quantum mechanics - Energies and numbers of bound states in finite potential well

Hello I understand how to approach finite potential well (I learned a lot in my other topic here). However i am disturbed by equation which describes number of states $N$ for a finite potential well ( $d$ is a width of a well and $W_p$ is potential ):

$$N \approx \dfrac{\sqrt{2m W_p}d}{\hbar \pi}$$

I am sure it has something to do with one of the constants $\mathcal L$ or $\mathcal K$ defined this way:

$$\begin{align} \mathcal L &\equiv \sqrt{\tfrac{2mW}{\hbar^2}} & \mathcal{K}&\equiv \sqrt{ \tfrac{ 2m(W_p-W) }{ \hbar^2 }} \end{align}$$

and the transcendental equations for ODD and EVEN solutions:

$$\begin{align} &\frac{\mathcal K}{\mathcal L} = \tan \left(\mathcal L \tfrac{d}{2}\right) &&-\frac{\mathcal L}{\mathcal K} = \tan \left(\mathcal L \tfrac{d}{2}\right)\\ &\scriptsize{\text{transc. eq. - EVEN}} &&\scriptsize{\text{transc. eq. - ODD}} \end{align}$$

QUESTION: Could anyone tell me where does 1st equation come from?

Answer

Inside the well, the wave functions of a bound state behave approximately like $\sin(kx)$ i.e. standing waves where $k=\pi M/d$ for an integer $M$. This contributes the kinetic energy $\hbar^2 k^2/ 2m$. The highest-lying bound states are those for which the energy left to the particle is $0-$, a small negative number, outside the well.

So the kinetic energy inside the well must be approximately equal to the height of the well $W_p$:

$$W_p = \frac{\hbar^2 k^2}{2m}$$ This implies $$k = \frac{\sqrt{2mW_p}}{\hbar}$$ for the maximum allowed $k$ but I have mentioned that the spacing between the eigenstates in the $k$ space is $\pi/d$. So one has to divide the maximum $k$ above by $\pi/d$ to get the formula for $N$ you asked about.