The energy of the photon given out when an electron drops between levels equals:

The difference in energy between the two levels

The Quantum Atom | HSC Physics Module 8

In short: In the quantum model of the atom, electrons can only have certain energy levels. An electron gives out or takes in a photon only when it jumps between levels, with $h f = E_{i} - E_{f}$ , which is why each element has its own line spectrum.

Bohr's model of the hydrogen atom

In 1913, Niels Bohr kept Rutherford's nucleus but added new rules, called postulates:

Electrons move in certain allowed orbits, called stationary states, without giving off radiation. Each has a fixed energy.
An electron gives off or absorbs a photon only when it jumps between allowed orbits. The photon's energy equals the difference in energy:

$E = h f = E_{i} - E_{f}$

Allowed orbits have angular momentum that is a whole number multiple of $\frac{h}{2 π}$ :

$m v r = \frac{nh}{2 π}$

The whole number $n$ is the principal quantum number. $n = 1$ is the ground state, the lowest energy.

Energy levels of hydrogen

For hydrogen, the allowed energies are:

$E_{n} = - \frac{13.6}{n ^{2}} eV$

$n$	Energy
1	−13.6 eV
2	−3.40 eV
3	−1.51 eV
4	−0.85 eV
∞	0 (the electron is free)

The energies are negative because the electron is bound to the nucleus. To free an electron from the ground state (ionise the atom) takes 13.6 eV.

Emission and absorption

Emission: an electron drops to a lower level and gives out a photon. A bigger drop means a higher-energy photon with a shorter wavelength.
Absorption: an atom absorbs a photon only if its energy exactly matches a jump up. This makes the dark lines in absorption spectra.

The Rydberg equation

The wavelengths of hydrogen's lines are given by:

$\frac{1}{λ} = R (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})$

$R = 1.097 \times 1 0^{7}$ m⁻¹ (the Rydberg constant)
$n_{f}$ = final level, $n_{i}$ = initial level

Lines are grouped into series by where the electron lands:

Series	Ends on	Type of light
Lyman	$n_{f} = 1$	ultraviolet
Balmer	$n_{f} = 2$	visible
Paschen	$n_{f} = 3$	infrared

Limits of Bohr's model

Bohr's model worked brilliantly for hydrogen, but:

It only worked for atoms with one electron.
It could not explain why some lines are brighter than others.
It could not explain the fine splitting of lines into close pairs, or their splitting in a magnetic field.
It mixed old physics (orbits) with new rules, without explaining why the orbits are allowed.

de Broglie: electrons are waves too

In 1924, Louis de Broglie suggested that if light waves can act like particles, then particles might act like waves. Any moving particle has a wavelength:

$λ = \frac{h}{m v}$

where $m v$ is the particle's momentum.

For everyday objects, $λ$ is far too small to notice.
For electrons, $λ$ is about the size of the gaps between atoms in a crystal.

In 1927, Davisson and Germer fired electrons at a nickel crystal and saw a diffraction pattern, just like X-rays. Electrons really do behave as waves.

Why Bohr's orbits are allowed

de Broglie explained Bohr's rule. An electron's wave must fit around its orbit as a standing wave, a whole number of wavelengths:

$2 π r = nλ$

Any other orbit would make the wave cancel itself out. Substituting $λ = \frac{h}{m v}$ gives exactly Bohr's rule $m v r = \frac{nh}{2 π}$ .

The Quantum Atom

Bohr's postulates

Stationary states
Photon: E = Ei − Ef
mvr = nh / 2π

Hydrogen levels

En = −13.6 / n² eV
Ground state n = 1
Ionisation 13.6 eV

Spectra

Emission: drop down
Absorption: exact jump up
Rydberg equation
Balmer series visible

de Broglie

λ = h / mv
Electron diffraction
Standing waves explain orbits

Schrödinger

Electron clouds
Probability

Year 12Schrödinger's model

Erwin Schrödinger (1926) wrote an equation for the electron's wave in three dimensions. Its solutions do not give neat orbits. Instead, they give regions called orbitals, where the electron is likely to be found. The electron is described by a probability cloud, not a path.

This model works for atoms with many electrons, explains the brightness of spectral lines and underpins all of chemistry. It removed the certainty about where an electron is: we can only say where it probably is.

Series limit. The shortest wavelength in a series comes from an electron falling from $n_{i} = \infty$ . For the Balmer series: $\frac{1}{λ} = R (\frac{1}{4} - 0)$ , so $λ = 365$ nm. Lines crowd together near this limit because the energy levels get closer together as $n$ grows.

Which transition? The bigger the energy gap, the shorter the wavelength. In hydrogen, jumps to $n = 1$ have the biggest gaps. The longest Lyman wavelength is the smallest jump, $n = 2 \to 1$ : $\frac{1}{λ} = R (1 - \frac{1}{4})$ , so $λ = 122$ nm.

Matching an absorbed wavelength. Work out the photon energy, then find the pair of levels with exactly that gap. Light at 103 nm carries 12.1 eV, the gap from $n = 1$ ( $- 13.6$ eV) to $n = 3$ ( $- 1.51$ eV). A 12.5 eV photon matches no gap from the ground state (12.09 eV or 12.75 eV are the nearest), so it is not absorbed and passes straight through.

Ionising from a higher level takes only the energy to reach zero: from $n = 2$ , 3.40 eV.

How many lines? Electrons falling from level $n$ to the ground state by every route give $\frac{n ( n - 1 )}{2}$ different lines: 6 from $n = 4$ .

de Broglie wavelength after acceleration. $q V = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}$ , so $λ = \frac{h}{2 m q V}$ . An electron through 150 V has $λ \approx 1.0 \times 1 0^{- 10}$ m. Going the other way, the kinetic energy for a wavelength $λ$ is $K = \frac{h ^{2}}{2 m λ ^{2}}$ .

Equal wavelengths mean equal momentum, $m v$ . An alpha particle (4 times a proton's mass) with the same wavelength moves at a quarter of the proton's speed. On a graph of speed against $\frac{1}{λ}$ , the gradient is $\frac{h}{m}$ , which identifies the particle.

Orbits and standing waves. The $n$ th orbit holds exactly $n$ wavelengths. Since $2 π r = nλ$ and $λ = \frac{h}{m v}$ , the ground-state radius is $r = \frac{h}{2 π m v}$ , about $5.3 \times 1 0^{- 11}$ m.

Quick summary

Bohr: electrons sit in fixed energy levels and only emit or absorb photons when they jump: $h f = E_{i} - E_{f}$ .
Hydrogen levels: $E_{n} = - \frac{13.6}{n ^{2}}$ eV.
Rydberg: $\frac{1}{λ} = R (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})$ . The Balmer series ( $n_{f} = 2$ ) is visible.
Bohr's model only worked for hydrogen-like atoms.
de Broglie: particles have a wavelength $λ = \frac{h}{m v}$ . Electron diffraction confirmed it.
Allowed orbits fit a whole number of electron wavelengths. Schrödinger replaced orbits with probability clouds.

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Try 3 questions now

In Bohr's model, an electron gives out light when it:
1. A Stays in the same orbit
2. B Drops to a lower energy level
3. C Moves to a higher energy level
4. D Leaves the nucleus
Show answer
B: Drops to a lower energy level
Dropping down releases the energy difference as a photon.
The lowest energy level of an atom is called the:
1. A Ground state
2. B Excited state
3. C Ionisation level
4. D Nucleus
Show answer
A: Ground state
n = 1 is the ground state.
The energy of the photon given out when an electron drops between levels equals:
1. A The energy of the higher level
2. B The difference in energy between the two levels
3. C 13.6 eV every time
4. D The sum of the two levels
Show answer
B: The difference in energy between the two levels
hf = E initial − E final.