A radioactive nucleus can decay by two differnet processess. The mean value period for the first process is `t_(1)` and that the second process is `t_(2)` .The effective mean value period for the two processes is .

Options

A. `(t_(1) + t_(2))/(2)`
B. `t_(1) + t_(2)`
C. `sqrt t_(1)t_(2)`
D. `(t_(1) + t_(2))/(t_(1) + t_2)`

Explanation

To find the effective mean value period for a radioactive nucleus that can decay by two different processes, we can follow these steps:

### Step 1: Understand the decay processes
We have two decay processes characterized by their mean value periods:
- For the first process, the mean value period is \( T_1 \).
- For the second process, the mean value period is \( T_2 \).

### Step 2: Relate mean value periods to decay constants
The decay constant (\( \lambda \)) is related to the mean value period (\( T \)) by the formula:
\[ \lambda = \frac{1}{T} \]
Thus, we can express the decay constants for both processes:
- For the first process:
\[ \lambda_1 = \frac{1}{T_1} \]
- For the second process:
\[ \lambda_2 = \frac{1}{T_2} \]

### Step 3: Write the equation for the total decay rate
The total decay rate (\( \frac{dn}{dt} \)) for the nucleus can be expressed as:
\[ \frac{dn}{dt} = -\lambda_1 n - \lambda_2 n \]
This can be rewritten as:
\[ \frac{dn}{dt} = -(\lambda_1 + \lambda_2) n \]

### Step 4: Substitute the decay constants
Substituting the expressions for \( \lambda_1 \) and \( \lambda_2 \):
\[ \frac{dn}{dt} = -\left(\frac{1}{T_1} + \frac{1}{T_2}\right) n \]

### Step 5: Find the effective decay constant
The effective decay constant (\( \lambda_{\text{effective}} \)) can be defined as:
\[ \lambda_{\text{effective}} = \lambda_1 + \lambda_2 = \frac{1}{T_1} + \frac{1}{T_2} \]

### Step 6: Relate effective decay constant to effective mean value period
The effective mean value period (\( T_{\text{effective}} \)) is related to the effective decay constant by:
\[ \lambda_{\text{effective}} = \frac{1}{T_{\text{effective}}} \]
Thus, we can express \( T_{\text{effective}} \) as:
\[ T_{\text{effective}} = \frac{1}{\lambda_{\text{effective}}} = \frac{1}{\left(\frac{1}{T_1} + \frac{1}{T_2}\right)} \]

### Step 7: Simplify the expression
To simplify the expression for \( T_{\text{effective}} \):
\[ T_{\text{effective}} = \frac{1}{\left(\frac{1}{T_1} + \frac{1}{T_2}\right)} = \frac{T_1 T_2}{T_1 + T_2} \]

### Final Result
The effective mean value period for the two processes is:
\[ T_{\text{effective}} = \frac{T_1 T_2}{T_1 + T_2} \]