th)` of its initial activity in 40 days. The half-life of the radioacctive substance expressed in days is

Options

A. 2.5
B. 5
C. 10
D. 20

Explanation

To solve the problem, we need to find the half-life of a radioactive substance that decays to \(\frac{1}{16}\) of its initial activity in 40 days. Here’s a step-by-step solution:

### Step 1: Understand the decay process
The initial activity of the radioactive substance is denoted as \(R_0\). After 40 days, the activity is \(\frac{R_0}{16}\).

### Step 2: Relate the decay to half-lives
The relationship between the activity at any time and the initial activity can be expressed as:
\[ R = R_0 \left(\frac{1}{2}\right)^n \]
where \(n\) is the number of half-lives that have passed.

### Step 3: Set up the equation
From the information given, we can set up the equation:
\[ \frac{R_0}{16} = R_0 \left(\frac{1}{2}\right)^n \]
We can cancel \(R_0\) from both sides (assuming \(R_0 \neq 0\)):
\[ \frac{1}{16} = \left(\frac{1}{2}\right)^n \]

### Step 4: Express \(\frac{1}{16}\) in terms of powers of 2
We know that:
\[ \frac{1}{16} = \left(\frac{1}{2}\right)^4 \]
Thus, we can equate the exponents:
\[ n = 4 \]

### Step 5: Relate the total time to half-lives
We know that the total time \(T\) is related to the number of half-lives and the half-life (\(t_{1/2}\)) by the formula:
\[ T = n \cdot t_{1/2} \]
Given that \(T = 40\) days and \(n = 4\):
\[ 40 = 4 \cdot t_{1/2} \]

### Step 6: Solve for the half-life
Now, we can solve for \(t_{1/2}\):
\[ t_{1/2} = \frac{40}{4} = 10 \text{ days} \]

### Conclusion
The half-life of the radioactive substance is **10 days**.

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