For a certain radioactive substance, it is observed that after `4h`, only `6.25%` of the original sample is left undeacyed. It follows that.

Options

A. the half-life of the sample is `1h`
B. the mean life of the sample is `(1)/(1n2)h`
C. the decay constant of the sample is `1n(2) h^(-1)`
D. after a further `4h`, the amount of the substance left over would by only 0.39% of the original amount

Explanation

To solve the problem step by step, we will use the concepts of radioactive decay and the relevant equations.

### Step 1: Understand the given data
We know that after 4 hours, only 6.25% of the original sample remains undecayed. This can be expressed mathematically as:
\[ N = N_0 \times \left( \frac{6.25}{100} \right) \]
where \( N \) is the remaining quantity, and \( N_0 \) is the initial quantity.

### Step 2: Convert the percentage to a fraction
Convert 6.25% to a fraction:
\[ \frac{6.25}{100} = \frac{1}{16} \]
This means:
\[ N = N_0 \times \frac{1}{16} \]

### Step 3: Use the radioactive decay formula
The radioactive decay can be described by the equation:
\[ N = N_0 e^{-\lambda t} \]
Substituting the values we have:
\[ N_0 \times \frac{1}{16} = N_0 e^{-\lambda \cdot 4} \]

### Step 4: Simplify the equation
We can cancel \( N_0 \) from both sides (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{16} = e^{-\lambda \cdot 4} \]

### Step 5: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ \ln\left(\frac{1}{16}\right) = -\lambda \cdot 4 \]

### Step 6: Simplify the logarithm
Using the property of logarithms:
\[ \ln\left(\frac{1}{16}\right) = \ln(1) - \ln(16) = 0 - \ln(16) = -\ln(16) \]
Thus, we have:
\[ -\ln(16) = -\lambda \cdot 4 \]
This simplifies to:
\[ \lambda = \frac{\ln(16)}{4} \]

### Step 7: Calculate \( \lambda \)
Since \( 16 = 2^4 \), we can express the logarithm:
\[ \lambda = \frac{\ln(2^4)}{4} = \frac{4 \ln(2)}{4} = \ln(2) \]

### Step 8: Calculate the half-life
The half-life \( t_{1/2} \) is given by the formula:
\[ t_{1/2} = \frac{0.693}{\lambda} \]
Substituting \( \lambda = \ln(2) \):
\[ t_{1/2} = \frac{0.693}{\ln(2)} \]

### Step 9: Calculate the mean life
The mean life \( \tau \) is given by:
\[ \tau = \frac{1}{\lambda} \]
Substituting \( \lambda = \ln(2) \):
\[ \tau = \frac{1}{\ln(2)} \]

### Step 10: Calculate the remaining amount after 8 hours
After 8 hours, which is 2 half-lives:
\[ N = N_0 \left(\frac{1}{2}\right)^2 = N_0 \times \frac{1}{4} \]
To find the percentage:
\[ \text{Remaining percentage} = \left(\frac{N}{N_0}\right) \times 100 = \frac{1}{4} \times 100 = 25\% \]

### Final Results
1. The decay constant \( \lambda = \ln(2) \).
2. The half-life \( t_{1/2} = 1 \) hour.
3. The mean life \( \tau = \frac{1}{\ln(2)} \).
4. After 8 hours, 25% of the original sample remains.