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 determine the activity of a radioactive element after a total of 20 years, given that its activity decreases to 1/5 of its initial value after the first 10 years.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
The initial activity of the radioactive element is denoted as \( A_0 \). After 10 years, the activity decreases to \( \frac{1}{5} A_0 \).

2. **Using the Exponential Decay Formula**:
The activity of a radioactive substance can be expressed using the formula:
\[ A(t) = A_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant and \( t \) is time.

3. **Setting Up the Equation for the First 10 Years**:
After 10 years, we can set up the equation:
\[ A(10) = A_0 e^{-\lambda \cdot 10} = \frac{1}{5} A_0 \]

4. **Dividing Both Sides by \( A_0 \)**:
Dividing both sides by \( A_0 \) gives:
\[ e^{-\lambda \cdot 10} = \frac{1}{5} \]

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

6. **Calculating Activity After Another 10 Years**:
Now we want to find the activity after another 10 years (i.e., after a total of 20 years). The initial activity for this next period is \( \frac{1}{5} A_0 \). We can set up the equation:
\[ A(20) = \left(\frac{1}{5} A_0\right) e^{-\lambda \cdot 10} \]

7. **Substituting the Value of \( e^{-\lambda \cdot 10} \)**:
From step 4, we know that \( e^{-\lambda \cdot 10} = \frac{1}{5} \). Substituting this into the equation gives:
\[ A(20) = \left(\frac{1}{5} A_0\right) \cdot \frac{1}{5} \]

8. **Final Calculation**:
Therefore, we have:
\[ A(20) = \frac{1}{5} A_0 \cdot \frac{1}{5} = \frac{1}{25} A_0 \]

### Conclusion:
The activity of the radioactive element after a total of 20 years will be \( \frac{1}{25} A_0 \).