Lecture 4. Light, quanta, and also atoms

Thursday 9 September 2021

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

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

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Summary

Since most of the keys of modern-day atomic theory have actually been revealed through cautious investigation the the interactions in between matter and energy, we continue our investigation of the nature of atom by very first considering some basic aspects that energy and energy changes. The many revealing phenomena concern the interplay in between light and atoms, for this reason we likewise discuss the nature of light and also the electromagnetic (EM) spectrum. Curiously, return light and also other creates of EM radiation have the right to be well described as power propagating as a wave, as shown by the phenomenon the interference, the wave summary fails in its capability to account for certain observations such together black human body radiation and also the photoelectric effect. A quite various model is noted quantum theory, which postulates a quantized nature the light. Ultimately, we find out that EM radiation manifests both wave and also particle properties, if matter(in the nano- and picoscale realms of atoms and also subatomic particles) additionally manifests nature of waves.

Electromagnetic radiation and the quantum nature that energy and also matter

Although timeless physics had explained most that its habits as a an outcome of its wave nature, Planck and also Einstein proved that electromagnetic (EM) radiationbehaves as if its power is carried at the atomic scale in small bundles the energy called photons with particle-like nature. In various other words, EM radiation has actually a quantum nature. Furthermore, Einstein showed that the energy E, of each of these tiny bundles, or quanta, of EM radiation of a offered frequency ν is offered by the following vital equation (often referred to as the Planck-Einstein relation):

E = hν

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

E = hc / λ

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

Example: how much energy (J) is carried by one photon of visible light through λ = 535 nm? after finding the power of one photon of every wavelength, to express the energy of a mole of this photons in kJ/mol.

Solution: We"ll use truncated worths of the complying with constants (which will certainly be sufficiently an exact for the number of significant figures the the calculate input):

c = 2.998 × 108 m/s (speed that 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 come m), us 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 lugged by a single photon the 535 nm light is 3.71 × 10−19 J.

The conversion to kJ/mol involves using Avogadro"s number and also 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 quantities using the value of ν for 535 nm radiation the was derived in the very first example above.

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

Solution: (a) usage 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 complete energy ceded per photon will certainly be calculated according to the second kind of the Einstein-Planck relation 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 present the conversion factor for nm → m in the over calculation. Next, using the regulation of preservation of energy, we reason that the power required come eject an electron from the metal plus the excess kinetic power of the ejected electron have to be equal to the energy delivered to each metal atom by the 225-nm light. There space several ways to to express tis as an equation. The energy required to eject the electron is called ionization energy - in the paper definition of the photoelectric impact this is sometimes labeled together W. We recognize from component (a) that

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

An expression for the conservation of energy is then

Ephoton, 225 nm = W + KEelectron

which we have the right to solve for 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 fixed of the electron and the formula because that kinetic energy, and also solve for velocity v of the electron:

velectron = (2KEelectron/ melectron)½

Using melectron = 9.1094 × 10−28 g, and the reality that 1 J = 1 kg·m2·s−2, we 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 practically 0.3% of the rate of light! In this example, we space simply applying the definition of kinetic power to the electron using its remainder mass, and therefore ignoring any type of relativistic effects.

Quantum mechanics and the atom

A etc string, or normally speaking, a vibrating string v two fixed finish points serves as a one-dimensional analogy to the therapy of an electron bound come a nucleus by the electrostatic force of attraction as a three-dimensional "electron-wave". A mathematical description of a guitar string with fixed size L is a "wave" equation with a set of solutions, or "wave" functions. Such wave functions must satisfy the boundary conditions that the amplitude is zero in ~ both ends of the string. Place the string on a horizontal "x" axis through one finish at x = 0 and also the other at x = L, the amplitude must always be zero at this two end points. This conditions enable a collection of standing wave services with an integral variety of half-wavelengths matching L, together the adhering to equation expresses: ½nλ = L, where n = 1, 2, 3, ...

This deserve to be rearranged as λ = 2L/n, wherein n = 1, 2, 3, ...

For the electron in an atom, the analogous tide equation services (wave functions, Ψ) room three-dimensional and also rather than a single quantum number n, these services are mentioned by a collection of 3 quantum number (see below). We have the right to think that the quantum numbers together labels, or indices, the the solutions to the Schrödinger equation, and additionally note that each solution (we"ll contact these services orbitals) is associated with a definite power E.

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Although the wave duty Ψ has no straight physical interpretation, that is square (Ψ2) is a probability distribution function that defines the probability of recognize the electron in any kind of given location relative come the nucleus.

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Orbitals for the hydrogen atom

The rules because that quantum numbers specify the to adjust of orbitals feasible for one electron bound to a nucleus. There room n2 orbitals for any given value of the primary quantum number n.In the case of the hydrogen atom, the energies are figured out only through the primary quantum number n, and also all n2 orbitals because that a given n space all same in power (and are as such termed degenerate). The power level chart for electronic orbitals in the hydrogen presented below. In this diagram, we present the chemistry convention because that naming orbitals according to their set of quantum numbers. The principal quantum number n is retained, when the angular momentum quantum number l is replaced with a collection of letter designations (s, p, d, ...).

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The unique energy levels for a hydrogen atom and also orbital degeneracies indicate that we have the right to think of teams of degenerate orbitals as organizing electrons in many-electron species into energy-level "shells".

Size, shape, and also orientation that the orbitals

The forms of the orbitals are established by the angular inert quantum number l. The orbitals for which l = 0 room spherically symmetric approximately the nucleus, and are labeled together s orbitals. Because that l = 1, a collection of three mutually perpendicular dumbbell-shaped p orbitals arise. As soon as l = 2, 5 so-called d orbitals with more complicated shapes space possible. The orientations the the p- and also d- and all higher l value orbitals room specified by the magnetic quantum number ml. Because that further exploration of the sizes, shapes, orientations, and also other attributes such as radial distribution functions, a trip to The Orbitron is very recommended. Be certain to examine out the exotic shapes of f and also g orbitals!