Atomic Structure Question 493

Question: An electron, a proton and an alpha particle have kinetic energies of 16E, 4E and E respectively. What is the qualitative order of their de-Broglie wavelengths-

Options:

A) $ {\lambda_{e}}>{\lambda_{p}}={\lambda_{\alpha }} $

B) $ {\lambda_{p}}={\lambda_{\alpha }}={\lambda_{e}} $

C) $ {\lambda_{p}}>{\lambda_{e}}>{\lambda_{\alpha }} $

D) $ {\lambda_{\alpha }}<{\lambda_{e}}>{\lambda_{p}} $

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Answer:

Correct Answer: A

Solution:

  • de-Broglie wavelength $ =\frac{h}{\sqrt{2m,(K6)}} $


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