Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy Functions.
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$\uparrow$ Fig.1: Potential $V(x)$ as a function of position $x$ in OP's example.
First let us include the metaplectic correction/Maslov index. The turning point at an infinitely hard wall and an inclined potential wall have Maslov index $2$ and $1$, respectively, cf. e.g. this Phys.SE post. In total $3$. We should then adjust the Bohr-Sommerfeld quantization rule with a fraction $\frac{3}{4}$.
$$ \int_{x_-}^{x_+} \! \frac{dx}{\pi} k(x)~\simeq~n+\frac{3}{4},\qquad n~\in~\mathbb{N}_0,\tag{1} $$
where
$$ k(x)~:=~\frac{\sqrt{2m(E-V(x))}}{\hbar}, \qquad
V(x)~:=~-V_0 \frac{L-x}{L}. \tag{2} $$
At the threshold, we can assume $n=0$ and $E=0$. The limiting values of the turning points are $x_-=0$ and $x_+=L$. Straightforward algebra yields that the
threshold between the existence of zero and one bound state is
$$V_0~\simeq~\frac{81}{128} \frac{\pi^2\hbar^2}{mL^2} \tag{3} .$$
$^1$ For comparison, the WKB approximation for the threshold of the corresponding square well problem yields
$$V_0~\simeq~\frac{\pi^2\hbar^2}{2m L^2} \tag{4} ,$$
while the exact quantum mechanical result is
$$V_0~=~\frac{\pi^2\hbar^2}{8m L^2} \tag{5} ,$$
cf. e.g. Alonso & Finn, Quantum and Statistical Physics, Vol 3, p. 77-78. Not impressive!
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$\uparrow$ Fig.2: Corresponding square well potential as a function of position $x$. Each of the 2 infinitely hard walls has Maslow index 2.
Yes you are long as the incident beam of electrons is not incoming from the right
Because the energy is greater than the potential we do not expect the particle to penetrate inside the x<0 region. This means the particle cannot exist on the right because the kinetic energy is not enough to overcome the potential.
When you get the solution
$\psi(x) = Fe^{-lx} + Ge^{lx}$
its a basic postulate of quantum mechanics that bound states must approach 0 as x goes to infinite and this can be proved mathematically. Remember that any smooth function that fails to approach zero at infinity will obviously have an infinite integral and the wavefunction must have an integral of 1.
so G must be 0 so that $Fe^{-lx} + Ge^{lx}$ does not go to infinity as x approaches infinite
Best Answer
you are right there is no way to use orthognality and this issue is mainly due to the scattering states. The scattering states are not orthogonal to the bound states and even to each other.