Question

A proton, an electron, and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:

• A. \(\lambda_e>\lambda_\alpha>\lambda_p\)
• B. \(\lambda_\alpha<\lambda_p<\lambda_e\)
• C. \(\lambda_p<\lambda_e<\lambda_\alpha\)
• D. \(\lambda_p>\lambda_e>\lambda_\alpha\)

Answer 1

The Correct Option is B

To compare the de-Broglie wavelengths of particles possessing identical energies, an understanding of the de-Broglie wavelength formula and its dependence on particle mass and energy is required.

The de-Broglie wavelength (\(\lambda\)) is defined by the equation:

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

wherein:

• \(h\) represents Planck's constant.
• \(p\) denotes the momentum of the particle.

For a particle with kinetic energy \(E\), its momentum can be expressed as:

\(p = \sqrt{2mE}\)

Substituting this expression into the de-Broglie equation yields:

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

This equation indicates that for particles with equivalent energy \(E\), the wavelength \(\lambda\) is inversely proportional to the square root of their mass:

\(\lambda \propto \dfrac{1}{\sqrt{m}}\)

The masses of the specified particles are compared as follows:

• Proton (\(m_p\)): Approximately \(1.67 \times 10^{-27}\) kg
• Electron (\(m_e\)): Approximately \(9.11 \times 10^{-31}\) kg
• Alpha particle (\(m_\alpha\)): Comprises 2 protons and 2 neutrons, approximately \(4 \times 1.67 \times 10^{-27}\) kg = \(6.68 \times 10^{-27}\) kg

Based on these mass values:

• The electron, being the least massive, will exhibit the longest wavelength.
• The proton, heavier than the electron but lighter than the alpha particle, will have an intermediate wavelength.
• The alpha particle, being the most massive, will possess the shortest wavelength.

Consequently, the de-Broglie wavelengths are ordered as follows:

\(\lambda_\alpha < \lambda_p < \lambda_e\)

The definitive order of the de-Broglie wavelengths is:

\(\lambda_\alpha<\lambda_p<\lambda_e\)