An electron and a proton are possessing same amount of kinetic energy. Which of it two has greater De-Broglie Wavelength ? Justify your answer.

To determine which particle, the electron or the proton, has a greater De Broglie wavelength when both possess the same amount of kinetic energy, we need to delve into the relationship between kinetic energy, mass, and wavelength. The De Broglie wavelength is given by the formula:

Understanding De Broglie Wavelength

The De Broglie wavelength (\( \lambda \)) is calculated using the equation:

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

where \( h \) is Planck's constant and \( p \) is the momentum of the particle. Momentum (\( p \)) can also be expressed in terms of kinetic energy (\( KE \)) and mass (\( m \)) as follows:

\( p = \sqrt{2m \cdot KE} \)

Relating Kinetic Energy and Mass

Given that both the electron and the proton have the same kinetic energy, we can substitute this into the momentum equation:

\( \lambda = \frac{h}{\sqrt{2m \cdot KE}} \)

From this equation, it becomes clear that the De Broglie wavelength is inversely proportional to the square root of the mass of the particle. This means that as the mass increases, the wavelength decreases.

Comparing Electron and Proton

• The mass of an electron (\( m_e \)) is approximately \( 9.11 \times 10^{-31} \) kg.
• The mass of a proton (\( m_p \)) is about \( 1.67 \times 10^{-27} \) kg.

Since the mass of the proton is significantly greater than that of the electron, we can conclude that:

\( m_e < m_p \)

Conclusion on Wavelength Comparison

As a result, when both particles have the same kinetic energy, the electron, having a much smaller mass, will have a greater De Broglie wavelength compared to the proton. This can be summarized as:

\( \lambda_e > \lambda_p \)

In essence, the lighter the particle, the longer its wavelength when kinetic energy is held constant. This principle is fundamental in quantum mechanics and illustrates the wave-particle duality of matter.