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The kinetic energy of the electron orbiting in the first excited state of hydrogen atom is 3.4 eV. Determine the de Broglie wavelength associated with it.



de Broglie wavelength, λ=h/(2mE)^1/2 where h=Plank's constant=6.626×10-34 J s m=Mass of the electron=9.1×10-31 kg E=Energy of the electron=3.4 eV=3.4×1.6×10-19
J⇒λ=6.626×10-342(9.1×10-31 )(3.4×1.6×10-19)⇒λ=0.67×10-9 m
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