If 20% of a radioactive sample decays in 10 days, then the amount of substance left after 20 days will be approximately

Options

A. 0.6
B. 0.61
C. 0.63
D. 0.7

Explanation

To solve the problem, we need to determine the amount of radioactive substance left after 20 days, given that 20% of the sample decays in the first 10 days.

### Step-by-Step Solution:

1. **Understanding the Decay**:
- If 20% of the sample decays in 10 days, then 80% of the sample remains after 10 days.
- Let’s assume the initial amount of the substance is \( N_0 = 100 \) units. After 10 days, the remaining amount \( N \) is:
\[ N = N_0 - 0.2N_0 = 0.8N_0 = 80 \text{ units} \]

2. **Using the Decay Formula**:
- The decay of a radioactive substance can be described by the formula:
\[ N = N_0 e^{-\lambda t} \]
- Rearranging gives:
\[ \frac{N}{N_0} = e^{-\lambda t} \]

3. **Finding the Decay Constant \( \lambda \)**:
- For the first 10 days, we have:
\[ \frac{80}{100} = e^{-\lambda \cdot 10} \]
- This simplifies to:
\[ 0.8 = e^{-10\lambda} \]
- Taking the natural logarithm of both sides:
\[ \ln(0.8) = -10\lambda \]
- Thus, we can solve for \( \lambda \):
\[ \lambda = -\frac{\ln(0.8)}{10} \]

4. **Calculating the Remaining Amount After 20 Days**:
- Now, we need to find the amount remaining after 20 days:
\[ N_{20} = N_0 e^{-\lambda \cdot 20} \]
- Substituting \( \lambda \):
\[ N_{20} = 100 e^{-20\lambda} = 100 e^{-20 \left(-\frac{\ln(0.8)}{10}\right)} = 100 e^{2\ln(0.8)} = 100 (0.8^2) \]
- Calculating \( 0.8^2 \):
\[ 0.8^2 = 0.64 \]
- Therefore:
\[ N_{20} = 100 \times 0.64 = 64 \text{ units} \]

### Final Answer:
The amount of substance left after 20 days will be approximately **64 units**.