41.6.1 Excitation by Absorption

Left to itself, an atom will be in its lowest-energy ground state. How does an atom get into an excited state? The process of getting an atom into an excited state is called excitation, and there are two basic mechanisms: absorption and collision. We’ll begin by looking at excitation by absorption.

One postulate of the Bohr model is that an atom can jump from one stationary state, of energy $E_{1}$ , to a higher-energy state $E_{2}$ by absorbing a photon of frequency

f = \frac{Δ E_{atom}}{h} = \frac{E_{2} - E_{1}}{h}

(41.15)

Because we are interested in spectra, it is more useful to write Equation 41.15 in terms of the wavelength:

λ = \frac{c}{f} = \frac{h c}{Δ E_{atom}} = \frac{1240 eV nm}{Δ E (in eV)}

(41.16)

The final expression, which uses the value $h c = 1240 eV nm$ , gives the wavelength in nanometers if $Δ E_{atom}$ is in electron volts.

A photograph of two beryllium atoms trapped in an ion trap emitting light in the form of two dots. — The dots of light are being emitted by two beryllium ions held in a device called an ion trap. Each ion, which is excited by an invisible ultraviolet laser, emits about $10^{6}$ visible-light photons per second.

Bohr’s idea of quantum jumps remains an integral part of our interpretation of the results of quantum mechanics. By absorbing a photon, an atom jumps from its ground state to one of its excited states. However, a careful analysis of how the electrons in an atom interact with a light wave shows that not every conceivable transition can occur. The allowed transitions must satisfy a selection rule: A transition (either absorption or emission) from a state in which the valence electron has orbital quantum number $l_{1}$ to another with orbital quantum number $l_{2}$ is allowed only if

Δ l = | l_{2} - l_{1} | = 1 (selection rule for emission and absorption)

(41.17)

That is, the electron’s orbital quantum number must change by exactly 1. Thus an atom in an $s$ state $(l = 0)$ can absorb a photon and be excited to a $p$ state $(l = 1)$ but not to another $s$ state or to a $d$ state. An atom in a $p$ state $(l = 1)$ can emit a photon by dropping to a lower-energy $s$ state or to a lower-energy $d$ state but not to another $p$ state.

EXAMPLE 41.5 Absorption in hydrogen

What is the longest wavelength in the absorption spectrum of hydrogen? What is the transition?

SOLVE The longest wavelength corresponds to the smallest energy change $Δ E_{atom}$ . Because the atom starts from the $1 s$ ground state, the smallest energy change occurs for absorption to the first $n = 2$ excited state. The energy change is

Δ E_{atom} = E_{2} - E_{1} = \frac{- 13.6 eV}{2^{2}} - \frac{- 13.6 eV}{1^{2}} = 10.2 eV

The wavelength of this transition is

λ = \frac{1240 eV nm}{10.2 eV} = 122 nm

This is an ultraviolet wavelength. Because of the selection rule, the transition is $1 s \to 2 p$ , not $1 s \to 2 s$ .

EXAMPLE 41.6 Absorption in sodium

What is the longest wavelength in the absorption spectrum of sodium? What is the transition?

SOLVE The sodium ground state is $[Ne] 3 s$ . The lowest excited state is the $3 p$ state. $3 s \to 3 p$ is an allowed transition $(Δ l = 1)$ , so this is the longest wavelength. You can see from the data in Figure 41.23 that $Δ E_{atom} = 2.104 eV$ for this transition.

The corresponding wavelength is

λ = \frac{1240 eV nm}{2.104 eV} = 589 nm

REVIEW This wavelength (yellow color) is a prominent feature in the spectrum of sodium. Because the ground state has $l = 0$ , absorption must be to a $p$ state. The $s$ states and $d$ states of sodium cannot be excited by absorption.