If `10%` of a radioactive material decays in `5` days, then the amount of original material left after `20` days is approximately.

Options

A. `60%`
B. `65%`
C. `70%`
D. `75%`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the decay process
We know that if 10% of a radioactive material decays in 5 days, then 90% of the original material remains after 5 days.

### Step 2: Set up the equation
Let \( N_0 \) be the initial amount of radioactive material. After 5 days, the remaining amount \( N \) can be expressed as:
\[ N = N_0 \times 0.9 \]

### Step 3: Use the decay formula
The decay of radioactive material can be described by the equation:
\[ N = N_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant and \( t \) is time.

### Step 4: Substitute known values
From Step 2, we can substitute into the decay formula:
\[ N_0 \times 0.9 = N_0 e^{-\lambda \cdot 5} \]
We can cancel \( N_0 \) from both sides (assuming \( N_0 \neq 0 \)):
\[ 0.9 = e^{-\lambda \cdot 5} \]

### Step 5: Solve for \( \lambda \)
Taking the natural logarithm on both sides:
\[ \ln(0.9) = -\lambda \cdot 5 \]
Thus,
\[ \lambda = -\frac{\ln(0.9)}{5} \]

### Step 6: Calculate remaining material after 20 days
Now we want to find the amount remaining after 20 days. Using the decay formula again:
\[ N_{20} = N_0 e^{-\lambda \cdot 20} \]

### Step 7: Substitute \( \lambda \) into the equation
Substituting \( \lambda \):
\[ N_{20} = N_0 e^{-20 \cdot \left(-\frac{\ln(0.9)}{5}\right)} \]
This simplifies to:
\[ N_{20} = N_0 e^{4 \ln(0.9)} = N_0 (0.9)^4 \]

### Step 8: Calculate \( (0.9)^4 \)
Calculating \( (0.9)^4 \):
\[ (0.9)^4 = 0.6561 \]

### Step 9: Final amount remaining
Thus, the amount of original material left after 20 days is approximately:
\[ N_{20} \approx 0.6561 N_0 \]

### Conclusion
Therefore, approximately 65.61% of the original material remains after 20 days.