We can determine the **smallest wavelength, λ_{min} **in the

$\overline{)\frac{1}{{\lambda}_{min}}{=}{R}{\times}\left(\frac{1}{{{n}_{f}}^{2}}-\frac{{\displaystyle 1}}{{\displaystyle {{n}_{i}}^{2}}}\right)}\phantom{\rule{0ex}{0ex}}$

*where: *

**λ _{min}**

R = 1.0974 x 10^{7}m^{-1} (Rydberg Constant) ***value can be found in textbooks or online *n

Recall that for the** ****Balmer series the final principal energy level n _{f} is always = 2.**

The smallest wavelength, ** λ_{min}** will be the

Recall that the **highest transition** releases the** highest energy, E.**

**Energy, E** is **inversely proportional** to the **wavelength, ****λ**:

What is the smallest wavelength in the Balmer's series?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Bohr and Balmer Equations concept. If you need more Bohr and Balmer Equations practice, you can also practice Bohr and Balmer Equations practice problems.