Half - life for radioactive `.^(14)C` is 5760 years. In how many years 200 mg of `.^(14)C` will be reduced to 25 mg ?

Options

A. 17280 years
B. 23040 years
C. 5760 years
D. 11524 years

Answer

To solve the problem of how long it takes for 200 mg of radioactive \(^{14}C\) to reduce to 25 mg, we will use the concept of half-life and the decay constant. Here are the steps to find the solution:

### Step 1: Understand the half-life concept
The half-life (\(t_{1/2}\)) of a radioactive substance is the time required for half of the substance to decay. For \(^{14}C\), the half-life is given as 5760 years.

### Step 2: Determine the decay constant (\(\lambda\))
The decay constant (\(\lambda\)) can be calculated using the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
Substituting the given half-life:
\[ \lambda = \frac{0.693}{5760} \approx 1.204 \times 10^{-4} \text{ years}^{-1} \]

### Step 3: Use the decay formula
The amount of substance remaining after time \(t\) can be expressed as:
\[ N = N_0 e^{-\lambda t} \]
Where:
- \(N_0\) is the initial amount (200 mg)
- \(N\) is the final amount (25 mg)
- \(t\) is the time we want to find

### Step 4: Rearranging the decay formula
We can rearrange the formula to solve for \(t\):
\[ t = -\frac{1}{\lambda} \ln\left(\frac{N}{N_0}\right) \]

### Step 5: Substitute the values into the equation
Substituting \(N = 25\) mg, \(N_0 = 200\) mg, and \(\lambda \approx 1.204 \times 10^{-4}\):
\[ t = -\frac{1}{1.204 \times 10^{-4}} \ln\left(\frac{25}{200}\right) \]

### Step 6: Calculate the logarithm
Calculating the fraction:
\[ \frac{25}{200} = 0.125 \]
Now, calculate the natural logarithm:
\[ \ln(0.125) \approx -2.0794 \]

### Step 7: Substitute back into the time equation
Now substitute back into the equation for \(t\):
\[ t = -\frac{1}{1.204 \times 10^{-4}} \times (-2.0794) \]
\[ t \approx \frac{2.0794}{1.204 \times 10^{-4}} \approx 17249.17 \text{ years} \]

### Step 8: Round the answer
Rounding to the nearest year, we find:
\[ t \approx 17249 \text{ years} \]

### Final Answer
The time required for 200 mg of \(^{14}C\) to reduce to 25 mg is approximately **17249 years**.

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