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Take QED, consider the spherical symmetric solutions for hydrogen, Add a condition that the mass,charge is zero for $r<r_0$. and at $r_0$, all mass,charge. Let the Photon be a standing wave. Assume $w$ constant (the eigen value) and $w$ proportional to $m_e c^2 + 2|V|$, e.g. assume $|V|=|K|$ (true for the Bohr model). To check yourself, lookup spherical wells for Dirac or Klein Gordon. Then you get Bohrs result. You essentially will need $j_0(r_0) = \sin(k_{\text{electron}} r_0)/(k r_0)=0$, Probably you can use the correspondence principle on the spherical shell (Hence $|K|=|V|$), Also note that you can solve the standard QED by having the same $k(r),w$ as this setup (hence the same energy) by stitching infinitely many infinitesimal thin spherical wells together e.g. you have a solution $A(r)j_0(kr)+B(r)\eta_0(kr)$, that approximately have the same eigen values, coincidence? you may need to fudge a $4\pi =$ 'surface area of spherical shell with radius 1'. So very very close because, $$ j_0(kr) = \int_{|\hat \xi|=1}\exp(i k (\hat \xi \cdot \hat r))\,dS(\hat\xi) $$.

Maybe things still are unclear but maybe the link below can make the thoughts clearer, Discussoinal paper

Take QED, consider the spherical symmetric solutions for hydrogen, Add a condition that the mass,charge is zero for $r<r_0$. and at $r_0$, all mass,charge. Let the Photon be a standing wave. Assume proportional to $m_e c^2 \pm |V|/2$, e.g. assume the classical $|V|/2 = |K|$ (true for the Bohr model). To check yourself, lookup spherical wells for Dirac or Klein Gordon. Then to get Bohrs result assume the whole QM setup with $h$ in stead of $\hbar$ and use the correspondence principle on the shell. You essentially will need $j_0(r_0) = \sin(k_{\text{electron}} r_0)/(k r_0)=0$, (The correspondence principle on the spherical shell (Hence $|K|=|V|/2$).

Maybe things still are unclear but maybe the link below can make the thoughts clearer, Discussoinal paper

Take QED, consider the spherical symmetric solutions for hydrogen, Add a condition that the mass,charge is zero for $r<r_0$. and at $r_0$, all mass,charge. Let the Photon be a standing wave. Assume $w$ constant (the eigen value) and $w$ proportional to $m_e c^2 + 2|V|$, e.g. assume $|V|=|K|$ (true for the Bohr model). To check yourself, lookup spherical wells for Dirac or Klein Gordon. Then you get Bohrs result. You essentially will need $j_0(r_0) = \sin(k_{\text{electron}} r_0)/(k r_0)=0$, Probably you can use the correspondence principle on the spherical shell (Hence $|K|=|V|$), Also note that you can solve the standard QED by having the same $k(r),w$ as this setup (hence the same energy) by stitching infinitely many infinitesimal thin spherical wells together e.g. you have a solution $A(r)j_0(kr)+B(r)\eta_0(kr)$, that approximately have the same eigen values, coincidence? you may need to fudge a $4\pi =$ 'surface area of spherical shell with radius 1'. So very very close because, $$ j_0(kr) = \int_{|\hat \xi|=1}\exp(i k (\hat \xi \cdot \hat r))\,dS(\hat\xi) $$.

Take QED, consider the spherical symmetric solutions for hydrogen, Add a condition that the mass,charge is zero for $r<r_0$. and at $r_0$, all mass,charge. Let the Photon be a standing wave. Assume $w$ constant (the eigen value) and $w$ proportional to $m_e c^2 + 2|V|$, e.g. assume $|V|=|K|$ (true for the Bohr model). To check yourself, lookup spherical wells for Dirac or Klein Gordon. Then you get Bohrs result. You essentially will need $j_0(r_0) = \sin(k_{\text{electron}} r_0)/(k r_0)=0$, Probably you can use the correspondence principle on the spherical shell (Hence $|K|=|V|$), Also note that you can solve the standard QED by having the same $k(r),w$ as this setup (hence the same energy) by stitching infinitely many infinitesimal thin spherical wells together e.g. you have a solution $A(r)j_0(kr)+B(r)\eta_0(kr)$, that approximately have the same eigen values, coincidence? you may need to fudge a $4\pi =$ 'surface area of spherical shell with radius 1'. So very very close because, $$ j_0(kr) = \int_{|\hat \xi|=1}\exp(i k (\hat \xi \cdot \hat r))\,dS(\hat\xi) $$.

Maybe things still are unclear but maybe the link below can make the thoughts clearer, Discussoinal paper

Take QED, consider the spherical symmetric solutions for hydrogen, Add a condition that the mass,charge is zero for $r<r_0$. and at $r_0$, all mass,charge. Let the Photon be a standing wave. Assume $w$ constant (the eigen value) and $w$ proportional to $m_e c^2 + 2|V|$, e.g. assume $E=K$ (true for the Bohr model). To check yourself, lookup spherical wells for Dirac or Klein Gordon. Then you get Bohrs result. You essentially will need $j_0(r_0) = \sin(k_{\text{electron}} r_0)/(k r_0)=0$, Probably you can use the correspondence principle on the spherical shell (Hence $|K|=|V|$), Also note that you can solve the standard QED by having the same $k(r),w$ as this setup (hence the same energy) by stitching infinitely many infinitesimal thin spherical wells together e.g. you have a solution $A(r)j_0(kr)+B(r)\eta_0(kr)$, that approximately have the same eigen values, coincidence? you may need to fudge a $4\pi =$ 'surface area of spherical shell with radius 1'. So very very close.

Take QED, consider the spherical symmetric solutions for hydrogen, Add a condition that the mass,charge is zero for $r<r_0$. and at $r_0$, all mass,charge. Let the Photon be a standing wave. Assume $w$ constant (the eigen value) and $w$ proportional to $m_e c^2 + 2|V|$, e.g. assume $|V|=|K|$ (true for the Bohr model). To check yourself, lookup spherical wells for Dirac or Klein Gordon. Then you get Bohrs result. You essentially will need $j_0(r_0) = \sin(k_{\text{electron}} r_0)/(k r_0)=0$, Probably you can use the correspondence principle on the spherical shell (Hence $|K|=|V|$), Also note that you can solve the standard QED by having the same $k(r),w$ as this setup (hence the same energy) by stitching infinitely many infinitesimal thin spherical wells together e.g. you have a solution $A(r)j_0(kr)+B(r)\eta_0(kr)$, that approximately have the same eigen values, coincidence? you may need to fudge a $4\pi =$ 'surface area of spherical shell with radius 1'. So very very close because, $$ j_0(kr) = \int_{|\hat \xi|=1}\exp(i k (\hat \xi \cdot \hat r))\,dS(\hat\xi) $$.