The half-life of a radioactive substance is 5000 years. In how many years, its activity will decay to 0.2 times of its initial value ? Given `"log"_10 5` = 0.6990.

Options

A. `1.16xx10^5` years
B. `1.16xx10^4` years
C. `1.16xx10^3` years
D. `1.16xx10^6` years

Answer

To find out how many years it will take for the activity of a radioactive substance to decay to 0.2 times its initial value, we can use the concept of half-life and the decay formula.

### Step-by-Step Solution:

1. **Understanding Half-Life**:
The half-life (t₁/₂) of a radioactive substance is the time taken for half of the substance to decay. In this case, the half-life is given as 5000 years.

2. **Decay Formula**:
The relationship between the remaining quantity (N) of a radioactive substance and its initial quantity (N₀) after time (t) can be expressed as:
\[ N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]
We want to find the time (t) when the activity decays to 0.2 times its initial value (N = 0.2 N₀).

3. **Setting Up the Equation**:
Substitute N = 0.2 N₀ into the decay formula:
\[ 0.2 N_0 = N_0 \left(\frac{1}{2}\right)^{\frac{t}{5000}} \]
We can cancel N₀ from both sides (assuming N₀ ≠ 0):
\[ 0.2 = \left(\frac{1}{2}\right)^{\frac{t}{5000}} \]

4. **Taking Logarithm**:
To solve for t, take the logarithm (base 10) of both sides:
\[ \log(0.2) = \frac{t}{5000} \log\left(\frac{1}{2}\right) \]

5. **Expressing Logarithms**:
We can express log(0.2) as:
\[ \log(0.2) = \log\left(\frac{1}{5}\right) = -\log(5) \]
Therefore, we have:
\[ -\log(5) = \frac{t}{5000} \log\left(\frac{1}{2}\right) \]

6. **Finding log(1/2)**:
We know that:
\[ \log\left(\frac{1}{2}\right) = -\log(2) \]
Using the change of base formula, we can express log(2) in terms of log(5):
\[ \log(2) = \log(10) - \log(5) \quad (\text{since } \log(10) = 1) \]
Thus:
\[ \log(2) = 1 - 0.6990 = 0.3010 \]
Therefore:
\[ \log\left(\frac{1}{2}\right) = -0.3010 \]

7. **Substituting Values**:
Now substitute back into the equation:
\[ -\log(5) = \frac{t}{5000} \cdot (-0.3010) \]
This simplifies to:
\[ \log(5) = \frac{t \cdot 0.3010}{5000} \]

8. **Solving for t**:
Rearranging gives:
\[ t = \frac{5000 \cdot \log(5)}{0.3010} \]
Substituting the value of log(5):
\[ t = \frac{5000 \cdot 0.6990}{0.3010} \]

9. **Calculating t**:
Performing the calculation:
\[ t \approx \frac{3495}{0.3010} \approx 11616.61 \text{ years} \]
Rounding gives:
\[ t \approx 1.16 \times 10^4 \text{ years} \]

### Final Answer:
The time required for the activity to decay to 0.2 times its initial value is approximately **11616.61 years** or **1.16 × 10^4 years**.