1. $n$
2. $n$
4. $n$

1. $i$
2. $i$
4. $i$

1. $E$
2. $E$
4. $E$

5. $$\color{blue}{\rightarrow}\color{black}\;$$$$E(eV) = -\frac{3,16}{n^2}$$

1. $E_{ni}$
2. $E_{ni}$
4. $E_{ni}$

5. $$\color{blue}{\rightarrow}\color{black}\;$$$$E_{ni}(eV) = (\frac{1}{i^2}-\frac{1}{n^2})\cdot 3.16 $$

1. $\nu_{ni}$
2. $\nu_{ni}$
4. $\nu_{ni}$

5. $$\color{blue}{\rightarrow}\color{black}\;$$$$\nu_{ni}(Hz) = \frac{E_{ni}}{h} $$

   $$\color{blue}{\rightarrow}\color{black}\;$$$$\nu_{ni}(Hz) = \frac{1}{h}(\frac{1}{i^2}-\frac{1}{n^2})\cdot 3.16 \cdot 1.60218 \cdot 10^{-19} $$

1. $\lambda_{ni}$
2. $\lambda_{ni}$
4. $\lambda_{ni}$

5. $$\color{blue}{\rightarrow}\color{black}\;$$$$\lambda_{ni}(nm) = \frac{c\cdot h}{E_{ni}} $$

   $$\color{blue}{\rightarrow}\color{black}\;$$$$\lambda_{ni}(nm) = \frac{c\cdot h\cdot 10^9}{(\frac{1}{i^2}-\frac{1}{n^2})\cdot 3.16 \cdot 1.60218 \cdot 10^{-19} } $$

Calculations on saturated solutions of pure ionic substances

$$\color{blue}{\rightarrow}\color{black}\; m\,=\,n $$ $$\color{blue}{\rightarrow}\color{black}\; m\,=1 \; \,n=2 $$ $$\color{blue}{\rightarrow}\color{black}\; m\,=1 \; \,n=3 $$ $$\color{blue}{\rightarrow}\color{black}\; m\,=2 \; \,n=1 $$ $$\color{blue}{\rightarrow}\color{black}\; m\,=3 \; \,n=1 $$ $$\color{blue}{\rightarrow}\color{black}\; m\,=2 \; \,n=3 $$ $$\color{blue}{\rightarrow}\color{black}\; m\,=3 \; \,n=2 $$