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5th Year HL · Chemistry
Structure of Matter · Atomic models and periodicity

Spectra and the Bohr model

Pass white light through a prism and you get a continuous rainbow with no gaps. Now do the same with the light from a hydrogen discharge tube and you see only a few coloured lines, separated by darkness. Classical physics could not explain this pattern. It was the first serious clue that the energy of an electron in an atom cannot take just any value.

Continuous and line spectra

Comparison of a continuous spectrum with the full range of colours, a line emission spectrum with thin bright lines and a line absorption spectrum with matching dark lines. continuous spectrum (bands of colour, no gaps) line emission spectrum line absorption spectrum (same positions, dark lines)

Flame tests and line spectra are used to identify elements. Each metal ion gives a characteristic flame colour: lithium crimson red, sodium yellow (a strong line at 589 nm), potassium lilac, copper blue-green, barium apple green and strontium red. The line spectra of sodium, strontium and copper are different "fingerprints".

Simplified line emission spectra of sodium, strontium and copper between 400 nm and 700 nm, each with a different pattern of bright lines. Sodium (strong line near 589 nm) Strontium (red, orange and blue lines) Copper (several green and blue-green lines) 400 nm 700 nm

The diagram is a simplified sketch of the pattern of lines, not a precise record. In practice you compare a sample's lines with reference data for known elements.

To identify an element from its line spectrum, compare the positions of the lines with those of known elements. For example, an unknown sample gives a flame test that is yellow, and its emission spectrum has a strong line at about 589 nm. Sodium has exactly this line, so the sample contains sodium. A red flame whose spectrum has several lines in the red and orange regions would point to strontium instead. A blue-green flame points to copper.

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The analysis of light emitted or absorbed by substances is a fundamental tool for understanding the electron shell of the atom. a) Explain the essential physical…

The analysis of light emitted or absorbed by substances is a fundamental tool for understanding the electron shell of the atom.

  1. a) Explain the essential physical difference between a continuous spectrum and a discontinuous or line spectrum.
  2. b) Justify why the atomic emission spectrum of a low-pressure gas consists of discrete lines, and explain how these lines correspond to those observed in its absorption spectrum.
Solve

Energy and frequency of a photon

Two electromagnetic waves: a long wavelength with low frequency and low energy, and a short wavelength with high frequency and high energy. long λlow f, low E short λhigh f, high E
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A laser pointer commonly used in classroom demonstrations emits visible red electromagnetic radiation with a wavelength of $656.3\,\mathrm{nm}$. From these data,…

A laser pointer commonly used in classroom demonstrations emits visible red electromagnetic radiation with a wavelength of $656.3\,\mathrm{nm}$. From these data, calculate:

  1. a) The frequency associated with this electromagnetic radiation, expressed in SI units.
  2. b) The energy carried by a single photon emitted by the laser device.

Data: speed of light in a vacuum, $c = 3.00 \times 10^8\,\mathrm{m\times s^{-1}}$; Planck constant, $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$.

Solve

The Bohr model: quantised energy levels

According to classical physics, an accelerating charge radiates energy continuously, so an orbiting electron should spiral into the nucleus in about $10^{-11}\ \text{s}$. Yet atoms are stable. Why do electrons not fall into the nucleus?

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In the atomic model proposed by Bohr for the hydrogen atom, the energy associated with each electronic level depends on the principal quantum number $n$ through the…

In the atomic model proposed by Bohr for the hydrogen atom, the energy associated with each electronic level depends on the principal quantum number $n$ through the expression:

$$E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\mathrm{J}$$
  1. a) Calculate the energy of the electron in the ground state ($n=1$) and in the first excited state ($n=2$).
  2. b) Explain the physical meaning of the negative sign assigned to these energy levels and what the state in which $n \rightarrow \infty$ represents.
Solve

Electronic transitions: absorption and emission

Heat sodium vapour in a flame and it gives out a very pure yellow light of wavelength 589.0 nm. Cool sodium vapour lit by white light absorbs exactly that same wavelength, leaving a dark line in the spectrum. What happens inside the atom to make an element emit and absorb photons of identical energy?

Energy level diagram of hydrogen: a photon is absorbed and the electron rises from n=1 to n=2, then falls back to n=1 and emits a photon. Rising and falling cover the same energy gap. n=1 n=2 n=3 n→∞ emission absorption

Spectral series

Transitions that end on the same level form a series. The Lyman series ends on $n_f = 1$ and lies in the ultraviolet. The Balmer series ends on $n_f = 2$ and lies mainly in the visible region.

To find the starting level $n_i$ from a measured wavelength, first get the photon energy from $E = hc/\lambda$, then solve $E = 2.18 \times 10^{-18}\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)$ for $n_i$.

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The following diagram shows the first three energy levels of the hydrogen atom and two electronic transitions labelled A and B: . Using the diagram and the data…

The following diagram shows the first three energy levels of the hydrogen atom and two electronic transitions labelled A and B:

Energy (eV) n = 3 (E₃ = −1.51 eV) n = 2 (E₂ = −3.4 eV) n = 1 (E₁ = −13.6 eV) Transition A Transition B

. Using the diagram and the data provided:

  1. a) Explain whether transitions A and B correspond to processes of energy absorption or emission.
  2. b) Calculate the energy (in $\mathrm{eV}$) and the wavelength (in $\mathrm{nm}$) of the photon involved in transition A ($3 \rightarrow 2$), and state, with reasoning, which of the two transitions (A or B) emits radiation of shorter wavelength.

Data: $E_1 = -13.6\,\mathrm{eV}$; $E_2 = -3.4\,\mathrm{eV}$; $E_3 = -1.51\,\mathrm{eV}$; conversion factor: $1\,\mathrm{eV} = 1,602 \times 10^{-19}\,\mathrm{J}$; $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $c = 3.00 \times 10^8\,\mathrm{m\times s^{-1}}$.

Solve
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In a discharge lamp containing monatomic hydrogen gas, a mass electronic transition takes place from the fourth quantised level ($n=4$) to the ground state ($n=1$). a)…

In a discharge lamp containing monatomic hydrogen gas, a mass electronic transition takes place from the fourth quantised level ($n=4$) to the ground state ($n=1$).

  1. a) Calculate the energy released by a single hydrogen atom in this de-excitation and determine the frequency of the emitted photon.
  2. b) Calculate the total energy emitted, expressed in $\mathrm{kJ\times mol^{-1}}$, if one whole mole of hydrogen atoms undergoes this de-excitation simultaneously.

Data: Bohr formula for hydrogen: $E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\mathrm{J}$; Planck constant, $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; Avogadro constant, $N_A = 6,022 \times 10^{23}\,\mathrm{mol^{-1}}$.

Solve
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In an optical spectrometer, an emission line of the hydrogen atom is recorded in the visible range (Balmer series, whose final level is $n_f = 2$) with an experimental…

In an optical spectrometer, an emission line of the hydrogen atom is recorded in the visible range (Balmer series, whose final level is $n_f = 2$) with an experimental wavelength of $\lambda = 434.1\,\mathrm{nm}$.

  1. a) Calculate the energy of the emitted photon in joules and determine the principal quantum number of the initial orbital level $n_i$ from which the electron came, using the Bohr expression $E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\mathrm{J}$.
  2. b) Justify rigorously, by means of the relevant energy balance, whether an incident photon of radiation with an energy of $3.00\,\mathrm{eV}$ could be absorbed by a hydrogen atom in the state $n=2$.

Data: $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $c = 3.00 \times 10^8\,\mathrm{m\times s^{-1}}$; $1\,\mathrm{eV} = 1,602 \times 10^{-19}\,\mathrm{J}$.

Solve
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A hydrogen discharge tube emits ultraviolet radiation corresponding to the Lyman series, in which all transitions end in the ground state $n_f = 1$. Two specific lines…

A hydrogen discharge tube emits ultraviolet radiation corresponding to the Lyman series, in which all transitions end in the ground state $n_f = 1$. Two specific lines are considered: transition 1 from $n=2$ and transition 2 from $n=3$.

  1. a) Calculate the energy of the photons emitted in each of the two transitions (in $\mathrm{J}$) and their associated wavelengths (in $\mathrm{nm}$).
  2. b) Explain why the Bohr equations approximate the principal levels and spectral lines of the singly ionised helium ion $\mathrm{He}^+$ by including the factor $Z^2$ in the energy ($Z=2$) and neglecting fine structure and mass corrections, but are unable to calculate the emission frequencies of the neutral helium atom ($\mathrm{He}$).

Data: $E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\mathrm{J}$ (for H); $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $c = 3.00 \times 10^8\,\mathrm{m\times s^{-1}}$.

Solve

Ionisation energy and limitations of the Bohr model

Energy level diagram showing the vertical jump from the ground state n=1 to n infinity, where the electron is free. n=1 n→∞ (E=0) ionisation energy

Limitations of the model

The Bohr model explains the hydrogen line spectrum very well, but it has important limitations:

  • It works only for hydrogen and other one-electron species such as $\text{He}^+$. In those cases $E_n = -2.18 \times 10^{-18}\,Z^2/n^2\ \text{J}$. For atoms with more than one electron, the repulsion between electrons is ignored and the model fails.

  • It cannot explain the fine structure of spectra: at high resolution, each line splits into several closer lines.

  • It gives the electron a fixed circular orbit at a definite distance from the nucleus. Real electrons are better described by the orbital model, which you will meet next. It replaces orbits with orbitals: regions of high probability of finding the electron.

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The Bohr atomic model was a milestone in introducing the concept of quantisation into atomic structure, successfully explaining the lines of the hydrogen spectrum;…

The Bohr atomic model was a milestone in introducing the concept of quantisation into atomic structure, successfully explaining the lines of the hydrogen spectrum; however, it soon revealed insurmountable limitations.

  1. a) Explain two specific shortcomings of the Bohr model: its inability to describe atomic systems with more than one electron, and the fine structure of spectra observed with high-resolution spectroscopes.
  2. b) Explain why attributing to the electron a trajectory with simultaneously defined position and momentum is incompatible with Heisenberg's principle. Distinguish this limitation from the fact that a de Broglie standing wave can indeed provide an interpretation of Bohr's quantisation.
Solve
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The ionisation energy of a gaseous chemical species is the minimum energy required to remove its outermost electron, taking it from its initial state to rest at…

The ionisation energy of a gaseous chemical species is the minimum energy required to remove its outermost electron, taking it from its initial state to rest at infinity.

  1. a) Using the Bohr model for hydrogen ($E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\mathrm{J}$), calculate the ionisation energy of a hydrogen atom in its ground state ($n=1$) in joules and the corresponding value for one mole of atoms in $\mathrm{kJ\times mol^{-1}}$.
  2. b) If a gaseous sample of hydrogen atoms is previously excited in level $n=3$, determine the maximum wavelength (in $\mathrm{nm}$) that a beam of monochromatic light must have to photoionise the atoms from that excited level, and identify, with reasoning, the region of the electromagnetic spectrum it belongs to.

Data: $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $c = 3.00 \times 10^8\,\mathrm{m\times s^{-1}}$; $N_A = 6,022 \times 10^{23}\,\mathrm{mol^{-1}}$.

Solve

Back to the start: The hydrogen tube gives only a few visible lines because its electrons fall from higher levels ($n = 3, 4, 5, 6$) to $n = 2$ (the Balmer series). Each coloured line is a photon of energy exactly $|E_2 - E_n|$ for one transition.

Summary

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