# Calculate the de Broglie wavelength for an electron in the first Bohr orbit of a hydrogen atom?

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According to deBroglie, λ = the circumference (2πr) of orbit for n=1

Finding this circumference ..

Total energy (E) of an electron in a hydrogen atom .. E = (-) ke² / 2r

For the ground state (n=1) this energy is -13.60 eV

E = (13.60 x 1.60^-19)J .. .. E = 2.176^-18 J

2.176^-18 = ke² / 2r

r = ke² / (2 x 2.176^-18)

r = (9.0^9)(1.60^-19)² / (4.352^-18) .. .. r = 5.29^-11 m

According to deBroglie 1λ = the circumference of orbit ..

So .. λ = 2πr = 2 x π x 5.29^-11m .. .. .. ► λ = 3.326^-10 m

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• From Standard formulation it is clear that the velocity of electron in Bhor's first orbit = 2.18*10^6 m/s

Now de Broglie wavelength is given by lamda = h/mv

h = Plank's Constant = 6.626*10^-34 Js

m= Mass of an electron = 9.1*10^-31 Kg

v = Velocity of an electron.

Using this lamda = 3.3366*10^-10 meter = 3.3366 Armstrong.