= Solution
A <travelling wave>, for example $A\cos(kx-\omega t)$, has moving constant-phase surfaces. A <standing wave>, for example $2A\sin(kx)\cos(\omega t)$, has stationary <nodes> and can be formed from two equal counterpropagating waves.
For the <half-line finite well> with $E>V_0$, define $k=\sqrt{2mE}/\hbar$ and $q=\sqrt{2m(E-V_0)}/\hbar$. The hard wall requires $\psi=0$ for $x<0$ and $\psi(0)=0$, so the general allowed forms are
$$
\boxed{\psi(x)=\begin{cases}
0,&x<0,\\
A\sin(kx),&0<x<a,\\
Be^{iq(x-a)}+Ce^{-iq(x-a)},&x>a.
\end{cases}}
$$
At the finite step both $\psi$ and $\psi'$ are continuous:
$$
A\sin(ka)=B+C,\qquad Ak\cos(ka)=iq(B-C).
$$
There is no requirement to match the derivative to zero across the infinite wall.
For a <bound state> $0<E<V_0$, put $\kappa=\sqrt{2m(V_0-E)}/\hbar>0$. Normalizability excludes the growing exponential, leaving $A\sin(kx)$ inside and $De^{-\kappa(x-a)}$ outside. Matching their logarithmic derivatives gives
$$
\boxed{k\cot(ka)=-\kappa,\qquad \tan(ka)=-k/\kappa.}
$$
The exterior expression is equivalent to $Ce^{-\kappa x}$ by rescaling its constant. Negative cotangent first occurs for $\pi/2<ka<\pi$; all higher allowed intervals also have $ka>\pi/2$. Thus $\lambda_1=2\pi/k<4a$. Since $k<\sqrt{2mV_0}/\hbar$, a normalizable <bound state> requires
$$
\boxed{V_0>\frac{\pi^2\hbar^2}{8ma^2}.}
$$
This is also sufficient: on the lowest negative-cotangent branch, the two matching sides cross whenever $a\sqrt{2mV_0}/\hbar>\pi/2$. \b[The printed non-strict inequality includes the threshold limit.] At equality $E=V_0$ and the exterior is not square-integrable, so it is not itself a <bound state>.
For the helium <mass> and <energy> scale given, the limiting width is
$$
\boxed{a_{\rm th}=\frac{\pi\hbar}{\sqrt{8m_{\rm He}V_0}}=1.14\times10^{-10}\,\mathrm m=0.114\,\mathrm{nm}.}
$$
Actual binding requires a width just larger than this threshold in the stated ideal potential.
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