Question

Calculate the ratio of de Broglie wavelengths of an electron \((\lambda_e)\) and a proton \((\lambda_p)\) moving with the same velocity.

• A. \( \dfrac{m_e}{m_p} \)
• B. \( \dfrac{m_p}{m_e} \)
• C. \(1\)
• D. \( \dfrac{m_e^2}{m_p^2} \)

Hint:

From the de Broglie relation \( \lambda = \frac{h}{mv} \): - If velocity is constant, wavelength is inversely proportional to mass. - A lighter particle always has a larger de Broglie wavelength.

Answer 1

The Correct Option is B

Concept: The de Broglie wavelength of a particle is given by \[ \lambda = \frac{h}{mv} \] where:
• \(h\) = Planck's constant
• \(m\) = mass of the particle
• \(v\) = velocity of the particle Thus, wavelength is inversely proportional to mass when velocity is constant.

Step 1: Write the wavelengths of electron and proton. \[ \lambda_e = \frac{h}{m_e v}, \qquad \lambda_p = \frac{h}{m_p v} \]

Step 2: Find the ratio. \[ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{m_e v}}{\frac{h}{m_p v}} \]

Step 3: Simplify the expression. \[ \frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e} \] \[ \boxed{\frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e}} \]