Question

A proton and an electron have the same de Broglie wavelength. If \(K_p\) and \(K_e\) be the kinetic energies of proton and electron respectively. Then choose the correct relation:

• A. \(K_p\) \(>\) \(K_e\)
• B. \(K_p = K_e\)
• C. \(K_p = K_e^2\)
• D. \(K_p\) \(<\) \(K_e\)

Answer 1

The Correct Option is D

To address this problem, understanding the de Broglie wavelength and its relationship with the kinetic energy of particles such as protons and electrons is essential.

The de Broglie wavelength (\(\lambda\)) for a particle is determined by the equation:

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

Here, \(h\) represents Planck's constant, and \(p\) denotes the particle's momentum.

Momentum (\(p\)) can also be expressed in terms of kinetic energy (\(K\)) as:

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

In this formula, \(m\) is the mass of the particle.

Given that a proton and an electron possess identical de Broglie wavelengths, we can establish the following equality:

\(\frac{h}{\sqrt{2m_p K_p}} = \frac{h}{\sqrt{2m_e K_e}}\)

By canceling Planck's constant (\(h\)) and rearranging, we obtain:

\(\sqrt{\frac{K_p}{m_p}} = \sqrt{\frac{K_e}{m_e}}\)

Squaring both sides of the equation yields:

\(\frac{K_p}{m_p} = \frac{K_e}{m_e}\)

Solving for \(K_p\), we find:

\(K_p = K_e \cdot \frac{m_p}{m_e}\)

Considering that the mass of a proton (\(m_p\)) significantly exceeds that of an electron (\(m_e\)), it follows that:

\(K_p \lt K_e\)

This demonstrates that when a proton and an electron share the same de Broglie wavelength, the proton's kinetic energy is lower than the electron's.

Therefore, the correct conclusion is:

\(K_p \lt K_e\)