Indium -112 is radioactive and has a very short half-life (`t_(1//2)=14` min). Its decay constant and average life are repectively:

Options

A. 0.0495 `"min"^(-1)`, 9.7 min
B. `0.495 "min"^(-1)`, 20.2 min
C. `9.7 "min"^(-1)`,20.2 min
D. `0.0495 "min"^(-1)`,20.2 min

Answer

To find the decay constant and average life of Indium-112, we can use the relationships between half-life, decay constant, and average life. Here’s a step-by-step solution:

### Step 1: Identify the half-life
The half-life (\( t_{1/2} \)) of Indium-112 is given as 14 minutes.

### Step 2: Calculate the decay constant
The decay constant (\( \lambda \)) is related to the half-life by the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
Substituting the value of half-life:
\[ \lambda = \frac{0.693}{14 \text{ min}} \]
Calculating this gives:
\[ \lambda = \frac{0.693}{14} \approx 0.0495 \text{ min}^{-1} \]

### Step 3: Calculate the average life
The average life (\( t_{avg} \)) is related to the decay constant by the formula:
\[ t_{avg} = \frac{1}{\lambda} \]
Substituting the value of decay constant:
\[ t_{avg} = \frac{1}{0.0495 \text{ min}^{-1}} \]
Calculating this gives:
\[ t_{avg} \approx 20.20 \text{ min} \]

### Final Results
- Decay constant (\( \lambda \)): \( 0.0495 \text{ min}^{-1} \)
- Average life (\( t_{avg} \)): \( 20.20 \text{ min} \)