The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is :
[mₑ = mass of electron = 9 × 10⁻³¹ kg, h = Planck constant = 6.6 × 10⁻³⁴ Js, k B = Boltzmann constant = 1.38 × 10⁻²³ JK⁻¹]

Solution

Physics sir · 4 steps

In 3D the electron has 3/2k_BT of kinetic energy: get its momentum, then λ = h/p.

1. Thermal energy. An ideal gas particle moving in 3 dimensions has average kinetic energy E = 3/2k_BT, that is 1/2k_BT for each direction.
2. Energy to momentum. Since E = p²/2m, the momentum is p = √2mE = √3mk_BT. Put this into λ = h/p.

   λ = h/√(3 mₑ k_B T)

3. Put in numbers. 3 mₑ k_B T = 3 × 9 × 10⁻³¹ × 1.38 × 10⁻²³ × 300 = 1.118 × 10⁻⁵⁰, so p = 1.057 × 10⁻²⁵ kg m s⁻¹. Then λ = (6.6 × 10⁻³⁴)/(1.057 × 10⁻²⁵) = 6.24 × 10⁻⁹ m.
4. So. λ ≈ 6.2 nm. The closest option is 6.26 nm, option (a).

The idea to keep. For a gas particle at temperature T in 3D, use λ = h/√3mk_BT. The factor 3 comes from 3/2k_BT. Do not drop it and use k_BT instead.

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