Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are `tau = 1// lambda` and `t_(1//2) = 0.693//lambda`. Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after `n` half lives of radioactive elements can be calculated using the relation:
`N = N_(0) ((1)/(2))^(n)`
Half life of `.^(60)Co` is 5.3 years, the time taken for 99.9% decay will be

Options

A. 0.53 years
B. 53 years
C. 530 years
D. 5300 years

Explanation

To solve the problem of determining the time taken for 99.9% decay of Cobalt-60, we can follow these steps:

### Step 1: Understand the Problem
We know that the half-life of Cobalt-60 (Co-60) is given as 5.3 years. We need to find the time taken for 99.9% of the substance to decay, which means only 0.1% of the original amount remains.

### Step 2: Calculate the Decay Constant (k)
The decay constant (k) for a first-order reaction can be calculated using the formula:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the half-life:
\[ k = \frac{0.693}{5.3 \text{ years}} \approx 0.130 \text{ year}^{-1} \]

### Step 3: Set Up the Equation for Time (t)
For first-order kinetics, the time taken for a certain amount of substance to decay can be calculated using the formula:
\[ t = \frac{2.303}{k} \log \left( \frac{N_0}{N} \right) \]
Where:
- \(N_0\) is the initial amount (100%),
- \(N\) is the remaining amount (0.1% after 99.9% decay).

### Step 4: Substitute Values into the Equation
Substituting the values into the equation:
\[ t = \frac{2.303}{0.130} \log \left( \frac{100}{0.1} \right) \]
Calculating the logarithm:
\[ \frac{100}{0.1} = 1000 \quad \text{and} \quad \log(1000) = 3 \]
Thus, we can rewrite the equation as:
\[ t = \frac{2.303}{0.130} \times 3 \]

### Step 5: Calculate the Time (t)
Now we can calculate:
\[ t = \frac{2.303 \times 3}{0.130} \approx \frac{6.909}{0.130} \approx 53.1 \text{ years} \]

### Final Answer
The time taken for 99.9% decay of Cobalt-60 is approximately **53 years**.

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