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:

Updated On: Apr 10, 2026
  • \( \sqrt{\frac{m_p}{m_e}} \)
  • \( \sqrt{\frac{m_e}{m_p}} \)
  • \( \frac{m_p}{m_e} \)
  • \( \left( \frac{m_p}{m_e} \right)^2 \)
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The Correct Option is A

Solution and Explanation


Step 1: Understanding the Concept:
Every moving particle has an associated wave called the de Broglie wave. For a particle of mass \( m \) and charge \( q \) accelerated from rest through a potential difference \( V \), its kinetic energy \( K = qV \) determines its de Broglie wavelength.

Step 2: Key Formula or Approach:
De Broglie wavelength: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}} \]

Step 3: Detailed Explanation:
For an electron: charge is \( e \), mass is \( m_e \).
\[ \lambda_e = \frac{h}{\sqrt{2m_e eV}} \]
For a proton: charge is \( e \), mass is \( m_p \).
\[ \lambda_p = \frac{h}{\sqrt{2m_p eV}} \]
Now, calculate the ratio \( \frac{\lambda_e}{\lambda_p} \):
\[ \frac{\lambda_e}{\lambda_p} = \frac{h / \sqrt{2m_e eV}}{h / \sqrt{2m_p eV}} = \frac{\sqrt{2m_p eV}}{\sqrt{2m_e eV}} \]
\[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} \]

Step 4: Final Answer:
The ratio of their de Broglie wavelengths is \( \sqrt{\frac{m_p}{m_e}} \).
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