Why do we use tin in extreme ultraviolet (EUV) lithography? I am in awe of what we have achieved with EUV light, and like to think about it in my free time. One question that I have is: What are the calculations that point us toward tin as the best EUV light source? How do we calculate its quantity and the right amount of energy to zap it with? I want to start off by plugging tin's elemental number into Rydberg's equation, which is $$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right).$$ However, I have to be careful, because that equation breaks down for p, d, and higher, more nuanced orbitals. So I started playing around with the electron affinity of tin, which is $-100\;\mathrm{kJ}/\mathrm{mol}$. I figure this can lead to how powerful the lasers that zap it need to be, because electron affinity shows us how much energy is needed to knock a valence electron out of its orbital, which then releases light. In order to get the amount of energy needed to do this, you just multiply $-100\;\mathrm{kJ}/\mathrm{mol}$ by $-1$ and divide by Avogadro's number. When I did though, I only got $1\;\mathrm{eV}$, which $\mathrm dE=h\nu$ says correlates to $1\cdot 10^3\;\mathrm{nm}$. light made by each electron. Clearly I am missing something since we are working with $<10\;\mathrm{nm}$ light these days. Please fill me in!
Best,
Anthony
 A: Lithium and Xeon were considered, but it turned out Tin had a better conversion efficiency.
From this open access paper

Near 13.5 nm wavelength, the EUV spectrum of Sn highly charged ions is dominated by intense unresolved transition arrays (UTAs ) arising mainly from the resonance transitions 4p6 4dm − 4p5 4dm+1 + 4dm−1 4f in Sn8+–Sn14+ although other more highly excited states may also significantly contribute.


High power densities are required to obtain the ∼20–40 eV plasma temperatures to produce the highly charged Sn ions. This temperature requirement can be understood through the Stefan–Boltzmann law j* = σSBT4 describing the energy emitted per second per unit surface by a black body j* as a function of temperature T; σSB is the Stefan–Boltzmann constant.

The droplet of tin needs to be small about 20 um otherwise even bigger lasers would need it.
The pre pulse that flattens the droplet has to be tailored so the droplet  doesn’t explode.
The pulse that creates the dense plasma has to be absorbed efficiently, and you want to do it in such a way that vaporized tin doesn’t coat your optics, so there i s a lot of science and spectroscopy to understand how the light is absorbed and what type of excited states are created.
To understand all the interactions the computational modeling is very intensive. The paper talks some about the team and methods used to calculate such a complex system and emphasis is placed on understanding the physics…
