Problem — The H-alpha line

Problem

A spectral line has wavelength $\lambda=656$ nm. Calculate the frequency and the photon energy; identify the region of the visible spectrum and the corresponding atomic transition.

Solution

Unit conversion. First convert the wavelength to metres: $\lambda=656\ \text{nm}=656\cdot 10^{-9}\ \text{m}=6{,}56\cdot 10^{-7}\ \text{m}$

Frequency. From $c=\lambda f$: $f=\frac{c}{\lambda}=\frac{3{,}00\cdot 10^8}{6{,}56\cdot 10^{-7}}\approx 4{,}57\cdot 10^{14}\ \text{Hz}$

Photon energy. $E=\frac{hc}{\lambda}=\frac{6{,}63\cdot 10^{-34}\cdot 3{,}00\cdot 10^8}{6{,}56\cdot 10^{-7}}\approx 3{,}03\cdot 10^{-19}\ \text{J}$

In electronvolts. Dividing by $1\;\text{eV}=1{,}6\cdot 10^{-19}$ J: $E=\frac{3{,}03\cdot 10^{-19}}{1{,}6\cdot 10^{-19}}\approx 1{,}89\ \text{eV}$

Identification. $\lambda=656$ nm falls in the red part of the visible spectrum. It is hydrogen’s $H_\alpha$ line, due to the electronic transition from level $n=3$ to level $n=2$ (Balmer series).

$\ev{f\approx 4{,}57\cdot 10^{14}\ \text{Hz},\qquad E\approx 3{,}03\cdot 10^{-19}\ \text{J}\approx 1{,}89\ \text{eV}}$

Connections

Topics: Electromagnetic waves Concepts: Photon Skills: Unit conversion