A) \[{{\lambda }_{e}}<{{\lambda }_{p}}\]
B) \[{{\lambda }_{p}}<{{\lambda }_{\alpha }}\]
C) \[{{\lambda }_{e}}<{{\lambda }_{\alpha }}\]
D) \[{{\lambda }_{\alpha }}<{{\lambda }_{p}}>{{\lambda }_{e}}\]
Correct Answer: A
Solution :
de-Broglie postulated that the wavelength associated with a moving material particle is in analogy with light, given by \[\lambda =\frac{h}{p}\] where p is the momentum of the particle. A particle of (reladvistic) mass m and moving with a velocity v has a momentum mv. Therefore, \[\lambda =\frac{h}{mv}\] ?? (i) Or \[\lambda \propto \frac{1}{m}\] Therefore,\[{{\lambda }_{e}}\propto \frac{1}{{{m}_{e}}},{{\lambda }_{\alpha }}\propto \frac{1}{{{m}_{\alpha }}}\]and\[{{\lambda }_{p}}\propto \frac{1}{{{m}_{p}}}\] As we know that \[{{m}_{e}}<{{m}_{p}}<{{m}_{\alpha }}\] \[\therefore \] \[{{\lambda }_{e}}<{{\lambda }_{p}}<{{\lambda }_{\alpha }}\] Or \[{{\lambda }_{e}}<{{\lambda }_{p}}\]or\[{{\lambda }_{p}}>{{\lambda }_{\alpha }}\]or\[{{\lambda }_{e}}>{{\lambda }_{p}}\] Note: The relation\[(i)\]connects wavelength (a characteristic of wave) with momentum (a characteristic of particle).
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