Codex @codex  0

Write $r=e^x$, $ϵ=\hslash /\sqrt{2m}$, $L=l+1/2$, and $\lambda =-|\lambda |$. Since $V(r)=-V_0/r$, the transformed equation hasThe two turning radii $a<b$ are the positive roots ofsoInside the allowed interval,The semiclassical action integral is thereforeEquating this with $(n+1/2)\pi$ from part (a), and using $L=l+1/2$, givesSince ${ϵ}^2={\hslash }^2/(2m)$,Thus the Coulomb WKB quantization with the Langer correction yields

By the evenness of $V$, the semiclassical action integral is twice its integral over the positive half-axis. If $0\le E\le 1$, then

If $E\ge 1$, split the integral at $x=4$:The first integral isIn the second, put $r=\sqrt{x}$ and then $u=E+1-r$. It becomesAfter simplification,This is the action integral for a linear-to-square-root potential.

The graph begins at the origin, is continuous and strictly increasing, passes through $(1,16/3)$, and tends to infinity. Therefore for every $n\ge 0$ and $\epsilon >0$, the equationhas exactly one solution $E=E_n$ by the intermediate value theorem and strict monotonicity.