Problem 6 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

At temperature \(\mathrm{T}\), the average kinetic energy of any particle is \(\frac{3}{2} \mathrm{kT}\). The de Broglie wavelength follows the order : [Main Online April 11, 2015] (a) Visible photon > Thermal neutron > Thermal electron (b) Thermal proton \(>\) Thermal electron \(>\) Visible photon (c) Thermal proton \(>\) Visible photon \(>\) Thermal electron (d) Visible photon \(>\) Thermal electron \(>\) Thermal neutron

Step-by-Step Solution

Answer

The correct order is: (d) Visible photon > Thermal electron > Thermal neutron.

Step 1: Understand the de Broglie Wavelength

The de Broglie wavelength \( \lambda \) of a particle is given by \( \lambda = \frac{h}{p} \), where \( h \) is the Planck's constant and \( p \) is the momentum of the particle. For particles at thermal equilibrium, momentum \( p \) can be linked to kinetic energy \( E_k = \frac{1}{2}mv^2 \) using \( p = mv \), leading to \( \lambda = \frac{h}{\sqrt{2mE_k}} \). The average kinetic energy \( E_k = \frac{3}{2}kT \).

Step 2: Apply the de Broglie Wavelength Formula

Substitute the kinetic energy for thermal particles in the de Broglie wavelength formula: \( \lambda = \frac{h}{\sqrt{3mkT}} \). This shows that \( \lambda \) is inversely proportional to the square root of the mass \( m \). Therefore, lighter particles have larger de Broglie wavelengths at the same temperature.

Step 3: Compare Particle Masses

The masses of particles in the options are: electron < photon (massless but related to wavelength by frequency) < neutron < proton. According to the relationship \( \lambda \propto \frac{1}{\sqrt{m}} \), electrons would have the longest wavelength among massive particles, followed by neutrons and then protons. Visible photons, being electromagnetic waves, generally have long wavelengths, longer than massive particles at thermal energy.

Step 4: Evaluate Each Option

- (a) Visible photon > Thermal neutron > Thermal electron: This is incorrect because thermal electrons have longer wavelengths than thermal neutrons. - (b) Thermal proton > Thermal electron > Visible photon: This is incorrect because visible photons should have longer wavelengths than electrons and protons. - (c) Thermal proton > Visible photon > Thermal electron: This is incorrect as photons typically have longer wavelengths than electrons and protons at the same temperature. - (d) Visible photon > Thermal electron > Thermal neutron: This is correct since visible photons have longer wavelengths than electrons, and electrons have longer wavelengths than neutrons.

Key Concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics that describes the energy an object possesses due to its motion. In this context, we are particularly interested in the kinetic energy of particles at thermal equilibrium. The average kinetic energy of any particle is given by the formula:

• \( E_k = \frac{3}{2}kT \)

Here, \( E_k \) represents the kinetic energy, \( k \) is the Boltzmann constant, and \( T \) is the temperature. This equation highlights that kinetic energy is directly proportional to the temperature of the system.

When particles are at thermal equilibrium, their kinetic energy acts in correlation with their motion, which affects their momentum and wavelength. Understanding this relationship helps in solving problems related to the de Broglie wavelength, as the wavelength depends inversely on the square root of kinetic energy.

Particle Momentum

Momentum is another key concept that plays a crucial role in understanding the behavior of particles. It is generally defined as the product of mass and velocity of an object:

• \( p = mv \)

Where \( p \) denotes momentum, \( m \) is the particle's mass, and \( v \) its velocity. Momentum is directly involved in determining the de Broglie wavelength. According to de Broglie's theory, every particle has an associated wavelength, given by:

• \( \lambda = \frac{h}{p} \)

In this formula, \( \lambda \) is the wavelength and \( h \) is Planck's constant. By rearranging and incorporating kinetic energy, we see:

• \( \lambda = \frac{h}{\sqrt{3mkT}} \)

This equation shows that as the momentum of a particle increases, the de Broglie wavelength decreases. Conversely, lower momentum means a longer wavelength. Therefore, understanding how momentum relates to kinetic energy and mass is vital when examining de Broglie wavelengths in thermal settings.

Particle Mass Comparison

In the context of de Broglie wavelengths, the mass of particles is an important factor to consider. Heavier particles have shorter wavelengths at the same kinetic energy compared to lighter particles. This is expressed through the relationship:

• \( \lambda \propto \frac{1}{\sqrt{m}} \)

Thus, the wavelength decreases as the mass increases. Let's compare the masses of the particles discussed:

• Electrons are the lightest among massive particles.
• Neutrons and protons are significantly heavier than electrons.
• Photons, while massless, have a wavelength determined by frequency, generally longer than thermal electrons and neutrons.

The mass hierarchy affects how we interpret de Broglie wavelength problems. For instance, electrons have longer de Broglie wavelengths than neutrons or protons at the same temperature. Visible photons, being light waves, often have longer wavelengths compared to massive particles at thermal energies. Understanding these mass differences is crucial for accurately comparing de Broglie wavelengths in various scenarios.