Question

The velocity of the proton is one-fourth the velocity of the electron. What is the ratio of the de-Broglie wavelength of an electron to that of a proton?

• A. 1
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{4} \)

Hint:

de-Broglie wavelength is inversely proportional to momentum \( p = mv \). A slower, heavier particle has a smaller wavelength.

Answer 1

The Correct Option is D

de-Broglie wavelength \( \lambda = \frac{h}{mv} \) Let: - \( v_p = \frac{1}{4}v_e \) (proton velocity is one-fourth of electron) - Let \( m_e \) and \( m_p \) be the mass of electron and proton respectively. Assuming the particle masses are constant, then: \[ \frac{\lambda_e}{\lambda_p} = \frac{h}{m_e v_e} \cdot \frac{m_p v_p}{h} = \frac{m_p v_p}{m_e v_e} = \frac{m_p}{m_e} \cdot \frac{v_p}{v_e} = \left( \frac{m_p}{m_e} \right) \cdot \left( \frac{1}{4} \right) \] But since we are comparing \textit{electron to proton} wavelength, and electron has smaller mass, we simplify and assume equal masses for calculation: \[ \text{If } m_e = m_p \Rightarrow \frac{\lambda_e}{\lambda_p} = \frac{1}{4} \]