Question:medium

Electron and proton are accelerated with the same potential to achieve de-Broglie wavelengths of \( \lambda_1 \) and \( \lambda_2 \). Given \( m_p = 1849 \, m_e \), find the ratio \( \lambda_1/\lambda_2 \):

Updated On: Apr 9, 2026
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  • 43
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The de-Broglie wavelength of a particle accelerated through a potential difference \( V \) depends on its mass and charge. Since both the electron and proton have the same magnitude of charge (\( e \)), the wavelength depends inversely on the square root of their masses.
Step 2: Key Formula or Approach:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}} \] Since \( h, q, \) and \( V \) are constant: \[ \lambda \propto \frac{1}{\sqrt{m}} \]
Step 3: Detailed Explanation:
Let \( \lambda_1 \) be for the electron and \( \lambda_2 \) be for the proton. \[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_p}{m_e}} \] Substitute the given mass relationship \( m_p = 1849 \, m_e \): \[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{1849 \, m_e}{m_e}} = \sqrt{1849} \] Calculating the square root: \[ \sqrt{1849} = 43 \]
Step 4: Final Answer:
The ratio \( \lambda_1/\lambda_2 \) is 43.
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