How do we realise the wave for an electron orbiting a nucleus in its first orbit as per bohr model?

How do we realise the wave for an electron orbiting a nucleus in its first orbit as per bohr model?

As per my textbook, we can account for as to why angular momentum is quantised. But I fail to understand what would the wave diagram for n=1 look like?? For example, for higher orbits.

As to my knowledge, such a diagram should not be possible for first orbit. Then how come we can account for quantisation for the first orbit?

For example, for higher orbits, the following diagrams follow... Maybe the term "orbiting" causes some problems. The electron is not moving around the nucleus as bohr assumed (he had no solution according to the problem that an electron on a circular orbit should emit electromagnetic waves and crash into the nucleus).

You should think about the elcetron as part of the whole system which can have some quantized energy eigenstates and so some quantum numbers. According to this point of view you need to solve the schrödinger equation and interpret the solution as probability-density for the elctron states.

The absolut value square of your wave function is a positiv number and thats what you call the orbital. How this orbitals look like depends on the quantum numbers of the electron state. In the picture below you can see a few orbitals for the lowest quantum numbers of the hydrogen atom. The electron wave for the $1^{st}$ orbit would be like a circle (not exactly a circle) moving up and down the orbit.

$mvr=\frac{h}{2\pi}$

If the motion of the wave was broken in 3 stages, it would look something like this :

Stage 1 : Stage 2 : Stage 3 : Why is this possible :

Bohr's model assumes that the electron wave is continuous over the stationary orbits circumference, i.e,

$n\lambda = 2\pi r$,

But according to de-Broglie, $\lambda=\frac{h}{mv}$,

$n\frac{h}{mv}=2\pi r$

Hence, $mvr=\frac{nh}{2\pi}$.

If you observe, in a electron wave over (say) $2_{nd}$ orbit there are 2 crests and 2 troughs. Similarly there are 3 sets of crests and troughs for the $3_{rd}$ orbit.

So its very much possible that in the $1_{st}$ orbit there is only 1 set of crest and trough.

Here's a link to a simulation for different models of Hydrogen: