The Bohr radius is approximately equal
to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state.
It is given by:
$$
a = \frac{4 \pi \varepsilon_0 \hbar^2}{m_e e^2} = \frac{\hbar}{m_e\,c}\frac{1}{\alpha}
$$
with the meaning of the constants as explained before. Suppose again that the elementary
particle (rest) mass (of the electron) shall be different / smaller in ints own past / greater in its own future,
say with a value $m$. Then we have for the Bohr radius $a$ at that time:
$$
a = \frac{\hbar}{m_e\,c\,\alpha} = \frac{m_0}{m}\frac{\hbar}{m_0\,c\,\alpha} \Longrightarrow
\quad \frac{a}{a_0} = \frac{m_0}{m}
$$
Here $a_0$ and $m_0$ are the values of the Bohr radius and (electron rest) mass here and now.
Conclusion: the (Bohr) radius of a hydrogen atom is inversely proportional with varying elementary
particle rest mass (of the electron).
Results remain valid even if Special Relativity is taken into account. The Bohr–Sommerfeld /
Dirac equation solution for the energy levels of the hydrogen atom is:
$$ E = m c^2 \left[\left\{ 1 + \frac{ (Z.\alpha)^2 }
{ \left[ \sqrt{ (j+ \frac{1}{2} )^2 - (Z.\alpha)^2 } + n \right] ^2 }
\right\}^{-1/2}-1\right] $$
Where: $ j+\frac{1}{2} \geq 1 $ en $ Z=1 $ , $ \alpha \approx 1/137 $ , $ m=m_e $.
There is an interesting consequence, worth to be noticed:
$$
\sqrt{ (j+ \frac{1}{2} )^2 - (Z.\alpha)^2 } = \mbox{real} \quad \Longleftrightarrow \quad
(1)^2 - (Z.\alpha)^2 \ge 0 \quad \Longleftrightarrow \quad Z \le 1/\alpha
$$
It follows that $\,Z \le 137$ : elements with more than $137$ protons are impossible,
but there is no nuclear physics involved with this equation.
It is seen again that
the energy levels are proportional with (the electron) rest mass. Because of $\,E=hc/\lambda\,$
there is a redshift with decreasing rest mass, just as with the non-relativistic approach.
Now the hydrogen atom stands as a quantum mechanical prototype for all other atoms in the universe. Let's make this argument more precise, starting with the helium atom as an example. The Schrödinger equation for the Helium atom is: $$ H\psi = E\psi \MET \psi(\vec{r}_1,\vec{r}_2) \EN \\ H=-{\frac {\hbar ^{2}}{2m_e}}(\nabla _{1}^{2}+\nabla _{2}^{2})+{\frac {e^{2}} {4\pi \epsilon _{0}}}{\Bigg [}{\frac {2}{r_{1}}}+{\frac {2}{r_{2}}}- {\frac {1}{|{\vec {r}}_{1}-{\vec {r}}_{2}|}}{\Bigg ]} $$ Assume for the moment that all size within the helium atom is scaled as $\vec{r'} = \vec{r}/L$, then we have, with a technique known as Nondimensionalization: $$ H\psi = \left\{-\frac{\hbar^2}{2m_e}(\nabla_1'^2+\nabla_2'^2)\frac{1}{L^2}+\frac{e^2}{4\pi\epsilon_0} \left[\frac{2}{r'_1}+\frac{2}{r'_2}-\frac{1}{|\vec{r'}_1-\vec{r'}_2|}\right]\frac{1}{L}\right\} \psi = E\psi \hieruit \\ \left\{-\half(\nabla_1'^2+\nabla_2'^2) + \left[\frac{2}{r'_1}+\frac{2}{r'_2}-\frac{1}{|\vec{r'}_1-\vec{r'}_2|}\right] \frac{m_e}{\hbar^2}L\frac{e^2}{4\pi\epsilon_0}\right\} \psi = \frac{m_e}{\hbar^2}L^2 E\psi $$ With: $$ \frac{m_e}{\hbar^2}L\frac{e^2}{4\pi\epsilon_0} = \frac{m_e c}{\hbar}L\frac{e^2}{4\pi\epsilon_0.\hbar c} = 1 \hieruit L = \frac{\hbar}{m_e\,c\,\alpha} = a $$ So the characteristic size for the helium atom is the Bohr radius of a hydrogen atom, i.e. the size of the helium atom must be of the same order of magnitude. But, what's far more important in the present context: that characteristic size is inversely proportional with varying elementary particle rest mass (of the electron). For the dimensionless energy $E'$ we define: $$ E' = \frac{2m_e}{\hbar^2}\left(\frac{\hbar}{m_e\,c\,\alpha}\right)^2 E \hieruit E = m_e c^2 \frac{\alpha^2}{2} E' = hc\,R\,E' \hieruit \\ \frac{hc}{\lambda} = hc\,R(E_1'-E_2') \hieruit \frac{1}{\lambda} = R(E_1'-E_2') $$ Where $\,R\,$ is the Rydberg constant again. This means that the energy levels of a helium atom are comparable with those of a hydrogen atom. But, what's far more important in the present context: the energy levels of a non-hydrogen atom are proportional too with varying elementary particle rest mass. And so the wavelength of the radiation emitted (or absorbed) by a non-hydrogen atom is inversely proportional with varying elementary particle rest mass, mainly the electron's mass.
Last but not least, let's consider the Lamb shift. From the Wikipedia reference we have: $$ \Delta E_{{\mathrm {Lamb}}}=\alpha ^{5}m_{e}c^{2}{\frac {k(n,0)}{4n^{3}}}\ {\mathrm {for}}\ \ell =0\,\\ \Delta E_{{\mathrm {Lamb}}}=\alpha ^{5}m_{e}c^{2}{\frac {1}{4n^{3}}}\left[k(n,\ell )\pm {\frac {1}{\pi (j+{\frac {1}{2}})(\ell +{\frac {1}{2}})}}\right]\ {\mathrm {for}}\ \ell \neq 0\ {\mathrm {and}}\ j=\ell \pm {\frac {1}{2}}, $$ where the proportionality between electron mass and energy is observed once again. Thus resulting in the same intrinsic redshift for all spectral phenomena, provided - of course - that the increasing elementary particle mass hypothesis is real.