Question:
An electron, helium ion (\( \text{He}^{++} \)) and proton having the same kinetic energy. The relation between their respective de-Broglie wavelengths \( \lambda_e, \lambda_{\text{He}^{++}} \) and \( \lambda_p \) is
An electron, helium ion (\( \text{He}^{++} \)) and proton having the same kinetic energy. The relation between their respective de-Broglie wavelengths \( \lambda_e, \lambda_{\text{He}^{++}} \) and \( \lambda_p \) is
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For same KE: wavelength depends only on mass $\longrightarrow$ lighter particle = larger wavelength.
Updated On: Apr 22, 2026
- \( \lambda_e>\lambda_p>\lambda_{\text{He}^{++}} \)
- \( \lambda_e>\lambda_{\text{He}^{++}}>\lambda_p \)
- \( \lambda_e<\lambda_p<\lambda_{\text{He}^{++}} \)
- \( \lambda_e<\lambda_{\text{He}^{++}} = \lambda_p \)
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The Correct Option is A
Solution and Explanation
Concept:
De-Broglie wavelength:
\[
\lambda = \frac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \frac{1}{\sqrt{m}}
\]
Step 1: Compare masses.
\[ m_e \ll m_p<m_{\text{He}^{++}} \]
Step 2: Apply relation.
Smaller mass \(\Rightarrow\) larger wavelength.
Step 3: Final order.
\[ \lambda_e>\lambda_p>\lambda_{\text{He}^{++}} \]
Step 1: Compare masses.
\[ m_e \ll m_p<m_{\text{He}^{++}} \]
Step 2: Apply relation.
Smaller mass \(\Rightarrow\) larger wavelength.
Step 3: Final order.
\[ \lambda_e>\lambda_p>\lambda_{\text{He}^{++}} \]
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