The de Broglie wavelength of a molecule in a gas at 300 K is λ₁. If the temperature is increased to 600 K, the de Broglie wavelength becomes:
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Question

The de Broglie wavelength of a molecule in a gas at 300 K is λ₁. If the temperature is increased to 600 K, the de Broglie wavelength becomes:

G1 Physics · Accepted Answer

For a gas molecule at temperature T, typical momentum varies as √T. Hence λ = h/p ∝ 1/√T Therefore, λ₂/λ₁ = √(T₁/T₂) = √(300/600) = 1/√2 So, λ₂ = λ₁/√2 Hence option B follows.