solution:-
de-broglie wavelength can be written as
λ=ph=mvh=2mKEh
therefore de-broglie wavelength for proton
λp=2mpKEph...........eq.1
de-brolie wavelength for electron can written as
λe=2meKEeh...........eq.2
from equation 1 and 2 we can write
λeλp=mpme
because kinetic energy is same for proton and electron
λeλp=1.67×10−279.1×10−31=10.02
therefore ratio of de-broglie waelength of proton and electron is 0.02:1
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