The average (mean) life of a radio nuclide which decays by parallel path is
`Aoverset(lambda_(1))(rarr)B` : `lambda_(1)=1.8xx10^(-2)sec^(-1)`
`2Aoverset(lambda_(2))(rarr)B` , `lambda_(2)=10^(-3)sec^(-1)`

Options

A. `52.63 sec`
B. `500 sec`
C. `50 sec`
D. none of these

Answer

To solve the problem of finding the average (mean) life of a radionuclide that decays by parallel paths, we can follow these steps:

### Step 1: Understand the Concept
In a parallel decay process, a radionuclide can decay through multiple pathways, each characterized by its own decay constant (λ). The average life (mean life) of the radionuclide can be calculated using the formula:

\[ \text{Average Life} = \frac{1}{\lambda_{\text{total}}} \]

where \( \lambda_{\text{total}} \) is the sum of the decay constants for all pathways.

### Step 2: Identify the Given Values
From the problem, we have two decay constants:
- \( \lambda_1 = 1.8 \times 10^{-2} \, \text{sec}^{-1} \)
- \( \lambda_2 = 10^{-3} \, \text{sec}^{-1} \)

### Step 3: Calculate the Total Decay Constant
Now we need to calculate the total decay constant \( \lambda_{\text{total}} \):

\[ \lambda_{\text{total}} = \lambda_1 + \lambda_2 \]

Substituting the values:

\[ \lambda_{\text{total}} = 1.8 \times 10^{-2} + 1.0 \times 10^{-3} \]

### Step 4: Perform the Addition
To add these two values, we can convert \( 1.0 \times 10^{-3} \) to the same exponent as \( 1.8 \times 10^{-2} \):

\[ 1.0 \times 10^{-3} = 0.1 \times 10^{-2} \]

Now we can add:

\[ \lambda_{\text{total}} = 1.8 \times 10^{-2} + 0.1 \times 10^{-2} = 1.9 \times 10^{-2} \, \text{sec}^{-1} \]

### Step 5: Calculate the Average Life
Now, we can find the average life using the total decay constant:

\[ \text{Average Life} = \frac{1}{\lambda_{\text{total}}} = \frac{1}{1.9 \times 10^{-2}} \]

### Step 6: Perform the Calculation
Calculating the average life:

\[ \text{Average Life} \approx \frac{1}{0.019} \approx 52.63 \, \text{seconds} \]

### Conclusion
Thus, the average (mean) life of the radionuclide is approximately **52.63 seconds**.

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