The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is λ₁. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes

Solution

Physics sir · 3 steps

Temperature sets the momentum, so λ falls as 1/√T: doubling T divides it by √2.

1. Energy from temperature. A gas molecule at temperature T has average kinetic energy 3/2kT. Its momentum is p = √2mE = √3mkT.

   λ = h/p = h/√3mkT ∝ 1/√T

2. Take the ratio. It is the same molecule, so m cancels: λ₂/λ₁ = √(T₁/T₂) = √(300/600) = 1/√2.
3. So. λ₂ = 1/√2λ₁, which is option (c).

The idea to keep. For thermal particles, λ = h/√3mkT. Hotter means faster, which means a shorter wavelength, and it goes as 1/√T, not 1/T.

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