20% of a radioactive substances decay in 10 days . Calculate the amount of the original material left after 30 days.

Options

A. `78%`
B. `62%`
C. `51%`
D. `48%`

Explanation

To solve the problem of how much of the original radioactive substance remains after 30 days, given that 20% decays in 10 days, we can follow these steps:

### Step 1: Determine the remaining fraction after 10 days
If 20% of the substance decays in 10 days, then 80% remains. We can express this as a fraction:
\[ \text{Remaining fraction after 10 days} = 1 - 0.20 = 0.80 \]

### Step 2: Use the decay formula
The decay of a radioactive substance can be described by the equation:
\[ N = N_0 e^{-\lambda t} \]
Where:
- \(N\) is the remaining quantity of the substance,
- \(N_0\) is the initial quantity,
- \(\lambda\) is the decay constant,
- \(t\) is the time elapsed.

From the first step, we know that after 10 days:
\[ 0.80 N_0 = N_0 e^{-\lambda \cdot 10} \]
Dividing both sides by \(N_0\) (assuming \(N_0 \neq 0\)):
\[ 0.80 = e^{-\lambda \cdot 10} \]

### Step 3: Solve for the decay constant \(\lambda\)
Taking the natural logarithm of both sides:
\[ \ln(0.80) = -\lambda \cdot 10 \]
Now, solving for \(\lambda\):
\[ \lambda = -\frac{\ln(0.80)}{10} \]

Calculating \(\ln(0.80)\):
\[ \lambda \approx -\frac{-0.2231}{10} \approx 0.02231 \text{ days}^{-1} \]

### Step 4: Calculate the half-life (\(T_{1/2}\))
The half-life of a radioactive substance is given by:
\[ T_{1/2} = \frac{0.693}{\lambda} \]
Substituting the value of \(\lambda\):
\[ T_{1/2} = \frac{0.693}{0.02231} \approx 31.1 \text{ days} \]

### Step 5: Calculate the remaining amount after 30 days
Now we can use the decay formula again to find the remaining amount after 30 days:
\[ N = N_0 e^{-\lambda \cdot 30} \]
Substituting the value of \(\lambda\):
\[ N = N_0 e^{-0.02231 \cdot 30} \]
Calculating the exponent:
\[ N = N_0 e^{-0.6693} \approx N_0 \cdot 0.511 \]

### Step 6: Convert to percentage
To find the percentage of the original material left:
\[ \text{Percentage remaining} = \left(\frac{N}{N_0}\right) \times 100 \approx 0.511 \times 100 \approx 51.1\% \]

Thus, the amount of the original material left after 30 days is approximately **51.1%**.