Quantum Theory and the Electronic Structure of Atoms Chapter 7 Part 1 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

The speed (u) of the wave = l x n Properties of Waves Wavelength (l) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = l x n

Speed of light (c) in vacuum = 3.00 x 108 m/s Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 108 m/s All electromagnetic radiation l x n = c

l x n = c l = c/n l = 3.00 x 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m A photon has a frequency of 6.0 x 104 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? l n l x n = c l = c/n l = 3.00 x 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m l = 5.0 x 1012 nm

The wavelength of the green light from a traffic signal is centered at 522 nm. What is the frequency of this radiation?

Problem 1 What is the wavelength (in meters) of an electromagnetic wave whose frequency is 3.64 x 107 Hz?

Mystery #1, “Heated Solids Problem” Solved by Planck in 1900 When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths. Radiant energy emitted by an object at a certain temperature depends on its wavelength. Energy (light) is emitted or absorbed in discrete units (quantum). E = h x n Planck’s constant (h) h = 6.63 x 10-34 J•s blackbody-spectrum

E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 5.00 x 10-5 (m) Calculate the energy (in joules) of a photons with a wavelength of 5.00 x 104 nm (infrared region). l = 5.00 x 104 (nm) x (1 x 10-9 (m)/(nm) = 5.00 x 10-5 (m) E = h x n E = h x c / l E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 5.00 x 10-5 (m) E = 3.98 x 10 -21 J

Calculate the energy (in joules) of a photons with a wavelength of 5 Calculate the energy (in joules) of a photons with a wavelength of 5.00 x 10-2 nm (x-ray region).

Problem 2 The energy of a photon is 5.87 x 10-20 J. What is its wavelength in nanometers?

Photon is a “particle” of light Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 hn Light has both: wave nature particle nature KE e- Photon is a “particle” of light hn = KE + W KE = hn - W where W is the work function and depends how strongly electrons are held in the metal photoelectric

E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m) When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm. E = h x n E = h x c / l E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m) E = 1.29 x 10 -15 J

The work function of cesium metal is 3. 42 x 10-19 J The work function of cesium metal is 3.42 x 10-19 J. Calculate the minimum frequency of light necessary to eject electrons from the metal.

The work function of cesium metal is 3. 42 x 10-19 J The work function of cesium metal is 3.42 x 10-19 J. Calculate the kinetic energy of the ejected electron if light of frequency 1.00 x 1015 s-1 is used for irradiating the metal.

Problem 3 The work function of titanium metal is 6.93 x 10-19 J. Calculate the energy of the ejected electrons if light of frequency 2.50 x 1015 s-1 is used to irradiate the metal. KE = _____x 10-19 J

Line Emission Spectrum of Hydrogen Atoms Mystery #3, “Emission Spectra” Line Emission Spectrum of Hydrogen Atoms

n (principal quantum number) = 1,2,3,… Bohr’s Model of the Atom (1913) e- can only have specific (quantized) energy values light is emitted as e- moves from one energy level to a lower energy level En = -RH ( ) 1 n2 n (principal quantum number) = 1,2,3,… RH (Rydberg constant) = 2.18 x 10-18J

E = hn E = hn

( ) ( ) ( ) Ephoton = DE = Ef - Ei 1 Ef = -RH n2 1 Ei = -RH n2 1 nf = 2 ni = 3 nf = 1 ni = 3 Ef = -RH ( ) 1 n2 f Ei = -RH ( ) 1 n2 i nf = 1 ni = 2 i f DE = RH ( ) 1 n2

Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. i f DE = RH ( ) 1 n2 Ephoton = Ephoton = 2.18 x 10-18 J x (1/25 - 1/9) Ephoton = DE = -1.55 x 10-19 J Ephoton = h x c / l l = h x c / Ephoton l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J l = 1280 nm

Problem 4 What is the wavelength (in nm) of a photon emitted during a transition from ni=6 to nf=4 in the H atom?

Mystery #1, “Heated Solids Problem” Solved by Planck in 1900 When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths. Radiant energy emitted by an object at a certain temperature depends on its wavelength. Energy (light) is emitted or absorbed in discrete units (quantum). E = h x n Planck’s constant (h) h = 6.63 x 10-34 J•s blackbody-spectrum

Photon is a “particle” of light Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 hn Light has both: wave nature particle nature KE e- Photon is a “particle” of light hn = KE + W KE = hn - W where W is the work function and depends how strongly electrons are held in the metal photoelectric

Line Emission Spectrum of Hydrogen Atoms Mystery #3, “Emission Spectra” Line Emission Spectrum of Hydrogen Atoms

n (principal quantum number) = 1,2,3,… Bohr’s Model of the Atom (1913) e- can only have specific (quantized) energy values light is emitted as e- moves from one energy level to a lower energy level En = -RH ( ) 1 n2 n (principal quantum number) = 1,2,3,… RH (Rydberg constant) = 2.18 x 10-18J

Why is e- energy quantized? De Broglie (1924) reasoned that e- is both particle and wave. 2pr = nl l = h mu u = velocity of e- m = mass of e-

Calculate the wavelength of the ‘particle’ in each of the following two cases: (a) The fastest serve ever recorded in tennis was about 150 miles per hour, or 68 m/s. Calculate the wavelength associated with a 6.0 x 10-2 kg tennis ball travelling at this speed. h mu l = kg m2 s2 J = = 6.63 x 10-34 J•s (6.0 x 10-2 kg x 68 m/s) l = 1.6 x 10-34 m = 1.6 x 10-25 nm Because the wavelength is so small compared to the size of the tennis ball, the wave properties cannot be detected by any existing measuring device.

Calculate the wavelength of the ‘particle’ in each of the following two cases: (b) Calculate the wavelength associated with an electron (9.1094 x 10-31kg) travelling at this speed.

Problem 5 Calculate the wavelength (in nm) of a H atom (mass = 1.674 x 10-27 kg) moving at 7.00 x 102 cm/s.

Chemistry in Action: Laser – The Splendid Light Light Amplification by Stimulated Emission of Radiation Laser light is (1) intense, (2) monoenergetic, and (3) coherent