Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength of nitrogen molecule is close to :
(Given : nitrogen molecule weight : 4.64 × 10 –26 kg,
Boltzman constant: 1.38 × 10 –23 J/K,
Planck constant : 6.63 × 10 –34 J.s)

Solution

Physics sir · 4 steps

Get the momentum straight from temperature: mvᵣₘₛ = √3mkT, then divide h by it.

1. Momentum from temperature. The r.m.s. speed is vᵣₘₛ = √(3kT/m). Multiplying by m gives the momentum p = mvᵣₘₛ = √3mkT.

   λ = h/p = h/√3mkT

2. Put in numbers. 3mkT = 3 × 4.64 × 10⁻²⁶ × 1.38 × 10⁻²³ × 400 = 7.68 × 10⁻⁴⁶, so p = √(7.68 × 10⁻⁴⁶) = 2.77 × 10⁻²³ kg m s⁻¹. That is a speed of about 597 m s⁻¹.
3. Find the wavelength. λ = (6.63 × 10⁻³⁴)/(2.77 × 10⁻²³) = 2.39 × 10⁻¹¹ m. Since 1 Å = 10⁻¹⁰ m, this is 0.239 Å.
4. So. The de Broglie wavelength is close to 0.24 Å, which is option (d).

The idea to keep. For a gas molecule at temperature T, write λ = h/√3mkT in one go. You never need to find the speed separately, so there is less room for arithmetic slips.

Answer from the NTA final answer key. Solution by PrepWiser.