The activity of a radioactive element decreases in 10 years to 1/5 of initial activity `A_(0)`. After further next 10 years, its activity will be

Options

A. `(A_(0))/(4)`
B. `(A_(0))/(10)`
C. `(A_(0))/(15)`
D. `(A_(0))/(25)`

Answer

To solve the problem, we need to find the activity of a radioactive element after a total of 20 years, given that its activity decreases to \( \frac{1}{5} \) of its initial activity \( A_0 \) in the first 10 years.

### Step-by-Step Solution:

1. **Understanding the decay of activity**:
The activity \( A(t) \) of a radioactive substance at time \( t \) can be expressed as:
\[ A(t) = A_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

2. **Finding the decay constant \( \lambda \)**:
From the problem, we know that after 10 years, the activity decreases to \( \frac{1}{5} A_0 \). Therefore, we can set up the equation:
\[ A(10) = A_0 e^{-\lambda \cdot 10} = \frac{1}{5} A_0 \]
Dividing both sides by \( A_0 \):
\[ e^{-\lambda \cdot 10} = \frac{1}{5} \]

3. **Taking the natural logarithm**:
To solve for \( \lambda \), take the natural logarithm of both sides:
\[ -\lambda \cdot 10 = \ln\left(\frac{1}{5}\right) \]
Thus,
\[ \lambda = -\frac{1}{10} \ln\left(\frac{1}{5}\right) \]

4. **Calculating the activity after the next 10 years**:
After another 10 years (total of 20 years), the initial activity for this period is \( A(10) = \frac{1}{5} A_0 \). We can find the activity at \( t = 20 \) years:
\[ A(20) = A(10) e^{-\lambda \cdot 10} \]
Substituting \( A(10) \):
\[ A(20) = \left(\frac{1}{5} A_0\right) e^{-\lambda \cdot 10} \]
We already know that \( e^{-\lambda \cdot 10} = \frac{1}{5} \):
\[ A(20) = \left(\frac{1}{5} A_0\right) \left(\frac{1}{5}\right) = \frac{1}{25} A_0 \]

5. **Final answer**:
Therefore, the activity of the radioactive element after 20 years will be:
\[ A(20) = \frac{1}{25} A_0 \]