Question

Calculate the de broglie wavelength for an electron in the first bohr orbit of a hydrogen atom

Answer 1

De Broglie's wavelength,

λ = the circumference (2πr) of orbit for n=1

That is, an electron in the first bohr orbit.

To find the circumference,

The total energy (E) of an electron in a hydrogen atom,

Energy of an electron,

E = (-) ke² / 2r  

For the ground state (n=1)

E = -13.60 eV  

E = (13.60 x 1.60^-19)J

E = 2.176^-18 J  

Substituting in the total energy formula,

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 De Broglie,

1λ = the circumference of orbit

So,

λ = 2πr = 2 x π x 5.29^-11 m

λ = 3.326^-10 m

The De Broglie wavelength for an electron in the first bohr orbit of a hydrogen atom is 3.326^-10 m

Answer 2

Explanation:

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