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}} $
Show Answer
Answer:
Correct Answer: A
Solution:
- de-Broglie wavelength $ =\frac{h}{\sqrt{2m,(K6)}} $
