Q.If the de-Broglie wavelength of a gas molecule at 0°C be λ, what will be its wavelength at 819°C?

A gas molecule's de Broglie wavelength varies as 1/sqrt(T); raising the temperature from 0°C (273 K) to 819°C (1092 K) is a fourfold increase in T, so the wavelength becomes lambda/2.

Step 1 - Relate wavelength to temperature. The de Broglie wavelength is lambda = h/p = h/sqrt(2mE), where the mean thermal kinetic energy of a gas molecule is E = (3/2)kT. Hence lambda = h/sqrt(2m x (3/2)kT) = h/sqrt(3mkT). Since h, m, k are constants, lambda is proportional to 1/sqrt(T), with T in kelvin.

Step 2 - Convert temperatures to kelvin: T1 = 0°C = 273 K; T2 = 819°C = 819 + 273 = 1092 K.

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