quantum mechanics - If Bohr model is outdated and we know that there is no such thing as an "electron orbital circumference" then how is $2pi r=nlambda$ still valid?

We know that Bohr model is outdated and we know that there is no such thing as an "electron orbital circumference" then how is $2\pi r=n\lambda$ still valid?

Edit :

If the electrons for higher orbitals are not moving in a circular path then how do we write $2\pi r=n\lambda$?

Answer

The Bohr model is a semi classical model, treating the electrons like satellites of the proton, in the successful hydrogen atom solution. The success relied that the Bohr assumptions reproduced the series that fitted the hydrogen emission spectra.

The solution of the Schrodinger equation for the hydrogen atom reproduces the success of the Bohr model in fitting the spectra, and gives a theoretical basis for quantum mechanics, with the interpretation of the $ψ*ψ$ of the ψsolutions as probability of finding the electron at x,y,z around the proton.

It has been shown that the most probable radius of the hydrogen ground state is the same as the Bohr radius, explaining the success of the Bohr model for this simple potential case.

Edit after edit of question:

If the electrons for higher orbitals are not moving in a circular path then how do we write 2πr=nλ?

It is a useful approximation for rule of thumb, not accurate. One would have to go through the calculations for each n. After all it is the most probable radius of the ground state that is identified with the lowest Bohr orbit radius. The expectation value of the radius ( the average) even for the ground state is 1.5 of the Bohr radius. After all it is a different mathematical model.