QJEEM23Ph5a20: JEE Main 2023 Physics - Class 12 JEEM23Ph5a20 Solution: The de Broglie wavelength of a molecule in a gas at room…

The de Broglie wavelength of a particle is given by:

$\lambda = \frac{h}{p}$

where h is Planck'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:

$p = \sqrt{2mK}$

where m is the mass of the molecule and K is the kinetic energy of the molecule.

At a given temperature T, the average kinetic energy of a molecule in a gas is given by:

$K = \frac{3}{2} kT$

where k is Boltzmann's constant.

Therefore, the de Broglie wavelength of a molecule in a gas is given by:

$\lambda = \frac{h}{\sqrt{2m(3/2)kT}} = \frac{h}{\sqrt{3mkT}}$

If the temperature of the gas is increased from T = 300 K to T = 600 K, the new de Broglie wavelength becomes:

$\lambda' = \frac{h}{\sqrt{3mk(2T)}} = \frac{h}{\sqrt{2} \sqrt{3mkT}} = \frac{1}{\sqrt{2}} \lambda$

So, the de Broglie wavelength of the gas molecule decreases by a factor of $\sqrt{2}$ when the temperature of the gas is doubled.

Therefore, the correct answer is $\lambda' = \frac{1}{\sqrt{2}} \lambda_1$