Question

What is the shortest wavelength present in the Brackett series of spectral lines?

Answer 1

Rydberg's formula for the Bracket series looks like this:

$\frac{1}{\lambda}=R(\frac{1}{4^2}-\frac{1}{m^2})$ , where $R=1.0974\cdot10^{7}\ (m^{-1})$ .

If $m\to\infin$ then

$\frac{1}{\lambda}=1.0974\cdot10^{7}\cdot(\frac{1}{4^2}-\frac{1}{\infin})\to\lambda=1458\ (nm)$ . Answer