# Why is the difference between atomic radii of Na and K greater than that of Rb and Cs?

I am aware that this question has already been asked here: Why does the "gap" between atomic radii get smaller down the periodic table?.

However, I was wondering if there is a way to answer this question using simple periodic trends instead of a formula to calculate atomic radius.

Since there is not only an s and p-block between Rb and Cs, but also a d-block, wouldn't the increase in shielding between Rb and Cs be higher than that of Na and K? Thus, shouldn't Rb and Cs have a larger difference in atomic radii?

I got this question in my chemistry class while learning about periodic trends, and it apparently also appeared on the 1993 AP Exam, so I would think that this problem could be explained using simple periodic trends.

Electronic orbitals do not come in fixed sizes. A given orbital will shrink due to greater nuclear charge as more protons are addded to the nucleus. Going from $\ce{Rb}$ to $\ce{Cs}$ you are adding more orbitals than when you go from $\ce{Na}$ to $\ce{K}$. But you are also adding more orbital-shrinking protons to the nucleus.
The shrinkage effect on each orbital with more protons gets stronger as we go to heavier atoms, when relativistic effects kick in. Put in very rough terms: when you solve the Schroedinger equation for electronic wavefunctions the size of the orbital would decrease if the mass of the electron were increased. In nonrelativistic quantum mechanics this does not occur, but in relativistic mechanics the energy needed to squeeze the wavefunctions (which in this case comes from electrostatic attraction) acts like mass, and thus allows for even more shrinkage. For alkaline earth metals we end up with almost no change at all going from $\ce{Ba}$ to $\ce{Ra}$. See https://periodictablegroups.wordpress.com/alkaline-earth-metals-2/.