A) electron
B) proton
C) both same
D) none of these
Correct Answer: A
Solution :
de-Broglie wavelength \[\lambda =\frac{h}{\sqrt{2m{{E}_{k}}}}\] Where h = Planck's constant Elutron and proton have same kinetic energy, So, \[\lambda \propto \frac{1}{m}\] Mass of electron is less than that of Proton, so electron will have greater wavelength.You need to login to perform this action.
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