Question 112: Dual Nature of Matter and Radiation

Question

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

Accepted Answer

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

Answer

2 $$\lambda_1$$

Answer

$$\frac{1}{2}$$$$\lambda_1$$

Answer

$$\frac{1}{\sqrt2}$$$$\lambda_1$$

Answer

$$\sqrt2~\lambda_1$$

Dual Nature of Matter and Radiation

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

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