Question:
The density of a gas is 1 molecule cm\(^{-3}\). If the molecular diameter is \(1 \times 10^{-8}\) cm, then the mean free path of the molecules is:
The density of a gas is 1 molecule cm\(^{-3}\). If the molecular diameter is \(1 \times 10^{-8}\) cm, then the mean free path of the molecules is:
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Mean free path depends inversely on number density and square of molecular diameter: \( \lambda = \frac{1}{\sqrt{2}n\pi d^2} \).
Updated On: Jun 19, 2026
- \(\frac{1}{\sqrt{2\pi}} \times 10^{14}\, m\)
- \(\frac{1}{2\pi} \times 10^{13}\, m\)
- \(\frac{1}{\sqrt{6\pi}} \times 10^{14}\, m\)
- \(\frac{1}{\sqrt{2\pi}} \times 10^{13}\, m\)
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The Correct Option is A
Solution and Explanation
Step 1: Use formula for mean free path.
\[ \lambda = \frac{1}{\sqrt{2} \, n \, \pi d^2} \]
Step 2: Convert given values into SI units.
Number density: \[ n = 1 \, \text{molecule cm}^{-3} = 10^{6} \, \text{molecules m}^{-3} \] Diameter: \[ d = 1 \times 10^{-8} \, cm = 10^{-10} \, m \]
Step 3: Compute \(d^2\).
\[ d^2 = 10^{-20} \, m^2 \]
Step 4: Substitute into formula.
\[ \lambda = \frac{1}{\sqrt{2} \times 10^{6} \times \pi \times 10^{-20}} \]
Step 5: Simplify powers of 10.
\[ \lambda = \frac{1}{\sqrt{2}\pi \times 10^{-14}} \]
Step 6: Final expression.
\[ \lambda = \frac{1}{\sqrt{2\pi}} \times 10^{14} \, m \]
Final Answer: \[ \boxed{\frac{1}{\sqrt{2\pi}} \times 10^{14}\, m} \]
\[ \lambda = \frac{1}{\sqrt{2} \, n \, \pi d^2} \]
Step 2: Convert given values into SI units.
Number density: \[ n = 1 \, \text{molecule cm}^{-3} = 10^{6} \, \text{molecules m}^{-3} \] Diameter: \[ d = 1 \times 10^{-8} \, cm = 10^{-10} \, m \]
Step 3: Compute \(d^2\).
\[ d^2 = 10^{-20} \, m^2 \]
Step 4: Substitute into formula.
\[ \lambda = \frac{1}{\sqrt{2} \times 10^{6} \times \pi \times 10^{-20}} \]
Step 5: Simplify powers of 10.
\[ \lambda = \frac{1}{\sqrt{2}\pi \times 10^{-14}} \]
Step 6: Final expression.
\[ \lambda = \frac{1}{\sqrt{2\pi}} \times 10^{14} \, m \]
Final Answer: \[ \boxed{\frac{1}{\sqrt{2\pi}} \times 10^{14}\, m} \]
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