A) \[{{\lambda }_{ph}}>{{\lambda }_{e}}\]
B) \[{{\lambda }_{ph}}<{{\lambda }_{e}}\]
C) \[{{\lambda }_{ph}}={{\lambda }_{e}}\]
D) \[\frac{{{\lambda }_{e}}}{{{\lambda }_{ph}}}=\]constant
Correct Answer: A
Solution :
Wavelength \[\lambda =\frac{hc}{p}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\{p=\sqrt{2mE}\}\] \[\therefore \] \[\lambda =\frac{hc}{\sqrt{2mE}}\] \[\therefore \] \[\lambda \propto \frac{1}{\sqrt{m}}\] Mass of photon is less than that of electron. \[\therefore \]wavelength of photon is greater than that of electron\[{{\lambda }_{ph}}>{{\lambda }_{e}}\]
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