Two species of radioactive atoms are mixed in equal number. The disintegration of the first species is `lambda` and of the second is `lambda`/3. After a long time the mixture will behave as a species with mean life of approximately

Options

A. `0.70 // lambda`
B. `2.10 // lambda `
C. `1.00 // lambda `
D. `0.52 // lambda `

Explanation

To solve the problem step by step, we need to find the mean life of a mixture of two radioactive species with different disintegration constants.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two species of radioactive atoms mixed in equal numbers. The disintegration constant (decay constant) for the first species is denoted as \( \lambda \), and for the second species, it is \( \frac{\lambda}{3} \).

2. **Mean Life Calculation**:
The mean life \( \tau \) of a radioactive species is related to its decay constant \( \lambda \) by the formula:
\[ \tau = \frac{1}{\lambda} \]
Therefore, the mean life for the first species is:
\[ \tau_1 = \frac{1}{\lambda} \]
And for the second species:
\[ \tau_2 = \frac{1}{\frac{\lambda}{3}} = \frac{3}{\lambda} \]

3. **Finding the Average Mean Life**:
Since both species are mixed in equal numbers, we can calculate the average mean life \( \tau_{avg} \) of the mixture. The average mean life can be calculated using the formula:
\[ \tau_{avg} = \frac{\tau_1 + \tau_2}{2} \]
Substituting the values we found:
\[ \tau_{avg} = \frac{\frac{1}{\lambda} + \frac{3}{\lambda}}{2} = \frac{\frac{4}{\lambda}}{2} = \frac{2}{\lambda} \]

4. **Final Result**:
Therefore, the mean life of the mixture will behave approximately as:
\[ \tau_{avg} \approx \frac{2}{\lambda} \]

### Conclusion:
The mean life of the mixture of the two radioactive species is approximately \( \frac{2}{\lambda} \).