6 Chapter 1.a) coherent radiation - EMR with identical 's or sets of 's with constant phase relationships b) dispersion of a transparent substance- transparent means not absorbed - i.e. question refers to normal dispersion - with . In normal dispersion, increases gradually and nonlinearly with increasing . c) anomalous dispersion- sharp with () due to absorption of EMR instead of transmission. d) work function - E required to remove an electron from a surface e) photoelectric effect - ejection of an electron from a surface due to absorption of h of E greater than or equal to work function. KE of ejected electron = h - work function f) ground state of a molecule - E levels (electronic, vibrational, and rotational) populated according to Boltzmann distribution which is a function of T and E between levels. At room T almost all molecules will be in ground electronic state, but will have some population in higher vibrational and rotational levels in ground electronic state. g) electronic excitation - e- in a lower E electronic state absorbs E to move to a higher E electronic state. h) blackbody radiation - continuous radiation emitted from a surface due to T of the surface. i) fluorescence - emission of h when an excited state returns to ground state after excitation by a photon. Atomic species usually exhibit resonance fluorescence (see k). Polyatomic species generally fluoresce at 's longer (lower E) than that of absorbed photon due to rapid vibrational relaxation in excited electronic state. j) phosphorescence - emission of h when an excited triplet electronic state returns to a singlet ground electronic state. k) resonance fluorescence - emitted photon has same as absorbed photon. l) photon - individual "particle" of light m) absorptivity - constant relating concentration and pathlength to absorbance. (Is molar absorptivity if concentration is M and pathlength is in cm). - n) wavenumber - inverse of wavelength (units: cm 1). Directly proportional to E. o) relaxation - processes, including fluorescence and phosphorescence, that allow an excited specie to return to ground state. p) Stokes shift - shift to longer (lower E) for emitted photon in fluorescence and phosphorescence relative to absorbed photon. 2. = c/ = 2.9979 x 108 m/s / 6.24 x 10-10 m = 4.804 x 1017 s-1 (s-1 = Hz) E = h = (6.626 x 10-34 J s) (4.804 x 1017 s-1) = 3.183 x 10-16 J E = (3.183 x 10-16 J) (6.2415 x 1018 eV/J) = 1.986 x 103 eV 3. = c/ = 2.9979 x 108 m/s / 3.517 x 10-6 m = 8.5240 x 1013 s-1 (s-1 = Hz) wavenumber = 1/ (in cm-1) = (1/3.517 x 10-4 cm) = 2,843 cm-1 E = nh = (6.626 x 10-34 Js) (8.5240 x 1013 s-1) = 5.6480 x 10-20 J/photon. This is E of one photon (n=1). Book did not give answer for E of one mole of photons 5.648 0 10 -20 J 6.022 10 23 photons 1 kJ kJ E of 1mol of photons = = 34.01 photon mol of photons 1000 J mol
4. = c/ = 2.9979 x 108 m/s/368 x 106 s-1 = 0.8146 m E = h = (6.626 x 10-34 Js)(368 x 106 s-1) = 2.438 x 10-25 J (Answer in book is incorrect) Questions 2, 3, and 4 assigned for you to get a feel for changes in E, , and associated with different regions of EMR spectrum. 5. Easiest way to find is to remember that is independent of medium. Determine it from c and in vacuum. = c/ = 2.9979 x 108 m/s / 589 x 10-9 m = 5.089 x 1014 s-1 = c/v so v = c/ = (2.9979 x 108 m/s)/1.09 = 2.75 x 108 m/s in = 1.09 material can be found from = v/ or 1 1 = 2 2 (derived from vi = i & i = c/vi), = 540 nm 6. sin 1/sin 2 = 2/ 1 sin 30o/sin 11.9o = 2/1.000 , 2 = 2.42 if E2 = 3E1 then 2 =1/3 1 779 nm/3 = 259.6 nm = 260 nm
7. E = hc/ or E = hc, so E1 1 = E2 2
8. need E of one photon to calculate from E = hc/. J/photon = 255 kJ/mol (1 mol/6.022 x 1023 photons)(1000J/1kJ) = (4.234 x 10-19 J/photon) E = hc/ ; = hc/E = (6.626 x 10-34 J s)(2.9979 x 108 m/s) / (4.234 x 10-19 J/photon) = 4.691 x 10-7 m = 469 9. nm a) 660 nm EMR is the longest l that will eject an e-, i.e. just enough E to overcome the work function = hc/ = (6.626 x 10-34 J s)(2.9979 x 108 m/s) / 660 x 10-9 m = 3.009 x 10-19 J eV0 = h - = hc/ - Note: If you have an older edition of the book, the signs in equations 6-16 and 6-18 are wrong. The sign in equation 6-19 is still opposite what it should be even in the new (6th) edition. 6.626 10 -34 Js 2.9979 10 8 m s - 3.00 10 -19 J = 3.57 10 -19 J - 3.00 10 -19 J = 5.70 10 - 20 J 9 9 9 1
555 10 -9 m Answer in book is wrong. They used 550 nm instead of 555 nm. 2 5.70 1 10 -20 kg
=
b)
KE =
1 2 KE mv 2 ; v = = 2 m
m2 s 2 = 1.25 10 11 = 3.53 10 5 m 1 7 s 9.10938 10 -31 kg
10. from Figure 6-18 max = approx. 1500 nm at 2000 K ; k = (2000 K)(1500 nm) = 3.0 x106 Knm (from question 7-3 k is given to 3 sig. figs. as 2.90 x 103 Km or 2.90 x 106 Knm) max = k/T = 3.0 x 106 Knm / 1800 K = 1.7 x 103 nm (or using 2.90 x 106 Knm, 1.61 x 103 nm) Answer in book is a little off. Note: This wavelength is just out of mid IR into near IR. This means Globar should be a good IR source. Look at Figure 7-3 and see that it is an IR source that can be used in all three regions of IR. 11. a) from question 5, 1 1 = 2 2 ; (589 nm )(1.000) / 1.50 = 393 nm b) (694.3 nm )(1.000) / 1.55 = 448 nm 12. There are four interfaces where light can be lost due to reflection: 1) Ir1/I0 = (1.55-1.00)2/(1.55+1.00)2 = 0.0465; Ir1 = 0.0465 I0 lost at first interface; light making it to 2nd surface is (1-.0465)I0 = 0.9535 I0 2) Ir2/0.9535I0 = (1.00-1.55)2/(1.00+1.55)2 ; Ir2 = (0.0465)(0.9535I0) = 0.0443 I0 lost at second interface, and amount moving on to next interface is 0.9535 I0 - 0.0443 I0 = 0.9092 I0 3) Ir3/0.9092I0 = (1.55-1.00)2/(1.55+1.00)2; Ir3 = (0.0465)(0.9092I0) = 0.0423I0 lost at third interface, and amount moving on to last interface is 0.9092 I0 - 0.0423 I0 = 0.8669 I0 4) Ir4/0.8669I0 = (1.00-1.55)2/(1.00+1.55)2 ; Ir4 = (0.0465)(0.8669I0) = 0.0403I0 lost at fourth interface amount actually transmitted after last interface is 0.8669 I0 - 0.0403 I0 = 0.8266 I0 (82.66 % transmitted and 17.34% lost to reflection) important note about reflection losses: the smaller the at each interface, the smaller the reflection loss will be (can put some numbers into formula and see) 13. From p. 146 1) ca...