Question

A proton, a neutron, an electron and an α-particle have same energy. If λ_p, λ_n, λ_e and λ_α are the de Broglie’s wavelengths of proton, neutron, electron and α particle respectively, then choose the correct relation from the following:

• A. λ_p = λ_n > λ_e > λ_α
• B. λ_α < λ_n < λ_p < λ_e
• C. λ_e < λ_p = λ_n > λ_α
• D. λ_e = λ_p = λ_n = λ_α

Answer 1

The Correct Option is A

 In this problem, we need to analyze the de Broglie wavelengths of different particles that have the same energy: a proton, a neutron, an electron, and an α-particle. 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. The momentum \(p\) can be expressed in terms of mass \(m\) and velocity \(v\) as \(p = mv\).

Since the kinetic energy \(E\) is the same for all particles, we use the relationship:

\(E = \frac{1}{2}mv^2\)

Solving for velocity, we find:

\(v = \sqrt{\frac{2E}{m}}\)

Substitute this into the momentum equation:

\(p = m \sqrt{\frac{2E}{m}} = \sqrt{2mE}\)

The de Broglie wavelength then becomes:

\(\lambda = \frac{h}{\sqrt{2mE}}\)

Since all particles have the same energy \(E\), the de Broglie wavelength depends on the mass of the particles:

• Proton and Neutron have relatively similar mass.
• Electron has a much smaller mass compared to protons and neutrons.
• α-particle is composed of 2 protons and 2 neutrons, making its mass significantly larger.

Thus, we consider the mass-related relationship for the wavelengths:

• Heavier particles will have a shorter wavelength.
• Therefore, the order of de Broglie wavelengths \(\lambda\) is inversely related to the mass:
• \(\lambda_{e} > \lambda_{p} = \lambda_{n} > \lambda_{\alpha}\)

Thus, the correct answer is:

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