A radioactive substance decays to `1//16^(th)` of its initial mass in 40 days. The half life of the substance, in days, is :

Options

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

Explanation

To solve the problem, we need to determine the half-life of a radioactive substance that decays to \( \frac{1}{16} \) of its initial mass in 40 days.

### Step-by-Step Solution:

1. **Understanding the Decay Formula**:
The decay of a radioactive substance can be described by the equation:
\[ n = n_0 e^{-\lambda t} \]
where:
- \( n \) is the remaining quantity of the substance,
- \( n_0 \) is the initial quantity,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

2. **Setting Up the Equation**:
According to the problem, the substance decays to \( \frac{1}{16} \) of its initial mass in 40 days. This can be expressed as:
\[ \frac{n}{n_0} = \frac{1}{16} \]
Substituting this into the decay formula gives:
\[ \frac{1}{16} = e^{-\lambda \cdot 40} \]

3. **Taking the Natural Logarithm**:
To solve for \( \lambda \), we take the natural logarithm of both sides:
\[ \ln\left(\frac{1}{16}\right) = -\lambda \cdot 40 \]
This simplifies to:
\[ -\ln(16) = -\lambda \cdot 40 \]
Therefore:
\[ \lambda = \frac{\ln(16)}{40} \]

4. **Finding the Half-Life**:
The half-life \( t_{1/2} \) of a radioactive substance is given by the formula:
\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]
Substituting the expression for \( \lambda \):
\[ t_{1/2} = \frac{\ln(2)}{\frac{\ln(16)}{40}} = \frac{40 \cdot \ln(2)}{\ln(16)} \]

5. **Simplifying \( \ln(16) \)**:
Since \( 16 = 2^4 \), we can express \( \ln(16) \) as:
\[ \ln(16) = \ln(2^4) = 4 \ln(2) \]
Substituting this back into the half-life equation gives:
\[ t_{1/2} = \frac{40 \cdot \ln(2)}{4 \ln(2)} = \frac{40}{4} = 10 \text{ days} \]

### Final Answer:
The half-life of the substance is **10 days**.