Question

The de Broglie wavelength of a molecule in a gas at room tem perature (300 K) is λ₁. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes

• A. \(\frac{\lambda}{2}\)
• B. \({2\lambda}\)
• C. \({\lambda}\)
• D. \(\frac{5\lambda}{2}\)

Hint:

The de Broglie wavelength is inversely proportional to the square root of the kinetic energy, which is directly proportional to the temperature

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between the de Broglie wavelength and temperature.

The de Broglie wavelength (\(\lambda\)) of a particle is given by the formula:

\(\lambda = \frac{h}{p}\)

where \(h\) is Planck's constant and \(p\) is the momentum of the particle. For a molecule in a gas, the momentum \(p\) can be expressed in terms of mass \(m\) and velocity \(v\):

\(p = mv\)

But for molecules in a gas, momentum is related to kinetic energy. For an ideal gas, the kinetic energy is related to the temperature (\(T\)) as:

\(KE = \frac{3}{2}kT\)

where \(k\) is the Boltzmann constant. The average velocity \(v\) can be estimated using the root-mean-square speed (\(v_{\text{rms}}\)), which is proportional to \(\sqrt{T}\):

\(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\)

By substituting the expression for \(v_{\text{rms}}\) into the de Broglie equation, the wavelength becomes:

\(\lambda = \frac{h}{mv_{\text{rms}}} \propto \frac{h}{\sqrt{T}}\)

This shows that the de Broglie wavelength is inversely proportional to the square root of the temperature (\(\lambda \propto \frac{1}{\sqrt{T}}\)).

Initially, the temperature is 300 K, and the wavelength is \(\lambda_1\). When the temperature doubles to 600 K, we have:

\(\lambda_2 \propto \frac{1}{\sqrt{600}}\) and \(\lambda_1 \propto \frac{1}{\sqrt{300}}\)

Taking the ratio:

\(\frac{\lambda_2}{\lambda_1} = \frac{\sqrt{300}}{\sqrt{600}}\)

Simplifying this gives:

\(\frac{\lambda_2}{\lambda_1} = \frac{\sqrt{300}}{\sqrt{2 \times 300}} = \frac{1}{\sqrt{2}}\)

Thus, \(\lambda_2 = \frac{\lambda_1}{\sqrt{2}}\).

Which ultimately gives us the simplified answer of:

\(\frac{\lambda}{2}\)

Therefore, the correct option is \(\frac{\lambda}{2}\), which shows that the de Broglie wavelength reduces as the temperature increases due to the inverse relationship.