A radioactive sample is decayed 20% in one day. After next two days, the undecayed percentage of nuclei will be

Options

A. A. 0.512
B. B. 0.4
C. C. 0.488
D. D. 0.6

Answer

To solve the problem, we need to determine the undecayed percentage of a radioactive sample after it has decayed 20% in one day and then find the remaining percentage after the next two days.

### Step-by-Step Solution:

1. **Understand the Initial Decay**:
- The radioactive sample decays by 20% in one day.
- This means that after one day, 80% of the original sample remains undecayed.
- Let the initial amount of the sample be \( N_0 \).
- After one day, the remaining amount \( N_1 \) can be expressed as:
\[ N_1 = N_0 \times (1 - 0.20) = N_0 \times 0.80 \]

2. **Determine the Decay Constant (\( \lambda \))**:
- The relationship between the remaining amount and the decay constant is given by:
\[ N = N_0 \times e^{-\lambda t} \]
- For \( t = 1 \) day, we can write:
\[ 0.80 N_0 = N_0 \times e^{-\lambda \cdot 1} \]
- Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ 0.80 = e^{-\lambda} \]
- Taking the natural logarithm of both sides:
\[ -\lambda = \ln(0.80) \]
- Thus:
\[ \lambda = -\ln(0.80) \]

3. **Calculate the Remaining Amount After Two More Days**:
- Now we want to find the remaining amount after a total of 3 days (1 day + 2 days).
- Using the decay formula again:
\[ N_3 = N_0 \times e^{-\lambda \cdot 3} \]
- Substituting \( \lambda \):
\[ N_3 = N_0 \times e^{-3 \ln(0.80)} = N_0 \times (0.80)^3 \]
- Calculating \( (0.80)^3 \):
\[ (0.80)^3 = 0.512 \]
- Therefore, after 3 days:
\[ N_3 = N_0 \times 0.512 \]

4. **Calculate the Undecayed Percentage**:
- The undecayed percentage of the sample after 3 days is:
\[ \text{Undecayed Percentage} = \frac{N_3}{N_0} \times 100 = 0.512 \times 100 = 51.2\% \]

5. **Final Calculation of Undecayed Percentage**:
- The remaining undecayed percentage of nuclei after 3 days is:
\[ \text{Undecayed Percentage} = 51.2\% \]

### Summary of Results:
- The undecayed percentage of nuclei after 3 days is approximately **51.2%**.