Mean Free Path NEET 2026: Number Density Ratio Explained

What does the mean free path NEET 2026 question ask?

In the mean free path NEET 2026 question, molecules of ideal gas A travel, on average, half as far between collisions as molecules of ideal gas B. Their molecular diameter is twice that of gas B. Find the relationship between the gases’ number densities.

A) $n_{A} = n_{B}$
B) $n_{A} = 2 n_{B}$
C)

n_{A} = \frac{n_{B}}{4}

n_{A} = \frac{n_{B}}{2}

The correct choice is D, obtained by keeping the diameter squared in the formula. The question belongs to Physics → Kinetic Theory of Gases → Mean Free Path.

The question bank labels this MCQ medium difficulty, with a 90-second solving target. These are question-bank metadata, not an official NTA difficulty rating or time allowance.

Which mean-free-path formula should I use?

Use the ideal-gas mean-free-path formula below. Track both molecular diameter and number density in its denominator:

λ = \frac{1}{\sqrt{2} π d^{2} n}

The symbols mean:

λ & : mean free path d & : molecular diameter n & = \frac{N}{V} : number density

Mean free path is the average distance travelled between successive collisions. Number density is the number of molecules per unit volume, not the total number of molecules or mass density.

The square-root-of-two factor and pi are common to both gases. Therefore, the official solution uses:

λ \propto \frac{1}{d^{2} n}

Translate the givens before cancelling:

λ_{A} = \frac{λ_{B}}{2}, d_{A} = 2 d_{B}

Neither diameter nor number density may be treated as unchanged. The diameter changes explicitly; the number-density relationship is what we must find.

How do I form the ratio and obtain option D?

Divide the mean free path of gas A by that of gas B, then substitute the givens. Because the formula is an inverse relation, the diameter-squared and number-density product for gas B moves to the numerator.

Start with the complete expressions:

\frac{λ_{A}}{λ_{B}} = \frac{\frac{1}{\sqrt{2} π d_{A}^{2} n_{A}}}{\frac{1}{\sqrt{2} π d_{B}^{2} n_{B}}}

Cancel the common constants. Dividing by a fraction means multiplying by its reciprocal:

\frac{λ_{A}}{λ_{B}} = \frac{d_{B}^{2} n_{B}}{d_{A}^{2} n_{A}}

Substitute the given mean-free-path ratio and molecular diameter:

\frac{1}{2} = \frac{d_{B}^{2} n_{B}}{(2 d_{B})^{2} n_{A}}

Square the entire diameter factor, including the numerical multiplier: $(2 d_{B})^{2} = 4 d_{B}^{2}$

Insert that result, then cancel the common diameter-squared term:

\frac{1}{2} = \frac{d_{B}^{2} n_{B}}{4 d_{B}^{2} n_{A}}

\frac{1}{2} = \frac{n_{B}}{4 n_{A}}

Cross-multiply explicitly so that the factor of four does not get lost: $4 n_{A} = 2 n_{B}$

Divide both sides by two: $2 n_{A} = n_{B}$

Finally, isolate the number density of gas A:

n_{A} = \frac{n_{B}}{2}

Option D: the number density of gas A is half that of gas B.

How can I check that option D is physically consistent?

Option D makes the denominator product twice as large, so mean free path becomes half as large. Doubling molecular diameter multiplies its square by four; with number density unchanged, this alone would reduce mean free path to one-quarter.

The derived result halves number density, offsetting part of that reduction. Check the combined denominator product:

\frac{d_{A}^{2} n_{A}}{d_{B}^{2} n_{B}} = \frac{(2 d_{B})^{2} (n_{B} / 2)}{d_{B}^{2} n_{B}} = 4 \times \frac{1}{2} = 2

Mean free path is inversely proportional to this product. Hence:

\frac{λ_{A}}{λ_{B}} = \frac{1}{2}

That exactly matches the question. The reusable check is to compare the whole product, not diameter alone: $d^{2} n$

Why does dropping the diameter square give option A?

Option A follows from treating diameter as a first-power term. The algebra can be correct while the starting formula is wrong, so checking only the cross-multiplication will not catch this error.

The incorrect relation is:

λ \propto \frac{1}{d n} (incorrect)

Using it gives the faulty substitution:

\frac{1}{2} = \frac{d_{B} n_{B}}{2 d_{B} n_{A}} = \frac{n_{B}}{2 n_{A}}

Cross-multiplication would then produce:

2 n_{A} = 2 n_{B} \Rightarrow n_{A} = n_{B} (wrong option A)

Collision cross-section scales with diameter squared. Doubling molecular diameter therefore introduces a factor of four, not two.

Prevent this error by writing the substituted diameter in brackets, with the square outside, before simplifying. Do not insert the factor two mentally: $(2 d_{B})^{2}$

How do I solve three related mean-free-path questions?

Use the same ratio method, keeping the diameter square visible until substitution is complete. These are original practice questions from Kinetic Theory of Gases, not additional verified PYQs. Attempt each prompt before reading its worked answer.

Question 1: If molecular diameter doubles while number density stays fixed, what happens to mean free path?

Write the new-to-original mean-free-path ratio. The unchanged number density cancels:

\frac{λ^{'}}{λ} = \frac{d^{2} n}{(2 d)^{2} n} = \frac{1}{4}

Answer: mean free path becomes one-quarter of its original value. The diameter-squared term makes the denominator four times larger.

Question 2: If number density halves while molecular diameter stays fixed, what happens to mean free path?

The diameter-squared terms cancel. Halving the denominator doubles its reciprocal:

\frac{λ^{'}}{λ} = \frac{d^{2} n}{d^{2} (n / 2)} = 2

Answer: mean free path doubles. Fewer molecules per unit volume allow a greater average distance between collisions.

Question 3: Gas X has half the molecular diameter of gas Y and twice its mean free path. Find their number-density ratio.

Keep the order X over Y for mean free path. The inverse formula reverses the corresponding products:

2 = \frac{d_{Y}^{2} n_{Y}}{(d_{Y} / 2)^{2} n_{X}} = \frac{4 n_{Y}}{n_{X}}

2 n_{X} = 4 n_{Y} \Rightarrow n_{X} = 2 n_{Y}

\frac{n_{X}}{n_{Y}} = 2

Answer: gas X has twice the number density of gas Y. Its diameter-squared factor is one-quarter, so doubling density makes the combined product one-half, giving twice the mean free path.

For your next attempt: write the formula, form the ratio, square the diameter factor, then check the product: $d^{2} n$

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Frequently asked questions

What is the mean free path formula for NEET?

Use λ = 1/(√2 × π × d² × n), where d is molecular diameter and n = N/V is number density. Mean free path is the average distance a molecule travels between successive collisions. Number density means molecules per unit volume, not mass density.

How do I solve the mean free path NEET 2026 question?

The givens are λ_A = λ_B/2 and d_A = 2d_B. Form the ratio λ_A/λ_B = (d_B² × n_B)/(d_A² × n_A), which gives 1/2 = n_B/(4n_A). Cross-multiplication gives n_A = n_B/2, so option D is correct.

Why is option A wrong in the mean free path question?

Option A comes from incorrectly using λ ∝ 1/(dn) instead of λ ∝ 1/(d²n). Collision cross-section scales with diameter squared, so doubling diameter contributes a factor of four, not two. Write (2d_B)² before simplifying to avoid this error.

What happens to mean free path if molecular diameter doubles?

If number density stays fixed, mean free path becomes one-quarter of its original value. This follows from λ ∝ 1/d²: doubling diameter makes the denominator four times larger.