The energy spectrum · Quantum Mechanics

Intuition

An atom that jumps from a higher level to a lower one emits a photon carrying the difference. Hydrogen’s levels crowd together under zero like the rungs of a ladder that narrows as it rises, so its lines come in series, one for each lower level: ultraviolet lines ending on $n=1$ , the visible Balmer lines ending on $n=2$ , infrared ones on $n=3$ . Each series crowds towards a limit, the energy needed to free the electron from that level.

A staircase whose steps get shorter and shorter towards a landing: jumping down from any step to the second one gives a family of drops that get closer together, towards the height of the whole flight.

Hydrogen’s levels $E_{n}=-13.6\ \text{eV}/n^{2}$ , crowding towards zero, the dashed line, where the electron is free. The drop from $n=2$ to emits the ultraviolet Lyman line at 122 nm; from to , the red Balmer line at 656 nm.

Lines and series

A transition from level $n_{2}$ to level $n_{1}$ emits a photon of energy $E_{n_{2}}-E_{n_{1}}$ . Written as a wavelength, this is the Rydberg formula.

\begin{aligned}\frac{hc}{\lambda}&=13.6\ \text{eV}\Bigl(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Bigr)\\ \frac{1}{\lambda}&=R_{\infty}\Bigl(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Bigr)\end{aligned}

The series

Lyman, $n_{1}=1$ : ultraviolet, from 122 nm down to the limit at 91 nm.
Balmer, $n_{1}=2$ : visible, 656 nm (red), 486, 434, 410 nm, down to 365 nm.

The Rydberg formula

E_{n}=-\frac{\text{Ry}}{n^{2}}\ \Longrightarrow\ \frac{1}{\lambda}=\frac{\text{Ry}}{hc}\Bigl(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Bigr)

The photon carries off the difference between the two levels, and a photon’s energy is $hc/\lambda$ . Divide by $hc$ .

Proof steps

E_{n}=-\frac{\text{Ry}}{n^{2}},\qquad \text{Ry}\approx13.6\ \text{eV}

The spectrum from the previous lesson.

\frac{hc}{\lambda}=E_{n_{2}}-E_{n_{1}}

Energy is conserved: the photon takes the difference, for $n_{2}>n_{1}$ .

E_{n_{2}}-E_{n_{1}}=\text{Ry}\Bigl(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Bigr)

Subtract the two levels.

\frac{1}{\lambda}=\frac{\text{Ry}}{hc}\Bigl(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Bigr)

Divide by $hc$ ; $\text{Ry}/hc$ is the Rydberg constant.

n_{2}\to\infty:\ \ \frac{1}{\lambda}\to\frac{\text{Ry}}{hc\,n_{1}^{2}}

Each series ends at a limit, set by the energy to free the electron from $n_{1}$ .

Applications

Practice

Photons From Jumps

A jump from level $n_{2}$ down to $n_{1}$ emits a photon carrying the difference of the energies. With $E=hc/\lambda$ , is about 1240 eV nm.

\frac{hc}{\lambda}=13.6\ \text{eV}\Bigl(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\Bigr)

Try it

Using $13.6$ eV and $hc=1240$ eV nm, what is the wavelength of the $n=3\to2$ line of hydrogen, in nm? Give the nearest whole number.