Chapter 7 Quantum Theory of Atom Bushra Javed

Contents The Wave Nature of Light Quantum Effects and Photons The Bohr Theory of the Hydrogen Atom Quantum Mechanics 5. Quantum Numbers and Atomic Orbitals 6. Quantum Numbers and Atomic Orbitals

A Theory that Explains Electron Behavior The quantum-mechanical model explains the manner in which electrons exist and behave in atoms It helps us understand and predict the properties of atoms that are directly related to the behavior of the electrons

The Nature of Light: Its Wave Nature Light is a form of electromagnetic radiation composed of perpendicular oscillating waves, one for the electric field and one for the magnetic field an electric field is a region where an electrically charged particle experiences a force a magnetic field is a region where a magnetized particle experiences a force All electromagnetic waves move through space at the same, constant speed 3.00 x 108 m/s in a vacuum = the speed of light, c

Electromagnetic Radiation

Electromagnetic Spectrum The range of frequencies and wavelengths of electromagnetic radiation is called the electromagnetic spectrum.

Characterizing Waves The amplitude is the height of the wave the distance from node to crest or node to trough the amplitude is a measure of how intense the light is – the larger the amplitude, the brighter the light The wavelength (l) is a measure of the distance covered by the wave the distance from one crest to the next

Wave Characteristics Tro: Chemistry: A Molecular Approach, 2/e

Characterizing Waves The frequency (n) is the number of waves that pass a point in a given period of time the number of waves = number of cycles units are hertz (Hz) or cycles/s = s−1 1 Hz = 1 s−1 The total energy is proportional to the amplitude of the waves and the frequency the larger the amplitude, the more force it has

Characterizing Waves The longer the wavelength of light, the lower the frequency. The shorter the wavelength of light, the higher the frequency.

Color The color of light is determined by its wavelength or frequency White light is a mixture of all the colors of visible light a spectrum RedOrangeYellowGreenBlueViolet When an object absorbs some of the wavelengths of white light and reflects others, it appears colored

Wavelength, Frequency & Speed of Light Wavelength and Frequency are related as: λ α 1/ v c = λ v Where c is the speed of light Light waves always move through empty space at the same speed. Therefore speed of light is a fundamental constant. c = 3.00 x 108 m/s

The Wave Nature of Light Example 1 Which of the following statements is incorrect? a. The product of wavelength and frequency of electromagnetic radiation is a constant. b. As the energy of a photon increases, its frequency decreases. c. As the wavelength of a photon increases, its frequency decreases. d. As the frequency of a photon increases, its wavelength decreases.

Units of Wavelength & Frequency nanometers, micrometers, meters and Angstrom 1 nm = 10-9m 1A° = 10-10m Units of Frequency: 1/second (per second) or 1s-1 Hertz = Hz, MHz and GHz 1-1s = 1 Hz

Wavelength & Frequency Example 2 What is the wavelength of blue light with a frequency of 6.4  1014/s? c = nl so l = c/n n = 6.4  1014/s c = 3.00  108 m/s l = 4.7  10−7 m

Wavelength & Frequency Example 3 What is the frequency of light having a wavelength of 681 nm? c = nl so n = c/l l = 681 nm = 6.81  10−7 m c = 3.00  108 m/s l = 4.41  1014 /s

Electromagnetic Spectrum Example 4 Which type of electromagnetic radiation has the highest energy? a. radio waves b. gamma rays c. blue light d. red light

Wave Theory of light One property of waves is that they can be diffracted—that is, they spread out when they encounter an obstacle about the size of their wavelength. In 1801, Thomas Young, a British physicist, showed that light could be diffracted. By the early 1900s, the wave theory of light was well established.

Diffraction

The Wave/Particle Nature of Light Max Planck worked on the light of various frequencies emitted by the hot solids at different temperatures. In 1900, Max Planck proposed that radiant energy is not continuous, but is emitted in small bundles. Planck theorized that the atoms of an heated object vibrate with a definite frequency. An individual unit of light energy is a photon.

Dual Nature of Light Einstein expanded on Planck’s work. Through Photoelectric Effect ,he showed that when light of particular frequency shines on a metal surface it knocks out electrons. This knocking out of electrons is dependent only on the frequency of the shining light.

The Photoelectric Effect

Einstein’s Explanation Einstein proposed that the light energy was delivered to the atoms in packets, called quanta or photons The energy of a photon of light is directly proportional to its frequency inversely proportional to its wavelength the proportionality constant is called Planck’s Constant, (h) and has the value 6.626 x 10−34 J∙s

Einstein’s Explanation Light, therefore, has properties of both waves and matter. Neither understanding is sufficient alone. This is called the particle–wave duality of light.

Kinetic Energy = Ephoton – Ebinding Ejected Electrons One photon at the threshold frequency gives the electron just enough energy for it to escape the atom binding energy, f When irradiated with a shorter wavelength photon, the electron absorbs more energy than is necessary to escape This excess energy becomes kinetic energy of the ejected electron Kinetic Energy = Ephoton – Ebinding KE = hn − f

The Photoelectric Effect Example 5: Suppose a metal will eject electrons from its surface when struck by yellow light. What will happen if the surface is struck with ultraviolet light? a. No electrons would be ejected. Electrons would be ejected, and they would have the same kinetic energy as those ejected by yellow light. c. Electrons would be ejected, and they would have greater kinetic energy than those ejected by yellow light. d. Electrons would be ejected, and they would have lower kinetic energy than those ejected by yellow light.

Photoelectric Effect Example 6 The blue–green line of the hydrogen atom spectrum has a wavelength of 486 nm. What is the energy of a photon of this light? E = hn and c = nl so E = hc/l l = 486 nm = 4.86  10−7 m c = 3.00  108 m/s h = 6.63  10−34 J  s E = 4.09  10−19 J

Problems with Rutherford’s Nuclear Model of the Atom Electrons are moving charged particles According to classical physics, moving charged particles give off energy Therefore electrons should constantly be giving off energy The electrons should lose energy, crash into the nucleus, and the atom should collapse!! but it doesn’t

Continuous Vs Line Spectra In addition, this understanding could not explain the observation of line spectra of atoms. A continuous spectrum contains all wavelengths of light. A line spectrum shows only certain colors or specific wavelengths of light. When atoms are heated, they emit light. This process produces a line spectrum that is specific to that atom.

Spectra When atoms or molecules absorb energy, that energy is often released as light energy fireworks, neon lights, etc. When that emitted light is passed through a prism, a pattern of particular wavelengths of light is seen that is unique to that type of atom or molecule – the pattern is called an emission spectrum non-continuous can be used to identify the material flame tests

Exciting Gas Atoms to Emit Light with Electrical Energy

Identifying Elements with Flame Tests Na K Li Ba

Emission Spectra

Emission Line Spectra When an electrical voltage is passed across a gas in a sealed tube, a series of narrow lines is seen. These lines are the emission line spectrum. The emission line spectrum for hydrogen gas shows three lines: 434 nm, 486 nm, and 656 nm.

The Bohr Model of the Atom The nuclear model of the atom does not explain what structural changes occur when the atom gains or loses energy Bohr developed a model of the atom to explain how the structure of the atom changes when it undergoes energy transitions Bohr’s major idea was that the energy of the atom was quantized, and that the amount of energy in the atom was related to the electron’s position in the atom quantized means that the atom could only have very specific amounts of energy

Bohr’s Model The electrons travel in orbits that are at a fixed distance from the nucleus Energy Levels therefore the energy of the electron was proportional to the distance the orbit was from the nucleus Electrons emit radiation when they “jump” from an orbit with higher energy down to an orbit with lower energy the emitted radiation was a photon of light the distance between the orbits determined the energy of the photon of light produced

Atomic Line Spectra

Bohr’s Model of Atom Example 7 Which of the following energy level changes in a hydrogen atom produces a visible spectral line? a. 3 → 1 b. 4 → 2 c. 5 → 3 d. all of the above

Rydberg’s Spectrum Analysis Rydberg analyzed the spectrum of hydrogen and found that it could be described with an equation that involved an inverse square of integers

Bohr’s Model of Atom Energy-Level Postulate An electron can have only certain energy values, called energy levels. Energy levels are quantized. For an electron in a hydrogen atom, the energy is given by the following equation: RH = 2.179  10−18 J n = principal quantum number

Bohr’s Model of Atom The energy of the emitted or absorbed photon is related to DE: We can now combine these two equations:

Bohr’s Model of Atom Light is absorbed by an atom when the electron transition is from lower n to higher n (nf > ni). In this case, DE will be positive. Light is emitted from an atom when the electron transition is from higher n to lower n (nf < ni). In this case, DE will be negative. An electron is ejected when nf = ∞.

Electron transitions for an electron in the hydrogen atom.

Example 8. What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes a transition from n = 6 to n = 3? ni = 6 nf = 3 RH = 2.179  10−18 J = −1.816  10−19 J 1.094  10−6 m

Bohr’s Model of Atom Example 9 What is the frequency (in kHz ) of light emitted when the electron in a hydrogen atom undergoes a transition from level n = 6 to level n = 2? h = 6.63 × 10-34 J . s), RH = 2.179 × 10-18 J) a. 5.49 × 105 kHz b. 7. 31× 1011 kHz c. 7. 31 × 1014 kHz d. 3.64 × 10–28 kHz

Wave properties of Matter In 1923, Louis de Broglie, a French physicist, reasoned that particles (matter) might also have wave properties. The wavelength of a particle of mass, m (kg), and velocity, v (m/s), is given by the de Broglie relation:

Wave properties of Matter Example 10 Compare the wavelengths of (a) an electron traveling at a speed that is one-hundredth the speed of light and (b) a baseball of mass 0.145 kg having a speed of 26.8 m/s (60 mph). Baseball m = 0.145 kg v = 26.8 m/s Electron me = 9.11  10−31 kg v = 3.00  106 m/s

Electron me = 9.11  10−31 kg v = 3.00  106 m/s 2.43  10−10 m Baseball m = 0.145 kg v = 26.8 m/s 1.71  10−34 m

Quantum (or Wave) mechanics Building on de Broglie’s work, Erwin Schrodinger devised a theory that could be used to explain the wave properties of electrons in atoms and molecules. The branch of physics that mathematically describes the wave properties of submicroscopic particles is called or wave mechanics.

Uncertainty Principle Heisenberg stated that the product of the uncertainties in both the position and speed of a particle was inversely proportional to its mass x = position, Dx = uncertainty in position v = velocity, Dv = uncertainty in velocity m = mass

Uncertainty Principle This means that the more accurately you know the position of a small particle, such as an electron, the less you know about its speed and vice-versa

Schrodinger’s Equation Schodinger’s Equation allows us to calculate the probability of finding an electron with a particular amount of energy at a particular location in the atom Solutions to Schödinger’s Equation produce many wave functions, A plot of distance vs. Y2 represents an orbital, a probability distribution map of a region where the electron is likely to be found

Solutions to the Wave Function, Y Calculations show that the size, shape, and orientation in space of an orbital are determined to be three integer terms in the wave function added to quantize the energy of the electron These integers are called quantum numbers principal quantum number, n angular momentum quantum number, l magnetic quantum number, ml

Principal Quantum Number, n Characterizes the energy of the electron in a particular orbital corresponds to Bohr’s energy level n can be any integer  1 The larger the value of n, the more energy the orbital has Energies are defined as being negative an electron would have E = 0 when it just escapes the atom The larger the value of n, the larger the orbital As n gets larger, the amount of energy between orbitals gets smaller

Angular Momentum Quantum Number, l The angular momentum quantum number determines the shape of the orbital l can have integer values from 0 to (n – 1) Each value of l is called by a particular letter that designates the shape of the orbital

Angular Momentum Quantum Number, s orbitals are spherical p orbitals are like two balloons tied at the knots d orbitals are mainly like four balloons tied at the Knot f orbitals are mainly like eight balloons tied at the knot

Magnetic Quantum Number, ml The magnetic quantum number is an integer that specifies the orientation of the orbital the direction in space the orbital is aligned relative to the other orbitals Values are integers from −l to +l including zero gives the number of orbitals of a particular shape when l = 2, the values of ml are −2, −1, 0, +1,+2 which means there are five orbitals with l = 2

Describing an Orbital Each set of n, l, and ml describes one orbital Orbitals with the same value of n are in the same principal energy level aka principal shell Orbitals with the same values of n and l are said to be in the same sublevel

Energy Shells and Subshells

Quantum Numbers Spin Quantum Number, ms This quantum number refers to the two possible orientations of the spin axis of an electron. It may have a value of either +1/2 or -1/2.

The Quantum Numbers Example 11 Give the notation (using letter designations for 1) of subshells for the following quantum numbers. n. = 3, l = 1 n. = 4, l = 0 n. = 5,1 = 3 n. = 6,1 = 2 n. = 2,1 = 1

The Quantum Numbers Example 12 Determine how many valid sets of quantum numbers exist for a 3p orbital ; Give two examples. Strategy: Write the the number of possible values for each quantum number. Quantum number n 1 m ms Possible values 3 1 1, 0, -1 +1/2, -1/2 # of possible values 1 1 3 2 Possible sets of values for a 3p electron: (1)(1)(3)(2) = 6

The Quantum Numbers There are six sets of quantum numbers that describe a 3p electron.   Examples: n = 3 1=1 m1= 0 ms = +1/2 n = 3 1=1 m1 = -1 ms = +1/2

The Quantum Numbers Example 13 Which of the following sets of quantum numbers (n, l, ml, ms) is not permissible? a. 1 0 0 +½ b. 4 0 0 -½ c. 2 2 1 +½ d. 3 1 0 -½

The Quantum Numbers Example 14 Determine how many valid sets of quantum numbers exist for 4d orbitals ;Give three examples.

The Quantum Numbers Example 15 Determine how many valid sets of quantum numbers exist for 5f orbitals ;Give three examples.

The Shapes of Orbitals s Orbitls : 1. All s orbitals are spherical 2. The probability of finding an s electron depends only on distance from the nucleus, not on direction 3. There is only one orientation of a sphere in space Therefore S orbital has ml = 0 The value of ψ2 for an s-orbital is greatest near the nucleus The size of the s orbital increases as the number of shell increases i.e. 3s > 2s>1s

S Orbitals

Probability Density Function The probability density function represents the total probability of finding an electron at a particular point in space Tro: Chemistry: A Molecular Approach, 2/e

p Orbitals p orbitlas are dumbbell –shaped. 2. The electron distribution is concentrated in identical lobes on either side of the nucleus and separated by a planar node cutting through the nucleus 3. Probability of finding a p electron near the nucleus is zero. 4. The two lobes of a p orbital have different phases of a wave (different algebraic signs) 5. The size of p orbitals increases in the order 4p>3p>2p

d Orbitals 1. The third and higher shells contain each contain five d orbitals 2. d orbitals have two shapes 3. Four of the five d orbitals are cloverleaf –shaped and have four lobes of maximum electron probability 4. The fifth d orbital is similar in shape to a pz orbital but has an additional donut –shaped region of electron probability centered in the xy plane 5. All five d orbitals have the same energy 6. Alternating lobes of the d orbitals have different phases

d Orbitals

f orbitals Tro: Chemistry: A Molecular Approach, 2/e