The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is $\lambda_1$. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes

<p>The de Broglie wavelength of a particle is given by:</p> <p>$\lambda = \frac{h}{p}$</p> <p>where h is Planck&#39;s constant and p is the momentum of the particle. The momentum of a gas molecule can be related to its kinetic energy (which is related to the temperature of the gas) by:</p> <p>$p = \sqrt{2mK}$</p> <p>where m is the mass of the molecule and K is the kinetic energy of the molecule. </p> <p>At a given temperature T, the average kinetic energy of a molecule in a gas is given by:</p> <p>$K = \frac{3}{2} kT$</p> <p>where k is Boltzmann&#39;s constant. </p> <p>Therefore, the de Broglie wavelength of a molecule in a gas is given by:</p> <p>$\lambda = \frac{h}{\sqrt{2m(3/2)kT}} = \frac{h}{\sqrt{3mkT}}$</p> <p>If the temperature of the gas is increased from T = 300 K to T = 600 K, the new de Broglie wavelength becomes:</p> <p>$$\lambda&#39; = \frac{h}{\sqrt{3mk(2T)}} = \frac{h}{\sqrt{2} \sqrt{3mkT}} = \frac{1}{\sqrt{2}} \lambda$$</p> <p>So, the de Broglie wavelength of the gas molecule decreases by a factor of $\sqrt{2}$ when the temperature of the gas is doubled. </p> <p>Therefore, the correct answer is $\lambda&#39; = \frac{1}{\sqrt{2}} \lambda_1$</p>