It is observerd that only `0.39%` of the original radioactive sample remains undecayed after eight hours. Hence,

Options

A. the half-life of that substance is `1h`
B. the mean-life of the substance is `[1//(log2)]h`
C. decay constant of the substance is `(log2)h^(-1)`
D. if the number of radioactive nuclei of the this substance at a given instant is `10`, then the number left after `30 min` would be `7.5`

Explanation

To solve the problem, we need to determine the half-life of a radioactive substance based on the information that only 0.39% of the original sample remains undecayed after 8 hours.

### Step-by-Step Solution:

1. **Understanding the Decay Formula**:
The number of undecayed nuclei \( N \) at time \( t \) can be expressed as:
\[ N = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial quantity, \( \lambda \) is the decay constant, and \( t \) is the time.

2. **Setting Up the Equation**:
Given that \( N = 0.39\% \) of \( N_0 \), we can write:
\[ 0.0039 N_0 = N_0 e^{-\lambda \cdot 8} \]
Dividing both sides by \( N_0 \):
\[ 0.0039 = e^{-\lambda \cdot 8} \]

3. **Taking the Natural Logarithm**:
To solve for \( \lambda \), take the natural logarithm of both sides:
\[ \ln(0.0039) = -\lambda \cdot 8 \]
Rearranging gives:
\[ \lambda = -\frac{\ln(0.0039)}{8} \]

4. **Calculating \( \lambda \)**:
Calculate \( \ln(0.0039) \):
\[ \ln(0.0039) \approx -5.55 \]
Therefore:
\[ \lambda = -\frac{-5.55}{8} \approx 0.69375 \, \text{h}^{-1} \]

5. **Finding the Half-Life**:
The half-life \( t_{1/2} \) is related to the decay constant by the formula:
\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]
Substituting the value of \( \lambda \):
\[ t_{1/2} = \frac{0.693}{0.69375} \approx 0.999 \, \text{hours} \approx 1 \, \text{hour} \]

6. **Verifying Other Options**:
- **Mean Life**: The mean life \( \tau \) is given by \( \tau = \frac{1}{\lambda} \):
\[ \tau = \frac{1}{0.69375} \approx 1.44 \, \text{hours} \]
- **Decay Constant**: The decay constant \( \lambda \) can also be expressed in terms of half-life:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \approx 0.69375 \]
- **Remaining Nuclei Calculation**: For \( N_0 = 10 \) after 30 minutes (0.5 hours):
\[ N = 10 e^{-0.69375 \cdot 0.5} \approx 10 e^{-0.346875} \approx 10 \cdot 0.707 \approx 7.07 \]

### Conclusion:
- The half-life of the substance is approximately 1 hour.
- The mean life is approximately 1.44 hours.
- The decay constant is correctly calculated.
- The number of nuclei remaining after 30 minutes is approximately 7.07, which does not match the option of 7.5.