The half life of a radioactive substance is 5 hours. In how much time will 15/16 of the material decay ?

Answer

To solve the problem, we need to determine the time it takes for 15/16 of a radioactive substance to decay, given that its half-life is 5 hours.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We know that the half-life (t_half) of the substance is 5 hours. This means that every 5 hours, half of the remaining substance will decay.
- We need to find out how long it takes for 15/16 of the substance to decay, which means only 1/16 of the substance will remain.

2. **Using the Decay Formula**:
- The formula for radioactive decay is given by:
\[ N = N_0 e^{-\lambda t} \]
where:
- \(N\) = remaining quantity of substance
- \(N_0\) = initial quantity of substance
- \(\lambda\) = decay constant
- \(t\) = time elapsed

3. **Finding the Decay Constant (\(\lambda\))**:
- The decay constant (\(\lambda\)) can be calculated using the half-life:
\[ \lambda = \frac{0.693}{t_{half}} \]
- Substituting \(t_{half} = 5\) hours:
\[ \lambda = \frac{0.693}{5} = 0.1386 \text{ hours}^{-1} \]

4. **Setting Up the Equation**:
- We want to find the time \(t\) when only 1/16 of the substance remains:
\[ N = \frac{N_0}{16} \]
- Plugging into the decay formula:
\[ \frac{N_0}{16} = N_0 e^{-\lambda t} \]
- Dividing both sides by \(N_0\) (assuming \(N_0 \neq 0\)):
\[ \frac{1}{16} = e^{-\lambda t} \]

5. **Taking the Natural Logarithm**:
- Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{16}\right) = -\lambda t \]
- We know that \(\ln\left(\frac{1}{16}\right) = -\ln(16)\), so:
\[ -\ln(16) = -\lambda t \]
- Thus:
\[ \ln(16) = \lambda t \]

6. **Substituting \(\lambda\) and Solving for \(t\)**:
- We know \(\lambda = 0.1386\) hours\(^{-1}\):
\[ \ln(16) = 0.1386 t \]
- The value of \(\ln(16)\) is approximately \(2.7726\):
\[ 2.7726 = 0.1386 t \]
- Solving for \(t\):
\[ t = \frac{2.7726}{0.1386} \approx 20 \text{ hours} \]

### Final Answer:
The time taken for 15/16 of the material to decay is approximately **20 hours**.