= Solution
Here $V(x)$ is <potential energy>, $E$ is the stationary-state total <energy eigenvalue>, and $\psi(x)$ is its spatial <wavefunction>. <Wavefunction normalization> gives $\int|\psi|^2dx=1$, and $|\psi|^2dx$ is the <probability> of finding the particle in $dx$.
A <bound state> requires $-V_0<E<0$. Inside the well the equation is $\psi''=-k^2\psi$, whereas outside it is $\psi''=q^2\psi$, with
$$
\boxed{k=\frac{\sqrt{2m(E+V_0)}}\hbar,\qquad q=\frac{\sqrt{-2mE}}\hbar.}
$$
The <ground state> is even and has no <node>, selecting $\cos(kx)$ inside. Normalizability selects decaying, not growing, exterior exponentials. Continuity of $\psi$ and $\psi'$ gives
$$
A\cos(ka/2)=Ce^{-qa/2},\qquad k\tan(ka/2)=q.
$$
The first <excited state>, if present, is odd: a <sine> inside and opposite-sign decaying tails outside. Matching gives $-k_1\cot(k_1a/2)=q_1>0$. On its first odd branch, $\pi/2<k_1a/2<\pi$, hence $k_1a>\pi$. Since $k_1^2<2mV_0/\hbar^2$, existence requires
$$
\boxed{V_0>\frac{\pi^2\hbar^2}{2ma^2}.}
$$
This is the <first odd bound-state threshold of a square well>. There is no excited <bound state> below that threshold; equality gives a non-normalizable zero-energy limiting state, not an extra <bound state>.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1-ia2012-well-states.png]
{title=Original finite-well wavefunctions: an even nodeless ground state and an odd excited state in a deeper illustrative well}
For the numerical well,
$$
V_0=\frac{\hbar^2(k^2+q^2)}{2m}=8.04\,\mathrm{eV}\simeq8.0\,\mathrm{eV},\qquad
\boxed{E=-\frac{\hbar^2q^2}{2m}=-4.95\,\mathrm{eV}.}
$$
The <exterior probability of a symmetric finite-well state> includes both tails:
$$
\boxed{P_{\rm out}=2\int_{a/2}^{\infty}C^2e^{-2qx}\,dx
=\frac{C^2}{q}e^{-qa}=0.178\simeq17.8\%.}
$$
The threshold is $9.40\,\mathrm{eV}$, exceeding the well depth, so \b[there is no excited <bound state>]. The rounded supplied coefficients integrate to $0.9973$ rather than exactly one; this rounding does not change the quoted <probability> precision materially.
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