Spectra and the Bohr Model
On the Fourth of July, fireworks burst in red, green, and blue. Each color comes from a different element heated in the shell: strontium for red, barium for green, copper for blue. Pass sunlight through a prism and you get a smooth rainbow. Do the same with the glow from a hydrogen gas tube and you see only a few thin lines of color separated by darkness. Why does each element glow in its own pattern of colors? The answer reveals that the energy of an electron in an atom cannot take just any value.
Continuous and line spectra
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.
- a) Explain the essential physical difference between a continuous spectrum and a discontinuous or line spectrum.
- 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.
Photons and the electromagnetic spectrum
The regions of the electromagnetic spectrum, from longest wavelength to shortest, are radio, microwave, infrared (IR), visible, ultraviolet (UV), X-ray, and gamma ray. Visible light runs from about 400 nm (violet) to 700 nm (red). Different regions interact with matter in different ways:
| Region | What it does to molecules and atoms |
|---|---|
| Microwave | changes molecular rotation |
| Infrared | changes molecular vibrations |
| Visible and UV | moves electrons between energy levels |
Absorbance and the Beer-Lambert law
Chemists also use light to measure concentration. A colored solution absorbs visible light, and the more concentrated it is, the more light it absorbs. The relationship is $A = \varepsilon b c$, where $A$ is absorbance, $\varepsilon$ is the molar absorptivity, $b$ is the path length (usually 1 cm), and $c$ is the concentration in mol/L. Because $A$ is proportional to $c$, a calibration graph of $A$ versus $c$ for known solutions is a straight line through the origin. Measure $A$ for an unknown, find it on the line, and read off its concentration.
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:
- a) The frequency associated with this electromagnetic radiation, expressed in SI units.
- 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, $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$.
The Bohr model: quantized energy levels
Classical physics says an orbiting electron should radiate energy continuously and spiral into the nucleus in about $10^{-11}\ \text{s}$. Yet atoms are stable. Why don't electrons collapse?
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:
$$E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\mathrm{J}$$- a) Calculate the energy of the electron in the ground state ($n=1$) and in the first excited state ($n=2$).
- b) Explain the physical meaning of the negative sign assigned to these energy levels and what the state in which $n \rightarrow \infty$ represents.
Electronic transitions: absorption and emission
Sodium vapor in a flame glows a pure yellow at $589.0\ \text{nm}$, and cool sodium vapor lit by white light absorbs exactly that same wavelength. Why do emission and absorption involve identical photon energies?
The following diagram shows the first three energy levels of the hydrogen atom and two electronic transitions labeled A and B: . Using the diagram and the data provided:…
The following diagram shows the first three energy levels of the hydrogen atom and two electronic transitions labeled A and B:
. Using the diagram and the data provided:
- a) Explain whether transitions A and B correspond to processes of energy absorption or emission.
- 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: $1\,\mathrm{eV} = 1,602 \times 10^{-19}\,\mathrm{J}$; $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $c = 3.00 \times 10^8\,\mathrm{m\times s^{-1}}$.
In a discharge lamp containing monatomic hydrogen gas, a mass electronic transition takes place from the fourth quantized 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 quantized level ($n=4$) to the ground state ($n=1$).
- a) Calculate the energy released by a single hydrogen atom in this de-excitation and determine the frequency of the emitted photon.
- 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, $N_A = 6,022 \times 10^{23}\,\mathrm{mol^{-1}}$.
Ionization energy and limits of the Bohr model
The ionization 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 ionization 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.
- a) Using the Bohr model for hydrogen ($E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\mathrm{J}$), calculate the ionization energy of a hydrogen atom in its ground state ($n=1$) in joules and the corresponding value for one mole of atoms in $\mathrm{kJ\times mol^{-1}}$.
- 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 photoionize the atoms from that excited level, and identify, with reasoning, the region of the electromagnetic spectrum it belongs to.
Data: $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $c = 3.00 \times 10^8\,\mathrm{m\times s^{-1}}$; $N_A = 6,022 \times 10^{23}\,\mathrm{mol^{-1}}$.
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 $\lambda = 434.1\,\mathrm{nm}$.
- 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 $E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\mathrm{J}$.
- 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: $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $c = 3.00 \times 10^8\,\mathrm{m\times s^{-1}}$; $1\,\mathrm{eV} = 1,602 \times 10^{-19}\,\mathrm{J}$.
Back to the start: A hydrogen tube shows only a few visible lines because its electrons fall from $n = 3, 4, 5, 6$ to $n = 2$ (the Balmer series). Each color is a photon with the exact energy $|E_2 - E_n|$. Fireworks work the same way: each element's electrons have their own energy gaps, so each gives its own colors.