Question

What is the ratio of the circumference of the first bohr orbit for the electron in the hydrogen atom to the de broglie wavelength of electron having the same velocity as the electron in the first bohr orbit of the hydrogen?

Answer 1

The angular momentum of an electron = mvr = nh/2π --------------1

Also according to de Broglie equation λ = h/mv

Or

mv = h/v -----------2

Putting 2 in 1

h/v =nh/2π or 2λr= nλ

Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

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