Problem 19 - Master Resource Book in JEE Main Physics

Question

A proton, a neutron, an electron and an \(\alpha\)-particle have same energy. Then their de-Broglie wavelengths compare as [NCERT Exemplar] (a) \(\lambda_{\mathrm{p}}=\lambda_{\mathrm{n}}>\lambda_{\mathrm{e}}>\lambda_{\alpha}\) (b) \(\lambda_{\alpha}=\lambda_{\mathrm{p}}>\lambda_{\mathrm{n}}>\lambda_{e}\) (c) \(\lambda_{e}=\lambda_{\mathrm{p}}>\lambda_{\mathrm{n}}>\lambda_{\alpha}\) (d) \(\lambda_{e}=\lambda_{p}>\lambda_{n}>\lambda_{a}\)

Step-by-Step Solution

Answer

(a) \(\lambda_{\mathrm{p}} = \lambda_{\mathrm{n}} > \lambda_{\mathrm{e}} > \lambda_{\alpha}\)

Step 1: Understanding de Broglie Wavelength

The de Broglie wavelength \(lambda\) of a particle is given by the equation \(lambda = \frac{h}{p}\), where \(h\) is the Planck constant and \(p\) is the momentum of the particle. When a particle has kinetic energy \(E\), its momentum \(p\) is \(p = \sqrt{2mE}\), where \(m\) is the mass.

Step 2: Comparing Masses and Correlating with Energy

Since all particles have the same energy, their momentum is dependent only on their mass. A proton and a neutron have similar masses, an electron is much lighter, and an \(\alpha\)-particle is heavier than a single proton or neutron. Therefore, the momentum increases with particle mass in this context.

Step 3: Relation Between Mass, Momentum and Wavelength

Since \(lambda = \frac{h}{\sqrt{2mE}}\), the wavelength is inversely proportional to the square root of the mass of the particles. Therefore, a lighter particle gives a longer wavelength. So the order of de Broglie wavelengths will be determined by the inverse order of their masses.

Step 4: Determining Order of Wavelengths

The order of de Broglie wavelengths, given same energy, is \(\lambda_e > \lambda_p = \lambda_n > \lambda_{\alpha}\) because electrons are lightest, protons and neutrons have similar masses, and \(\alpha\)-particles have the largest mass.

Step 5: Matching with Given Options

The correct order \(\lambda_e > \lambda_p = \lambda_n > \lambda_{\alpha}\) matches with option (a): \(\lambda_{\mathrm{p}} = \lambda_{\mathrm{n}} > \lambda_{\mathrm{e}} > \lambda_{\alpha}\).

Key Concepts

Planck constant

The Planck constant, represented by the symbol \( h \), is a fundamental quantity in quantum mechanics. It relates the energy of a photon to its frequency and plays a crucial role in the equations that govern quantum systems.

• Value: The value of the Planck constant is approximately \( 6.626 \times 10^{-34} \) Joule seconds.
• Importance: It sets the scale of quantum effects. This means, in simple terms, how quantum mechanical we expect a system to behave.

When applied to the de Broglie wavelength equation \( \lambda = \frac{h}{p} \), it connects the wave-like properties and particle-like properties of matter, highlighting that every particle has a wavelength related to its momentum. Thus, the Planck constant is crucial for illustrating the quantum nature of particles, even if they possess macroscopic mass.

momentum

Momentum, represented by \( p \), is a measure of the motion of a particle and is calculated as the product of its mass and velocity. For subatomic particles, it plays an integral role in defining properties such as the de Broglie wavelength.

• Equation: \( p = mv \), where \( m \) is mass and \( v \) is velocity.
• Special Case: When a particle has a specific kinetic energy \( E \), its momentum can be calculated as \( p = \sqrt{2mE} \).

This momentum equation reveals that for particles with the same kinetic energy, their momentum is influenced primarily by their mass. This directly affects their de Broglie wavelength, following the relation \( \lambda = \frac{h}{\sqrt{2mE}} \). In simple terms, a larger mass results in higher momentum, yielding a shorter wavelength. Understanding momentum in the quantum framework is vital in evaluating how particles like protons, neutrons, electrons, and alpha particles behave under similar energy conditions.

kinetic energy

Kinetic energy is the energy a particle possesses due to its motion. It is an essential factor in calculating the de Broglie wavelength and understanding how different particles behave when endowed with similar energy levels.

• Formula: The kinetic energy \( E \) of a particle is given by \( E = \frac{1}{2} mv^2 \).
• Relation to Momentum: Can be rearranged to \( v = \sqrt{\frac{2E}{m}} \) showing velocity in terms of energy and mass.

In a situation where particles have the same kinetic energy, it's crucial to consider their mass to understand phenomena like the de Broglie wavelength. Heavier particles need more momentum to have the same kinetic energy as lighter ones. Hence, for the same kinetic energy:

• Lighter particles like electrons exhibit longer wavelengths due to their smaller mass.
• Heavier particles, such as alpha particles, show shorter wavelengths.

Kinetic energy provides a lens through which we can see the dual wave-particle nature of particles and how they exhibit quantum characteristics as described by de Broglie's hypothesis.