Why do heavier isotopes of the same element have smaller atomic radii than lighter isotopes of the same element? I have been trying to figure out why higher-mass isotopes have higher melting and boiling points than lower-mass isotopes of the same element. 
A Quora answer on this topic explored the idea that electron orbits in atoms with smaller nuclei behave as if the electron is lighter, contributing to larger orbits, whereas atoms with larger nuclei behave as if the electron is heavier, contributing to smaller orbits. 
Could someone explain why the size of the nucleus affects electron orbits?
 A: To make it clear: I don't know the answer. However, here is how I would try to answer the why question:


*

*The nuclear radius increases with the number of nuclei. Assuming that the nucleus is a sphere, we would expect that the radius of the nucleus scales like $R_n = R_0 \cdot A^{1/3}$, where $A$ is the mass of the nucleus. You could cross-check this by looking at the so called liquid droplet model, but I am pretty sure that this is correct.

*If the electrons would go around the nucleus in circles, the radius of the nucleus should not influence the energy levels of the electrons. However, electrons do not merely circle the nucleus, but they have certain probability distributions depending on their orbits (s, p, d, ...). Some even have a non-zero probability to be within the nucleus. Therefore, the radius of the nucleus will affect the binding energy of the electrons. I would expect that the electrons are bound tighter if the nucleus is enlarged. However, I don't have a proper argument and you should validate this. Side mark: As can be seen from the muonic hydrogen experiments, a larger binding energy is equivalent to a heavier "electron".


Good luck.
A: Hints: 


*

*Heavier isotopes have higher reduced mass $\mu$.

*For a hydrogen-like atom, the energy levels $E$ are proportional (& the radius $r$ is inversely proportional) to the reduced mass $\mu$. 

*More generally, it can be deduced from dimensional analysis alone that the conclusions of pt. 2 hold for any spinless non-relativistic multi-electron atom governed by the Schrödinger equation:
$$r~~\propto~~\frac{\hbar^2}{\mu ~k_e e^2}~~\propto~~\frac{1}{\mu}\qquad\text{and}\qquad E~~\propto~~\mu\frac{ (k_e e^2)^2}{\hbar^2}~~\propto~~\mu.$$
This is because that that model has only 3 dimensionful constants: 


*

*The reduced mass: $[\mu]~=~M.$

*The Planck constant: $[\hbar]~=~ML^2/T~=~ET.$

*The Coulomb constant times the square of the elementary charge:  $[k_e e^2]~=~ML^3/T^2~=~EL.$


