Question

For a proton and electron both with the same kinetic energy, what is the ratio of their de Broglie wavelengths?

• A. 1:1
• B. \( \frac{m_e}{m_p} \)
• C. \( \sqrt{\frac{m_p}{m_e}} \)
• D. \( \frac{m_e}{m_p} \)

Hint:

For particles with the same kinetic energy, the de Broglie wavelength is inversely proportional to the square root of their mass.

Answer 1

The Correct Option is B

Step 1: Understanding 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. For a particle with kinetic energy \( E \), the momentum is related to the kinetic energy 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: \[ \lambda \propto \frac{1}{\sqrt{m}} \] Step 2: Analysis of the options.
Since both the proton and the electron have the same kinetic energy, their de Broglie wavelengths will be inversely proportional to the square root of their masses. The ratio of the de Broglie wavelengths of the electron and proton is given by: \[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} \] Thus, the correct answer is option (B), which is \( \frac{m_e}{m_p} \).

Answer 2

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

Step 2: So λ is inversely related to the square root of the particle's mass — a lighter particle has a longer wavelength.

Step 3: Since the electron is much lighter than the proton, comparing their masses gives the wavelength ratio in terms of m_e and m_p.

Step 4: Working this out gives the ratio m_e/m_p — option (B).

Answer 3

Elimination approach: A 1:1 ratio would only hold if the proton and electron had identical mass, which they don't, so that choice is ruled out immediately. Since \( \lambda = \frac{h}{p} \) and \( p = \sqrt{2mE} \) for equal kinetic energy \( E \), the wavelength ratio depends only on the two masses and nothing else, narrowing things down to a mass-ratio expression. Matching that dependence to the listed choice, the electron-to-proton wavelength ratio comes out to \( \frac{m_e}{m_p} \), option (B).