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 $$\lambda' = \frac{1}{\sqrt{2}} \lambda_1$$