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Quantum theory and atomic structure The Wave Nature of Light Frequency and Wavelength c=ln Amplitude (intensity) of a wave. Regions of the electromagnetic spectrum. Interconverting Wavelength and Frequency PROBLEM: PLAN: o A dental hygienist uses x-rays (l= 1.00A) to take a series of dental radiographs while the patient listens to a radio station (l = 325 cm) and looks out the window at the blue sky (l= 473 nm). What is the frequency (in s-1) of the electromagnetic radiation from each source? (Assume that the radiation travels at the speed of light, 3.00x108 m/s.) Use c = ln wavelength in units given o 1 A = 10-10 m 1 cm = 10-2 m 1 nm = 10-9 m wavelength in m n = c/l SOLUTION: o -10 -10 1.00A 10 o m = 1.00x10 m 1A = 3x1018 s-1 n= -2 325 cm 10 m = 325x10-2 m 1 cm n= = 9.23x107 s-1 325x10-2 m 473nm frequency (s-1 or Hz) 3x108 m/s = 6.34x1014 s-1 n= 473x10-9 m Different behaviors of waves and particles. The diffraction pattern caused by light passing through two adjacent slits. Blackbody radiation E=nhn DE = Dn h n DE = h n when n = 1 smoldering coal electric heating element light bulb filament Hot metal gives both longer and shorter wavelenghts compared to visible light. Moreover the maximum intensity shifts toward lower wavelenghts when temperature is increased. If that was true as the temperature raises the light will decrease the wavelenght and increases the energy (ultraviolet catastrophe) Planck explained this behaviour by assuming that light was formed by photons bearing energy units E = n hn n = positive integer Calculating the Energy of Radiation from Its Wavelength PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20cm. What is the energy of one photon of this microwave radiation? PLAN: After converting cm to m, we can use the energy equation, E = hn combined with n = c/l to find the energy. SOLUTION: E= E = hc/l 6.626X10-34J*s 1.20cm 3x108m/s 10-2m cm = 1.66x10-23J The line spectra of several elements. Rydberg equation 1 l = R 1 n12 1 n22 R is the Rydberg constant = 1.096776x107 m-1 Three series of spectral lines of atomic hydrogen. for the visible series, n1 = 2 and n2 = 3, 4, 5, ... Bohr theory • The quantum number n defines the energies of the allowed orbits in the H atom • The energy of an electron in an orbit has a negative value • An atom in the lowest possible energy level is said to be in the GROUND state. State for the hydrogen atom with higher energies are called EXCITED states • Introduce the concept of quantization. Quantum staircase. En = Rhc/n2 The Bohr explanation of the three series of spectral lines. DE = E final – E initial = - NARhc(1/n2final- 1/n2initial) DE = hn Flame tests. strontium 38Sr copper 29Cu Emission and absorption spectra of sodium atoms. The main components of a typical spectrophotometer. Lenses/slits/collimaters narrow and align beam. Source produces radiation in region of interest. Must be stable and reproducible. In most cases, the source emits many wavelengths. Sample in compartment absorbs characteristic amount of each incoming wavelength. Monochromator (wavelength selector) disperses incoming radiation into continuum of component wavelengths that are scanned or individually selected. Computer converts signal into displayed data. Detector converts transmitted radiation into amplified electrical signal. Particle wave duality • Photoelectric effect demonstrated that light, usually considered to be a wave, can also have the properties of particles , albeit without mass • De Broglie proposed that a free electron having a mass m moving with a velocity v should have an associated wavelenght l, calculated by the equation: l = h/mv l = h /mu Table 7.1 The de Broglie Wavelengths of Several Objects Substance Mass (g) Speed (m/s) l (m) slow electron 9x10-28 fast electron 9x10-28 5.9x106 1x10-10 alpha particle 6.6x10-24 1.5x107 7x10-15 one-gram mass 1.0 0.01 7x10-29 baseball 142 25.0 2x10-34 6.0x1027 3.0x104 4x10-63 Earth 1.0 7x10-4 Comparing the diffraction patterns of x-rays and electrons. x-ray diffraction of aluminum foil electron diffraction of aluminum foil Wave motion in restricted systems. Standing waves Standing waves features • Standing waves can have points of zero amplitude (called nodes) • They can have only certain vibration possible that is when L = n(l/2) where n = an integer ( n= 1,2, 3….. • Then, standing waves are quantized, vibration are quantized and the integer n is a quantum number. • By defining electron as a standing wave, quantization is introduced into the description of electronic structure • Schrodinger equation for an electron in a three dimensional space requires three quantum numbers : n, l, ml all whole numbers. • Each wave function is associated to an allowed energy value. The energy is quantized because only certain values of energy are possible for the electron. • The value of the wave function at a given point in space is the amplitude of the wave • The square of the defines the probability to find the electron in a rewgion of the space. • Schrodinger equation defines exactly the energy of an electron . It does not define its position orbitals • Define the region of the space within an electron of a given energy is most likely to be located • With Schrodinger equation, we describe the probability of a given electron to be within a certain region in space. Calculating the de Broglie Wavelength of an Electron PROBLEM: Find the deBroglie wavelength of an electron with a speed of 1.00x106m/s (electron mass = 9.11x10-31kg; h = 6.626x10-34 kg*m2/s). PLAN: Knowing the mass and the speed of the electron allows to use the equation l = h/mu to find the wavelength. SOLUTION: l= 6.626x10-34kg*m2/s 9.11x10-31kg 1.00x106m/s = 7.27x10-10m CLASSICAL THEORY Matter particulate, massive Energy continuous, wavelike Summary of the major observations and theories leading from classical theory to quantum theory. Since matter is discontinuous and particulate perhaps energy is discontinuous and particulate. Observation blackbody radiation Theory Planck: photoelectric effect Energy is quantized; only certain values allowed Einstein: Light has particulate behavior (photons) atomic line spectra Bohr: Energy of atoms is quantized; photon emitted when electron changes orbit. Figure 7.15 continued Since energy is wavelike perhaps matter is wavelike Observation Davisson/Germer: electron diffraction by metal crystal Theory deBroglie: All matter travels in waves; energy of atom is quantized due to wave motion of electrons Since matter has mass perhaps energy has mass Observation Compton: photon wavelength increases (momentum decreases) after colliding with electron Theory Einstein/deBroglie: Mass and energy are equivalent; particles have wavelength and photons have momentum. QUANTUM THEORY Energy same as Matter particulate, massive, wavelike The Heisenberg Uncertainty Principle Dx * mDu ≥ h 4p Applying the Uncertainty Principle PROBLEM: An electron moving near an atomic nucleus has a speed 6x106 ± 1% m/s. What is the uncertainty in its position (Dx)? PLAN: The uncertainty (Dx) is given as ±1%(0.01) of 6x106m/s. Once we calculate this, plug it into the uncertainty equation. SOLUTION: Du = (0.01)(6x106m/s) = 6x104m/s Dx * mDu ≥ h 4p Dx ≥ 6.626x10-34kg*m2/s 4p (9.11x10-31kg)(6x104m/s) = 9.52x10-9m The Schrödinger Equation HY = EY wave function mass of electron potential energy at x,y,z 8p2mQ d2Y d2Y d2Y + + + (E-V(x,y,z)Y(x,y,z) = 0 2 2 2 2 h dx dy dz how changes in space total quantized energy of the atomic system Electron probability in the ground-state H atom. Quantum Numbers and Atomic Orbitals An atomic orbital is specified by three quantum numbers. n the principal quantum number - a positive integer l the angular momentum quantum number - an integer from 0 to n-1 ml the magnetic moment quantum number - an integer from -l to +l Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals Name, Symbol (Property) Allowed Values Quantum Numbers Principal, n Positive integer (size, energy) (1, 2, 3, ...) 1 Angular momentum, l 0 to n-1 (shape) 0 0 0 0 Magnetic, ml -l,…,0,…,+l (orientation) 2 3 1 0 1 2 0 -1 0 +1 -1 0 +1 -2 -1 0 +1 +2 Determining Quantum Numbers for an Energy Level PROBLEM: What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3? PLAN: Follow the rules for allowable quantum numbers found in the text. l values can be integers from 0 to n-1; ml can be integers from -l through 0 to + l. SOLUTION: For n = 3, l = 0, 1, 2 For l = 0 ml = 0 For l = 1 ml = -1, 0, or +1 For l = 2 ml = -2, -1, 0, +1, or +2 There are 9 ml values and therefore 9 orbitals with n = 3. Determining Sublevel Names and Orbital Quantum Numbers PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers: (a) n = 3, l = 2 (b) n = 2, l = 0 (c) n = 5, l = 1 (d) n = 4, l = 3 PLAN: Combine the n value and l designation to name the sublevel. Knowing l, we can find ml and the number of orbitals. SOLUTION: n l sublevel name possible ml values # of orbitals (a) 3 2 3d -2, -1, 0, 1, 2 5 (b) 2 0 2s 0 1 (c) 5 1 5p -1, 0, 1 3 (d) 4 3 4f -3, -2, -1, 0, 1, 2, 3 7 The 2p orbitals. The 3d orbitals. One of the seven possible 4f orbitals.