The mean free path of a molecule of He gas is `alpha`. Its mean free path along any arbitrary coordinates axis will be

Options

A. `alpha`
B. `alpha/3`
C. `alpha/sqrt3`
D. `3 alpha`

Answer

To find the mean free path of a molecule of helium gas along any arbitrary coordinate axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Mean Free Path**:
The mean free path (λ) is the average distance a molecule travels between collisions. For a gas, this distance can be represented in three dimensions along the x, y, and z axes.

2. **Defining Mean Free Paths Along Each Axis**:
Let the mean free paths along the x, y, and z axes be denoted as λ_x, λ_y, and λ_z respectively.

3. **Relationship Between Mean Free Paths**:
The total mean free path (λ) in three-dimensional space can be expressed using the Pythagorean theorem:
\[ λ = \sqrt{λ_x^2 + λ_y^2 + λ_z^2} \]

4. **Assuming Equal Mean Free Paths**:
For an isotropic gas, we can assume that the mean free paths are equal along each axis:
\[ λ_x = λ_y = λ_z = λ_a \]

5. **Substituting into the Equation**:
Substituting λ_x, λ_y, and λ_z into the equation gives:
\[ λ = \sqrt{λ_a^2 + λ_a^2 + λ_a^2} = \sqrt{3λ_a^2} = λ_a \sqrt{3} \]

6. **Relating to Given Mean Free Path**:
We know from the problem that the mean free path of the helium gas is given as α. Therefore, we can set:
\[ λ = α \]
Thus, we have:
\[ α = λ_a \sqrt{3} \]

7. **Solving for λ_a**:
Rearranging the equation to find λ_a gives:
\[ λ_a = \frac{α}{\sqrt{3}} \]

8. **Conclusion**:
Therefore, the mean free path along any arbitrary coordinate axis (x, y, or z) is:
\[ λ_a = \frac{α}{\sqrt{3}} \]

### Final Answer:
The mean free path along any arbitrary coordinate axis will be \( \frac{α}{\sqrt{3}} \).