A) \[{{\lambda }_{e}}>{{\lambda }_{p}}={{\lambda }_{\alpha }}\]
B) \[{{\lambda }_{p}}={{\lambda }_{\alpha }}={{\lambda }_{e}}\]
C) \[{{\lambda }_{p}}>{{\lambda }_{e}}>{{\lambda }_{\alpha }}\]
D) \[{{\lambda }_{\alpha }}<{{\lambda }_{e}}>{{\lambda }_{p}}\]
Correct Answer: A
Solution :
[a] de-Broglie wavelength \[=\frac{h}{\sqrt{2m\,(K6)}}\]You need to login to perform this action.
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