Question

A proton (\( p \)) and an electron (\( e \)) will have the same de-Broglie wavelength when the ratio of their momenta is 
(Assume \( m_p = 1849 m_e \)):

• A. \( 1:43 \)
• B. \( 43:1 \)
• C. \( 1:1849 \)
• D. \( 1:1 \)

Hint:

For particles to have the same de-Broglie wavelength, their momentum must be equal, regardless of their mass difference.

Answer 1

The Correct Option is D

Step 1: Understanding de-Broglie wavelength formula. The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{p} \] where: - \( \lambda \) is the wavelength, - \( h \) is Planck’s constant, - \( p \) is the momentum. For two particles to have the same wavelength, their momentum must be equal: \[ \lambda_p = \lambda_e \Rightarrow \frac{h}{p_p} = \frac{h}{p_e} \] \[ p_p = p_e \] Step 2: Interpreting the given mass ratio. Given: \[ m_p = 1849 m_e \] Since momentum is \( p = m v \), for equal de-Broglie wavelengths, the momentum must be the same for both: \[ p_p = p_e \] Thus, the ratio of their momenta: \[ \frac{p_p}{p_e} = 1:1 \] Final Answer: \[ \boxed{1:1} \]