Electromagnetic Radiation: Wavelength, Frequency, and Energy, Study notes of Chemistry

Download Electromagnetic Radiation: Wavelength, Frequency, and Energy and more Study notes Chemistry in PDF only on Docsity! Electromagnetic Radiation ¥ WAVELENGTH (λ) - The distance between identical points on successive waves. ¥ FREQUENCY (ν) - The number of waves that pass through a particular point per second. ¥ AMPLITUDE - The vertical distance from the midline to a peak, or trough in the wave. λ = c ν Fig 7.1 (P 260) Fig 7.2 (P 260) The Spectrum of Electromagnetic Radiation ¥ The Wavelength of Visible light is between 400 and 700 nanometers ¥ Radio, TV , Microwave and Infrared radiation have longer wavelengths (shorter frequencies), and lower energies than visible light. ¥ Gamma rays and X-rays have shorter wavelengths (larger frequencies), and higher energies than visible light! Calculation of Frequency from Wavelength Problem: a) The wavelength of an x-ray is 1.00 x10 -9 m or 1 nm, what is the frequency of this x-ray? b) If the wavelength of Long wavelength electromagnetic radiation is 7.65 x 104 m, what is the frequency of this Long wavelength radiation used to contact submerged nuclear submarines at sea? Solution: frequency(s-1 or Hz) = speed of light (m/s) wavelength(m) frequency = = 3.00 x 1017 s-1 (Hz)3.00 x 10 8 m/s 1.00 x 10 - 9 m a) b) frequency = = 3.92 x 103 s-1 (Hz) 3.00 x 10 8 m/s 7.65 x 104 m Max Planck and Blackbody Radiation Blackbody radiation: When a solid object is heated, it emits light. If light is treated as a wave, one can show that E α 1/λ4 but that does not fit the observations. Planck proposed that light is only emitted as fixed quantities or quanta: E = nhν where h = 6.626 x 10-34 J s and n is an integer. Photoelectric Effect ¥ Since the threshold energy of Potassium is = 3.7 x 10 - 19 J, the red light will not have enough energy to knock an electron out of the potassium, but the blue light will eject an electron ! ¥ E Total = E Threshold + EKinetic Energy of Electron ¥ E Electron = ETotal - E Threshold ¥ E Electron = 5.0 x 10 - 19 J - 3.7 x 10 - 19 J = 1.3 x 10 - 19 J = hν ¥ and ν = 1.96 x 1014 Hz (infrared) The emission spectrum of hydrogen atoms The wavelengths observed are explained by: The Rydberg equation 1/λ = R(1/n12 - 1/n22) where R = 1.0968 x 107 m-1 and n = integers with n2 > n1 Niels Bohr proposed that hydrogen atoms only have certain fixed energy levels, which are associated with specific orbits. An atom can only absorb or emit a photon whose energy equals the difference between two energy states (orbits). Light and Atoms ¥ When an atom gains a photon, it enters an excited state. ¥ This state has too much energy - the atom must lose it and return back down to its ground state, the most stable state for the atom. ¥ An energy level diagram is used to represent these changes. Energy Level Diagram Light Emission Light Emission Light Emission Excited States photon’s path Ground State E Hydrogen atom energy states Bohr derived an equation for the energy of an electron in orbit using classical physics: E = - 2.18x10-18 J (Z2/n2) where n is an integer and Z is the charge of the nucleus. The emission spectrum for Hydrogen should be: ∆E = Efinal - Einitial = - 2.18x10-18 J (1/nf2 - 1/ni2) Since ∆E = hν = hc/λ, 1/λ = ∆E/hc, which gives: 1/λ = - 2.18x10-18 J (1/nf2 - 1/ni2) = 1.10x107 m (1/nf2 - 1/ni2) hc Emission Energetics Problem: A sodium vapor light street light emits bright yellow light of wavelength = 589 nm. What is the energy change for a sodium atom involved in this emission? How much energy is emitted per mole of sodium atoms? Solution: Ephoton = hν = = h x c λ ( 6.626 x 10 -34J s)( 3.00 x 10 8m/s) 589 x 10 -9m Ephoton = Eatom = 3.37 x 10 -19J/atom Energy per mole requires that we multiply by Avogadro s number. Emole = 3.37 x 10 -19J/atom x 6.022 x 1023 atoms/mole = Emole = 2.03 x 105 J/mol = 203 kJ / mol Emission Energetics Problem: A compact disc player uses light with a frequency of 3.85 x 1014 Hz. What is this light s wavelength? What portion of the electromagnetic spectrum does this wavelength fall? What is the energy of one mole of photons of this frequency? Solution: λ = c / ν = = 7.78 x 10 -7 m = 778 nm3.00 x 10 8m/s 3.85 x 1014/s 778 nm is in the Infrared region of the electromagnetic spectrum Ephoton = hν = (6.626 x 10 -34Js) x ( 3.85 x 1014 /s) = 2.55 x 10 -19 J Emole = (2.55 x 10 -19J) x (6.022 x 1023 / mole) = 1.54 x 105 J/mole Emission Energetics Problem: Find the energy change when an electron changes from the n=4 level to the n=2 level in the hydrogen atom? What is the wavelength of this photon? Solution: Ephoton = -2.18 x10 -18J - = 1 n12 1 n22 Ephoton = -2.18 x 10 -18J - = - 4.09 x 10 -19J 1 2 2 1 4 2 λ = = = h x c E (6.626 x 10 -34Js)( 3.00 x 108 m/s) 4.09 x 10 -19J λ = 4.87 x 10 -7 m = 487 nm The duality of matter Problems with the Bohr Model: ¥does not work for atoms with more than 1 electron ¥the classical physics predicts that the electron should rapidly crash into the nucleus Louis de Broglie: ¥reasoned that perhaps matter is wave-like ¥and that the electron is dimensionally restricted (i.e., a single or bit allows only certain wavelengths) ¥Since E = h¥c/λ = mc2, proposed that for any particle: λ = h/m¥u where u is the speed of a particle of mass u Quantum Numbers 2) Angular momentum quantum number (l ) ¥ is related to the sh ape of the orbital. ¥ Positive integer from 0 to ( n-1 ) ¥ n = 1 , l = 0; n = 2 , l = 0, 1; n = 3 , l = 0 , 1 , 2 Quantum Numbers 3) Magnetic Quantum Number - ml ¥ denotes the orientation of the orbital around the nucleus, describes the magnetic moment vector ¥ integers from -l through 0 to +l ¥ l = 0 , ml = 0; l =1 , ml = -1, 0, +1; l = 2 , ml = -2, -1, 0, +1, +2 Determining Quantum Numbers for an Energy Level Problem: What values of l and ml are allowed for a principal quantum number (n) of 4? How many orbitals are allowed for n = 4? Solution: The l values go from 0 to (n - 1), and for n = 4 they are: l = 0, 1, 2, 3. The values for ml go from -l to zero to +l For l = 0, ml = 0 l = 1, ml = -1, 0, +1 l = 2, ml = -2, -1, 0, +1, +2 l = 3, ml = -3, -2, -1, 0, +1, +2, +3 There are 16 ml values, so there are 16 orbitals for n=4 The total number of orbitals for a given value of n is n2, so for n = 4 there are 42 or 16 orbitals! l = 0 l = 1 l = 2 Energy Diagram of Orbitals (Approximate) l = 0 l = 1 l = 2 He Ne Ar Kr Xe Rn ZnCu Cd Hg Ag Au Ni Pd Pt Co Rh Ir Fe Ru Os Mn Tc Re Cr Mo W V Nb Ta Ti Zr Hf Sc Y La Ac The Periodic Table of the Elements S Orbita ls P Orbital s f Orbi tals Rf Ha Electronic Structure F Cl Br I At O S Se Te Po N P As Sb Bi C Si Ge Sn Pb B Al Ga In Tl H Li Be NaMg K Ca Rb Sr Cs Ba Fr Ra d Orbita ls Ce Pr Nd Sg PmSm EuGd Tb DyHo Er Tm Yb Lu Th Pa U Np PuAmCm Bk Cf Es FmMd NoLr