Lecture 4. Light, quanta, and atoms

Thursday 9 September 2021

Wave-particle duality. The quantum nature that the atom. Quantum numbers and also electronic orbitals.

Reading: Tro NJ. Chemistry: Structure and also Properties (2nd ed.) - Ch.2, pp.85-96, 99-103.

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Since many of the secrets of modern atomic theory have actually been revealed through careful investigation of the interactions in between matter and energy, we proceed our investigation of the nature of atoms by first considering some basic aspects of energy and energy changes. The many revealing phenomena problem the interplay in between light and also atoms, thus we likewise discuss the nature that light and the electromagnetic (EM) spectrum. Curiously, back light and other develops of EM radiation have the right to be well described as power propagating together a wave, as displayed by the phenomenon the interference, the wave summary fails in its capability to account for specific observations such as black body radiation and also the photoelectric effect. A quite different model is noted quantum theory, i beg your pardon postulates a quantized nature of light. Ultimately, we learn that EM radiation manifests both wave and also particle properties, when matter(in the nano- and also picoscale worlds of atoms and subatomic particles) additionally manifests nature of waves.

Electromagnetic radiation and the quantum nature the energy and also matter

Although classic physics had explained most of its behavior as a an outcome of its wave nature, Planck and also Einstein proved that electromagnetic (EM) radiationbehaves as if its power is lugged at the atomic scale in little bundles that energy called photons through particle-like nature. In other words, EM radiation has a quantum nature. Furthermore, Einstein verified that the energy E, of every of these little bundles, or quanta, of EM radiation the a offered frequency ν is provided by the following vital equation (often referred to as the Planck-Einstein relation):

E = hν

The factor h is well-known as Planck"s constant ( h = 6.62606931 × 10–34 J·s ), an important fundamental continuous of nature. If we combine the equation relating frequency, wavelength, and wave rate with the Planck-Einstein relationship above, we obtain

E = hc / λ

an alternate form of the Planck-Einstein relation, useful for converting between EM wavelength and energy.

Example: exactly how much power (J) is lugged by one photon of clearly shows light with λ = 535 nm? after ~ finding the energy of one photon of every wavelength, to express the power of a mole of these photons in kJ/mol.

Solution: We"ll usage truncated worths of the adhering to constants (which will be sufficiently precise for the number of significant figures the the calculate input):

c = 2.998 × 108 m/s (speed the light)h = 6.626 × 10−34 J·s (Planck"s constant) NA = 6.022 × 1023 mol−1 (Avogadro"s number)

Using the relationship E = hc/λ (and converting nm to m), we obtain

E = (6.626 × 10−34 J·s)(2.998 × 108 m/s) / (535 nm)(1 × 10−9 nm/m) = 3.713 × 10−19 J

Thus the energy brought by a single photon that 535 nm irradiate is 3.71 × 10−19 J.

The conversion come kJ/mol entails using Avogadro"s number and converting J come kJ:

E (kJ/mol) = (3.713 × 10−19 J/photon)(6.022 × 1023 photon/mol)(10−3kJ/J) = 224 kJ/mol.

Exercise: calculate these same amounts using the worth of ν because that 535 nm radiation the was derived in the very first example above.

Example: (a) In the photoelectric effect, because that a specific metal, the threshold frequency ν0, or minimum frequency the EM radiation that leads to production of a present upon illumination that the metal as result of ejected electron ("photoelectrons") is 6.41 × 1014 s−1. Calculate the energy per photon linked with light of this frequency. (b) expect this steel is illuminated v light having actually a wavelength λ = 225 nm. Just how much kinetic energy will photoelectrons produced possess? (c) What is the magnitude of electron velocity because that these photoelectrons?

Solution: (a) use the Einstein-Planck relation:

Ephoton = hν0= (6.626 × 10−34 J·s)(6.41 × 1014 s−1) = 4.247 × 10−19 J = 4.25 × 10−19 J

(b) In this scenario, the full energy yielded per photon will certainly be calculated follow to the second type of the Einstein-Planck relationship above,

Ephoton = hc / λ = (6.626 × 10−34 J·s)(2.998 × 108 m/s) / (225 nm)(10−9 m/nm) = 8.828 × 10−19 J

Ephoton = 8.83 × 10−19 J

Note we essential to introduce the conversion aspect for nm → m in the above calculation. Next, using the regulation of preservation of energy, we reason that the power required to eject an electron native the steel plus the overabundance kinetic power of the ejected electron need to be equal to the energy yielded to each steel atom by the 225-nm light. There are several methods to refer tis as an equation. The power required to eject the electron is called ionization energy - in the context of the photoelectric impact this is occasionally labeled together W. We know from component (a) that

W = Ephoton, threshold freq = hν0= 4.25 × 10−19 J

An expression because that the conservation of power is then

Ephoton, 225 nm = W + KEelectron

which we deserve to solve because that the electron"s kinetic energy

KEelectron = Ephoton, 225 nm − Ephoton, threshold freq = 8.83 × 10−19 J − 4.25 × 10−19 J

KEelectron = 4.58 × 10−19 J

(c) right here we usage the massive of the electron and the formula for kinetic energy, and solve because that velocity v of the electron:

velectron = (2KEelectron/ melectron)½

Using melectron = 9.1094 × 10−28 g, and also the truth that 1 J = 1 kg·m2·s−2, us calculate the electron velocity

velectron =(2)(4.25 × 10−19 kg·m2·s−2) / (9.1094 × 10−28 g)(10−3 kg/g)½

velectron =9.66 × 105 m/s

which is almost 0.3% the the speed of light! In this example, we space simply applying the meaning of kinetic energy to the electron making use of its rest mass, and also therefore ignoring any kind of relativistic effects.

Quantum mechanics and the atom

A guitar string, or normally speaking, a vibrating string through two fixed finish points serves together a one-dimensional analogy come the therapy of one electron bound come a cell core by the electrostatic pressure of attraction together a three-dimensional "electron-wave". A mathematical description of a guitar string v fixed length L is a "wave" equation with a set of solutions, or "wave" functions. Together wave features must meet the boundary conditions that the amplitude is zero at both end of the string. Place the string on a horizontal "x" axis with one finish at x = 0 and the other at x = L, the amplitude must always be zero at this two end points. These conditions permit a collection of standing wave solutions with an integral number of half-wavelengths equivalent L, together the following equation expresses: ½nλ = L, whereby n = 1, 2, 3, ...

This can be rearranged as λ = 2L/n, where n = 1, 2, 3, ...

For the electron in an atom, the analogous tide equation services (wave functions, Ψ) space three-dimensional and also rather than a solitary quantum number n, these remedies are mentioned by a set of 3 quantum number (see below). We can think that the quantum numbers as labels, or indices, the the services to the Schrödinger equation, and additionally note the each systems (we"ll speak to these remedies orbitals) is linked with a definite power E.


Although the wave function Ψ has actually no straight physical interpretation, that square (Ψ2) is a probability distribution role that explains the probability of finding the electron in any given place relative to the nucleus.

See more: How Many Limbs Do We Have Been Possible That Instead Of 4, We Developed 8 Limbs?

Orbitals for the hydrogen atom

The rules for quantum numbers define the sets of orbitals feasible for one electron bound to a nucleus. There are n2 orbitals for any given worth of the primary quantum number n.In the case of the hydrogen atom, the energies are determined only by the primary quantum number n, and also all n2 orbitals for a given n are all equal in energy (and are as such termed degenerate). The energy level diagram for digital orbitals in the hydrogen presented below. In this diagram, we introduce the chemistry convention for naming orbitals according to their collection of quantum numbers. The principal quantum number n is retained, if the angular momentum quantum number l is changed with a set of letter designations (s, p, d, ...).


The unique energy levels for a hydrogen atom and orbital degeneracies indicate that we can think of teams of degenerate orbitals as arranging electrons in many-electron species into energy-level "shells".

Size, shape, and orientation that the orbitals

The forms of the orbitals are identified by the angular momentum quantum number l. The orbitals because that which l = 0 are spherically symmetric roughly the nucleus, and also are labeled together s orbitals. Because that l = 1, a collection of 3 mutually perpendicular dumbbell-shaped p orbitals arise. As soon as l = 2, 5 so-called d orbitals with more complex shapes space possible. The orientations of the p- and also d- and also all higher l value orbitals are specified by the magnetic quantum number ml. For further expedition of the sizes, shapes, orientations, and also other features such as radial distribution functions, a expedition to The Orbitron is very recommended. Be sure to inspect out the exotic forms of f and also g orbitals!