# Problem: What is the smallest wavelength in the Balmer's series?

###### FREE Expert Solution

We can determine the smallest wavelength, λmin in the Balmer series using the Balmer Equation shown below:

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

where:

λmin  = wavelength, m at the highest initial principal energy level: ↑ni ↓λ

R = 1.0974 x 107m-1 (Rydberg Constant)      **value can be found in textbooks or online
ni = initial principal energy level
nf = final principal energy level = 2 for Balmer Series

Recall that for the Balmer series the final principal energy level nf is always = 2.

The smallest wavelength, λmin will be the minimum wavelength corresponding to the highest initial energy level, ni possible or approaching infinity,  (a symbol used to denote a super high number)  for a Hydrogen atom.

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

Energy, E is inversely proportional to the wavelength, λE, niλ

100% (64 ratings) ###### Problem Details

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