A sample of a radioactive substance undergoes 80% decomposition in 345 minutes. Its half life is …………minutes

Options

A. (a) `(In2)/(In5)xx345`
B. (b) `(In5)/(In2)xx345`
C. (c) `(In5)/(In4)xx345`
D. (d) `(In4)/(In5)xx345`

Answer

To find the half-life of a radioactive substance that undergoes 80% decomposition in 345 minutes, we can follow these steps:

### Step 1: Understand the Problem
We know that the substance undergoes 80% decomposition, which means that 20% of the original substance remains after 345 minutes. If we start with 100 grams of the substance, after 345 minutes, we have 20 grams left.

### Step 2: Set Up the First-Order Reaction Equation
Radioactive decay is a first-order reaction. The relationship between the rate constant (k) and the half-life (T_half) for a first-order reaction is given by the formula:
\[ T_{half} = \frac{\ln(2)}{k} \]

### Step 3: Calculate the Rate Constant (k)
We can use the first-order kinetics formula:
\[ k = \frac{1}{t} \ln\left(\frac{[A_0]}{[A_t]}\right) \]
Where:
- \( [A_0] \) is the initial concentration (100 grams)
- \( [A_t] \) is the concentration at time t (20 grams)
- \( t \) is the time (345 minutes)

Substituting the values:
\[ k = \frac{1}{345} \ln\left(\frac{100}{20}\right) \]
\[ k = \frac{1}{345} \ln(5) \]

### Step 4: Substitute k into the Half-Life Formula
Now, we substitute the value of k into the half-life formula:
\[ T_{half} = \frac{\ln(2)}{k} \]
Substituting for k:
\[ T_{half} = \frac{\ln(2)}{\frac{1}{345} \ln(5)} \]
\[ T_{half} = 345 \cdot \frac{\ln(2)}{\ln(5)} \]

### Step 5: Final Expression for Half-Life
Thus, the half-life of the radioactive substance is:
\[ T_{half} = 345 \cdot \frac{\ln(2)}{\ln(5)} \text{ minutes} \]