Question

What is the ratio of the de Broglie wavelengths for an electron and a proton if both have equal kinetic energy?

Hint:

For two particles with the same kinetic energy, the ratio of their de Broglie wavelengths is inversely proportional to the square root of their masses.

Answer 1

Step 1: Understanding the de Broglie wavelength.
The de Broglie wavelength \( \lambda \) of a particle is given by the equation: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) is related to the kinetic energy \( E \) by the equation: \[ p = \sqrt{2mE} \] Thus, the de Broglie wavelength is inversely proportional to the square root of the mass \( m \) of the particle.
Step 2: Comparing the wavelengths of the proton and electron.
Since both the electron and proton have the same kinetic energy, their momentum is proportional to the square root of their masses. Therefore, the ratio of their de Broglie wavelengths is: \[ \frac{\lambda_{\text{proton}}}{\lambda_{\text{electron}}} = \sqrt{\frac{m_{\text{proton}}}{m_{\text{electron}}}} \]

Answer 2

Step 1: The de Broglie wavelength is λ = h/p, and momentum from kinetic energy is p = √(2mE).

Step 2: So for the same kinetic energy, a heavier particle has a shorter wavelength — λ is inversely proportional to √mass.

Step 3: Since the proton is about 1836 times heavier than the electron, the electron's wavelength is longer.

Step 4: The ratio of wavelengths works out to λ_proton/λ_electron = √(m_electron/m_proton).

Answer 3

Numerical comparison.
For the same kinetic energy \( E \), momentum is \( p = \sqrt{2mE} \), so a heavier particle simply carries more momentum and, since \( \lambda = h/p \), a shorter wavelength.
The proton is roughly \( 1836 \) times heavier than the electron, so \[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} \approx \sqrt{1836} \approx 42.8 \] meaning the electron's de Broglie wavelength is about 43 times longer than the proton's at equal kinetic energy.