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: For equal kinetic energy \( E \), momentum is \( p = \sqrt{2mE} \), so the lighter particle carries less momentum.
Step 2: Since \( \lambda = \frac{h}{p} \), the wavelength ratio between the electron and the proton reduces to an expression in their two masses.
Step 3: Working this through gives \[ \boxed{\frac{\lambda_e}{\lambda_p} = \frac{m_e}{m_p}} \], matching option (B).