Question

The de Broglie wavelength associated with an electron accelerated through a potential difference V is \( \lambda_e \) and the de Broglie wavelength associated with a proton accelerated through the same potential difference is \( \lambda_p \). If their corresponding masses are \( m_e \) and \( m_p \), respectively, then the ratio of their de Broglie wavelengths \( \frac{\lambda_e}{\lambda_p} \) is:

• A. \( \sqrt{\frac{m_p}{m_e}} \)
• B. \( \sqrt{\frac{m_e}{m_p}} \)
• C. \( \frac{m_p}{m_e} \)
• D. \( \left( \frac{m_p}{m_e} \right)^2 \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When a charged particle accelerates through a potential difference $V$, it gains kinetic energy equal to $qV$. The de Broglie wavelength connects this kinetic energy to the particle's momentum and mass.
Step 2: Key Formula or Approach:
Kinetic Energy gained: $K = qV$.
Momentum: $p = \sqrt{2mK} = \sqrt{2mqV}$.
de Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$.
Step 3: Detailed Explanation:
Both the electron and the proton carry the same magnitude of elementary charge, so $q_e = q_p = e$.
Both are accelerated through the exact same potential difference $V$.
For the electron:
\[ \lambda_e = \frac{h}{\sqrt{2m_eeV}} \] For the proton:
\[ \lambda_p = \frac{h}{\sqrt{2m_peV}} \] Take the ratio of the two wavelengths:
\[ \frac{\lambda_e}{\lambda_p} = \frac{ \frac{h}{\sqrt{2m_eeV}} }{ \frac{h}{\sqrt{2m_peV}} } \] Cancel out the common terms $h$, $\sqrt{2}$, $\sqrt{e}$, and $\sqrt{V}$:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{m_p}}{\sqrt{m_e}} = \sqrt{\frac{m_p}{m_e}} \] Step 4: Final Answer:
The ratio is $\sqrt{\frac{m_p}{m_e}}$.