A) \[{{\lambda }_{e}}>{{\lambda }_{p}}>{{\lambda }_{He}}\]
B) \[{{\lambda }_{He}}>{{\lambda }_{p}}>{{\lambda }_{e}}\]
C) \[{{\lambda }_{He}}>{{\lambda }_{e}}>{{\lambda }_{p}}\]
D) \[{{\lambda }_{p}}>{{\lambda }_{e}}>{{\lambda }_{He}}\]
Correct Answer: A
Solution :
de-Broglie wavelength, \[\lambda =\frac{h}{\sqrt{2mE}}\] Since, E is the same, Hence \[\lambda \propto \frac{h}{\sqrt{m}}\] Since, \[{{m}_{He}}>{{m}_{p}}>{{m}_{e}}\] \[\therefore \,\,\,{{\lambda }_{e}}>{{\lambda }_{p}}>{{\lambda }_{He}}\]
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