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 of how much radioactive material is left after 20 days, given that 10% decays in 5 days, we can follow these steps:

### Step 1: Understand the Decay Information
We know that 10% of the material decays in 5 days. This means that 90% of the original material remains after 5 days.

### Step 2: Use the Exponential Decay Formula
The formula for radioactive decay is given by:
\[ N(t) = N_0 e^{-\lambda t} \]
where:
- \( N(t) \) is the amount of material remaining at time \( t \),
- \( N_0 \) is the initial amount of material,
- \( \lambda \) is the decay constant,
- \( t \) is the time.

### Step 3: Set Up the Equation for 5 Days
After 5 days, we have:
\[ N(5) = N_0 e^{-5\lambda} \]
Since 90% of the material remains, we can write:
\[ N(5) = 0.9 N_0 \]
Thus, we have:
\[ 0.9 N_0 = N_0 e^{-5\lambda} \]

### Step 4: Simplify the Equation
We can cancel \( N_0 \) from both sides (assuming \( N_0 \neq 0 \)):
\[ 0.9 = e^{-5\lambda} \]

### Step 5: Take the Natural Logarithm
Taking the natural logarithm of both sides gives:
\[ \ln(0.9) = -5\lambda \]
From this, we can solve for \( \lambda \):
\[ \lambda = -\frac{\ln(0.9)}{5} \]

### Step 6: Set Up the Equation for 20 Days
Now, we want to find the amount remaining after 20 days:
\[ N(20) = N_0 e^{-20\lambda} \]

### Step 7: Substitute for \( \lambda \)
Substituting our expression for \( \lambda \):
\[ N(20) = N_0 e^{-20 \left(-\frac{\ln(0.9)}{5}\right)} \]
This simplifies to:
\[ N(20) = N_0 e^{4\ln(0.9)} \]
Using the property of logarithms \( e^{\ln(a^b)} = a^b \):
\[ N(20) = N_0 (0.9)^4 \]

### Step 8: Calculate \( (0.9)^4 \)
Now we calculate \( (0.9)^4 \):
\[ (0.9)^4 = 0.6561 \]

### Step 9: Find the Percentage Remaining
To find the percentage of the original material left:
\[ \text{Percentage remaining} = 0.6561 \times 100\% \approx 65.61\% \]

### Conclusion
After 20 days, approximately **65.61%** of the original radioactive material remains.

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