CHAPTER 7 QUANTUM THEORY AND ATOMIC STRUCTURE

Transcription

1 CHAPTER 7 QUANTUM THEORY AND ATOMIC STRUCTURE Te value for te speed of ligt will be 3.00x0 8 m/s except wen more significant figures are necessary, in wic cases,.9979x0 8 m/s will be used. TOOLS OF THE LABORATORY BOXED READING PROBLEMS B7. Plan: Plot absorbance on te y-axis and concentration on te x-axis. Since tis is a linear plot, te grap is of te type y = mx + b, wit m = slope and b = intercept. Any two points may be used to find te slope, and te slope is used to find te intercept. Once te equation for te line is known, te absorbance of te solution in part b) is used to find te concentration of te diluted solution, after wic te dilution equation is used to find te molarity of te original solution. a) Absorbance vs. Concentration: Absorbance Concentration (M) Tis is a linear plot, tus, using te first and last points given: y y m = = = 3,50 =.3x0 4 /M x x x0.0x0 M Using te slope just calculated and any of te data points, te value of te intercept may be found. b = y mx = (3,50/M)(3.0x0 5 M) = = 0.00 (absorbance as no units) b) Use te equation just determined: y = (.3x0 4 /M) x x = (y 0.00)/(.3x0 4 /M) = (0.36/.3x0 4 ) M =.8538x0 5 M =.8x0 5 M Tis value is M f in a dilution problem (M i V i )= (M f V f ) wit V i = 0.0 ml and V f = 50. ml. M f V x0 M 50. ml f M i = = =.36538x0 4 =.4x0 4 M V 0.0 ml i 7-

2 B7. Plan: Te color of ligt associated wit eac wavelengt can be found from Figure 7.3. Te frequency of eac wavelengt can be determined from te relationsip c = or = c. Te wavelengt in nm must be converted to meters x0 m/s nm a) red = 9 67 nm 0 m = x0 4 = 4.47x0 4 s b) blue = c) yellow-orange = END OF CHAPTER PROBLEMS x0 m/s nm 453 nm 0 m x0 m/s nm 589 nm 0 m 9 = 6.65x0 4 = 6.6x0 4 s 9 = x0 4 = 5.09x0 4 s 7. All types of electromagnetic radiation travel as waves at te same speed. Tey differ in bot teir frequency, wavelengt, and energy. 7. Plan: Recall tat te sorter te wavelengt, te iger te frequency and te greater te energy. Figure 7.3 describes te electromagnetic spectrum by wavelengt and frequency. a) Wavelengt increases from left (0 nm) to rigt (0 nm) in Figure 7.3. Te trend in increasing wavelengt is: x-ray < ultraviolet < visible < infrared < microwave < radio wave. b) Frequency is inversely proportional to wavelengt according to te equation c = λν, so frequency as te opposite trend: radio wave < microwave < infrared < visible < ultraviolet < x-ray. c) Energy is directly proportional to frequency according to te equation E = ν. Terefore, te trend in increasing energy matces te trend in increasing frequency: radio wave < microwave < infrared < visible < ultraviolet < x-ray. 7.3 a) Refraction is te bending of ligt waves at te boundary of two media, as wen ligt travels from air into water. b) Diffraction is te bending of ligt waves around an object, as wen a wave passes troug a slit about as wide as its wavelengt. c) Dispersion is te separation of ligt into its component colors (wavelengts), as wen ligt passes troug a prism. d) Interference is te bending of ligt troug a series of parallel slits to produce a diffraction pattern of brigter and darker spots. Note: Refraction leads to a dispersion effect and diffraction leads to an interference effect. 7.4 Evidence for te wave model is seen in te penomena of diffraction and refraction. Evidence for te particle model includes te potoelectric effect and blackbody radiation. 7.5 a) Frequency: C < B < A b) Energy: C < B < A c) Amplitude: B < C < A d) Since wave A as a iger energy and frequency tan B, wave A is more likely to cause a current. e) Wave C is more likely to be infrared radiation since wave C as a longer wavelengt tan B. 7.6 Radiation (ligt energy) occurs as quanta of electromagnetic radiation, were eac packet of energy is called a poton. Te energy associated wit tis poton is fixed by its frequency, E = ν. Since energy depends on frequency, a tresold (minimum) frequency is to be expected. A current will flow as soon as a poton of sufficient energy reaces te metal plate, so tere is no time lag. 7-

3 7.7 Plan: Wavelengt is related to frequency troug te equation c = ν. Recall tat a Hz is a reciprocal second, or /s = s. Assume tat te number 950 as tree significant figures. c = ν 8 (m) = c = 3.00x0 m/s = = 36 m 3 0 Hz s 950. khz khz Hz nm m 0 m = x0 = 3.6x0 nm (nm) = c = 9 Å m 0 m = 3.58x0 = 3.6x0 Å (Å) = c = Wavelengt and frequency relate troug te equation c =. Recall tat a Hz is a reciprocal second, or /s = s. (m) = c 8 = 3.00x0 m/s = = 3. m 6 0 Hz s 93.5 MHz MHz Hz nm m 0 m = x09 = 3.x0 9 nm (nm) = c = 9 Å m = x0 0 = 3.x0 0 Å 0 m (Å) = c = Plan: Frequency is related to energy troug te equation E =. Note tat Hz = s. E = E = (6.66x0 34 J s)(3.8x0 0 s ) =.5788x0 3 =.5x0 3 J 7.0 E = c 34 8 = 6.66x0 J s 3.00x0 m/s Å 0 =.59x0 5 =.5x0 5 J.3 Å 0 m 7. Plan: Energy is inversely proportional to wavelengt ( E = c ). As wavelengt decreases, energy increases. In terms of increasing energy te order is red < yellow < blue. 7. Since energy is directly proportional to frequency (E = ): UV ( = 8.0x0 5 s ) > IR ( = 6.5x0 3 s ) > microwave ( = 9.8x0 s ) or UV > IR > microwave. 7.3 Plan: Wavelengt is related to frequency troug te equation c = ν. Recall tat a Hz is a reciprocal second, or /s = s. 9 = (s 0 Hz s ) =.35 GHz GHz =.35x0 Hz 0 s (nm) = c 8 =.9979x0 m/s nm x0 s 0 m = x0 7 =.3483x0 7 nm 7-3

4 (Å) = c 8 =.9979x0 m/s Å x0 s 0 m = x08 =.3483x0 8 Å 7.4 Frequency and wavelengt can be calculated using te speed of ligt: c = ν. a) = c 8 = 3.00x0 m/s m m 0 m = 3.5x0 3 = 3.x0 3 s b) (m) = c = x0 Hz x0 m/s m 6 = = m s 0 m Hz 7.5 Frequency and energy are related by E =, and wavelengt and energy are related by E = c/ ev.60x0 J (Hz) = E.33 MeV MeV ev = Hz x0 J s s = 3.56x00 = 3.x0 0 Hz 34 8 E = 6.66x0 J s 3.00x0 m/s 6 (m) = c = x0 3 = 9.33x0 3 m 0 ev 9.60x0 J.33 MeV MeV ev Te wavelengt can also be found using te frequency calculated in te equation c = 7.6 Plan: Te least energetic poton in part a) as te longest wavelengt (4 nm). Te most energetic poton in part b) as te sortest wavelengt (00 Å). Use te relationsip c = to find te frequency of te potons and relationsip E = c to find te energy. a) c = = c 8 = 3.00x0 m/s nm 9 4 nm 0 m =.39669x0 5 =.4x0 5 s 34 8 = E = c 6.66x0 J s 3.00x0 m/s nm 9 4 nm 0 m = 8.40x0 9 = 8.x0 9 J b) = c 8 = 3.00x0 m/s Å 0 00 Å 0 m =.3636x0 5 =.4x0 5 s E = c 34 8 = 6.66x0 J s 3.00x0 m/s Å 0 00 Å 0 m = x0 9 = 9.0x0 9 J 7.7 n in te Rydberg equation is equal to a Bor orbit of quantum number n were n =,, 3, Bor s key assumption was tat te electron in an atom does not radiate energy wile in a stationary state, and te electron can move to a different orbit by absorbing or emitting a poton wose energy is equal to te difference in energy between two states. Tese differences in energy correspond to te wavelengts in te known spectra for te ydrogen atoms. A Solar System model does not allow for te movement of electrons between levels. 7-4

5 7.9 An absorption spectrum is produced wen atoms absorb certain wavelengts of incoming ligt as electrons move from lower to iger energy levels and results in dark lines against a brigt background. An emission spectrum is produced wen atoms tat ave been excited to iger energy emit potons as teir electrons return to lower energy levels and results in colored lines against a dark background. Bor worked wit emission spectra. 7.0 Plan: Te quantum number n is related to te energy level of te electron. An electron absorbs energy to cange from lower energy (lower n) to iger energy (iger n), giving an absorption spectrum. An electron emits energy as it drops from a iger energy level (iger n) to a lower one (lower n), giving an emission spectrum. a) Te electron is moving from a lower value of n () to a iger value of n (4): absorption b) Te electron is moving from a iger value of n (3) to a lower value of n (): emission c) Te electron is moving from a iger value of n (5) to a lower value of n ():emission d) Te electron is moving from a lower value of n (3) to a iger value of n (4): absorption 7. Te Bor model works only for a one-electron system. Te additional attractions and repulsions in many-electron systems make it impossible to predict accurately te spectral lines. 7. Te Bor model as successfully predicted te line spectra for te H atom and Be 3+ ion since bot are oneelectron species. Te energies could be predicted from E n = were Z is te atomic number 8 Z.8x0 J n for te atom or ion. Te line spectra for H would not matc te line spectra for Be 3+ since te H nucleus contains one proton wile te Be 3+ nucleus contains 4 protons (te Z values in te equation do not matc); te force of attraction of te nucleus for te electron would be greater in te beryllium ion tan in te ydrogen atom. Tis means tat te pattern of lines would be similar, but at different wavelengts. 7.3 Plan: Calculate wavelengt by substituting te given values into Equation 7.3, were n = and n = 5 because n > n. Altoug more significant figures could be used, five significant figures are adequate for tis calculation. R R = x0 n n 7 m n = n = 5 R = n n x0 m =,303,9.6 m 5 nm (nm) = 9 = = nm,303,9.6 m 0 m 7.4 Calculate wavelengt by substituting te given values into te Rydberg equation, were n = and n = 3 because n > n. Altoug more significant figures could be used, five significant figures are adequate for tis calculation. R = n n x0 m = 9,749,0 m 3 Å (Å) = 0 = = 05.7 Å 9, 749,0 m 0 m 7.5 Plan: Te Rydberg equation is needed. For te infrared series of te H atom, n equals 3. Te least energetic spectral line in tis series would represent an electron moving from te next igest energy level, n = 4. Altoug more significant figures could be used, five significant figures are adequate for tis calculation. R = n n x0 m = 533,55 m

6 nm (nm) = 9 = = nm 533,55 m 0 m 7.6 Plan: Te Rydberg equation is needed. For te visible series of te H atom, n equals. Te least energetic spectral line in tis series would represent an electron moving from te next igest energy level, n = 3. Altoug more significant figures could be used, five significant figures are adequate for tis calculation. R = n n x0 m =,53,300 m 3 nm (nm) = 9 = = nm,53,300 m 0 m 7.7 Plan: To find te transition energy, use te equation for te energy of an electron transition and multiply by Avogadro s number to convert to energy per mole. 8 E =.8x0 J nfinal n initial 8 E =.8x0 J 5 = 4.578x0 9 J/poton x0 J 6.0x0 potons E = =.75687x0 poton mol 5 =.76x0 5 J/mol Te energy as a negative value since tis electron transition to a lower n value is an emission of energy. 7.8 To find te transition energy, use te equation for te energy of an electron transition and multiply by Avogadro s number. 8 E =.8x0 J nfinal n initial 8 E =.8x0 J 3 =.93778x0 8 J/poton x0 J 6.0x0 potons E = =.669x0 poton mol 6 =.7x0 6 J/mol 7-6

7 7.9 Plan: Determine te relative energy of te electron transitions. Remember tat energy is directly proportional to frequency (E = ). Looking at an energy cart will elp answer tis question. n = 5 n = 4 (a) (c) (d) n = n = 3 (b) n = Frequency is proportional to energy so te smallest frequency will be d) n = 4 to n = 3; levels 3 and 4 ave a smaller E tan te levels in te oter transitions. Te largest frequency is b) n = to n = since levels and ave a larger E tan te levels in te oter transitions. Transition a) n = to n = 4 will be smaller tan transition c) n = to n = 5 since level 5 is a iger energy tan level 4. In order of increasing frequency te transitions are d < a < c < b b > c > a > d 7.3 Plan: Use te Rydberg equation. Since te electron is in te ground state (lowest energy level), n =. Convert te wavelengt from nm to units of meters. 9 0 m = 97.0 nm = 9.70x0 nm m ground state: n = ; n =? = x0 m n n 8 9.7x0 m = x0 7 m = n = = n n = 6.4 n = = 9 0 m 8 nm nm =.8x0 6 m = x0 m n n 6.8x0 m = x0 7 m n n 5 7-7

8 0.078 = n 5 = = 0.8 n n = n = E = c 34 8 = 6.66x0 J s 3.00x0 m/s nm nm 0 m = x0 9 = 4.56x0 9 J 7.34 a) Absorptions: A, C, D; Emissions: B, E, F b) Energy of emissions: E < F < B c) Wavelengt of absorption: D < A < C 7.35 If an electron occupies a circular orbit, only integral numbers of wavelengts (= nr) are allowed for acceptable standing waves. A wave wit a fractional number of wavelengts is forbidden due to destructive interference wit itself. In a musical analogy to electron waves, te only acceptable guitar string wavelengts are tose tat are an integral multiple of twice te guitar string lengt ( L) De Broglie s concept is supported by te diffraction properties of electrons demonstrated in an electron microscope Macroscopic objects ave significant mass. A large m in te denominator of = /mu will result in a very small wavelengt. Macroscopic objects do exibit a wavelike motion, but te wavelengt is too small for umans to see it Te Heisenberg uncertainty principle states tat tere is fundamental limit to te accuracy of measurements. Tis limit is not dependent on te precision of te measuring instruments, but is inerent in nature Plan: Use te de Broglie equation. Mass in lb must be converted to kg and velocity in mi/ must be converted to m/s because a joule is equivalent to kg m /s. kg a) Mass (kg) = 3 lb = kg.05 lb mi km 0 m Velocity (m/s) = 0.6 mi km = m/s 3600 s 34 = mu = 6.66x0 J s kg m /s m J = x0 37 = 7.0x0 37 m kg s 3 0. mi km 0 m b) Uncertainty in velocity (m/s) = 0.6 mi km = m/s 3600 s x mv 4 x 4 mv x0 J s kg m /s m J.855x0 35 x0 35 m kg s 7-8

9 a) = mu = 6.66x0 J s 3 kg m /s 0 g 0.6 mi km 3600 s mi J kg km 0 m 6.6x0 g 3.4x0 = x0 5 = 6.6x0 5 m b) x mv x0 J s 3 kg m /s 0 g 0.6 mi km 3600 s x0 mi J kg km 0 m x 4 mv 46.6x0 g.78366x0 4 x0 4 m 7.4 Plan: Use te de Broglie equation. Mass in g must be converted to kg and wavelengt in Ǻ must be converted to m because a joule is equivalent to kg m /s. kg Mass (kg) = 56.5 g 3 = kg 0 g Wavelengt (m) = 5400 Å = u = 7.4 = u = mu 34 m = kg5.4x0 m mu 0 0 m Å = 5.4x0 7 m 6.66x0 J s kg m /s J =.77x0 6 =.x0 6 m/s 34 m = 6.66x0 J s 3 kg m /s 0 g pm 4 g00. pm J kg 0 m = x0 3 = 4.67x0 3 m/s 7.43 Plan: Te de Broglie wavelengt equation will give te mass equivalent of a poton wit known wavelengt and velocity. Te term mass equivalent is used instead of mass of poton because potons are quanta of electromagnetic energy tat ave no mass. A ligt poton s velocity is te speed of ligt, 3.00x0 8 m/s. Wavelengt in nm must be converted to m. 9 0 m Wavelengt (m) = 589 nm nm = 5.89x0 7 m = mu m = 34 u = x0 m3.00x0 m/s 6.66x0 J s kg m /s J = x0 36 = 3.75x0 36 kg/poton 7-9

10 7.44 = mu m = 34 u = 8 67 nm3.00x0 m/s 6.66x0 J s kg m /s nm 9 J = 3.96x0 36 kg/poton 0 m x0 kg 6.0x0 potons =.98x0 poton mol =.98x0 kg/mol 7.45 Te quantity expresses te probability of finding an electron witin a specified tiny region of space Since is te probability of finding an electron witin a small region or volume, electron density would represent a probability per unit volume and would more accurately be called electron probability density A peak in te radial probability distribution at a certain distance means tat te total probability of finding te electron is greatest witin a tin sperical volume aving a radius very close to tat distance. Since principal quantum number (n) correlates wit distance from te nucleus, te peak for n = would occur at a greater distance from te nucleus tan 0.59 Å. Tus, te probability of finding an electron at 0.59 Å is muc greater for te s orbital tan for te s a) Principal quantum number, n, relates to te size of te orbital. More specifically, it relates to te distance from te nucleus at wic te probability of finding an electron is greatest. Tis distance is determined by te energy of te electron. b) Angular momentum quantum number, l, relates to te sape of te orbital. It is also called te azimutal quantum number. c) Magnetic quantum number, m l, relates to te orientation of te orbital in space in tree-dimensional space Plan: Te following letter designations correlate wit te following l quantum numbers: l = 0 = s orbital; l = = p orbital; l = = d orbital; l = 3 = f orbital. Remember tat allowed m l values are l to + l. Te number of orbitals of a particular type is given by te number of possible m l values. a) Tere is only a single s orbital in any sell. l = and m l = 0: one value of m l = one s orbital. b) Tere are five d orbitals in any sell. l = and m l =,, 0, +, +. Five values of m l = five d orbitals. c) Tere are tree p orbitals in any sell. l = and m l =, 0, +. Tree values of m l = tree p orbitals. d) If n = 3, l = 0(s), (p), and (d). Tere is a 3s ( orbital), a 3p set (3 orbitals), and a 3d set (5 orbitals) for a total of nine orbitals ( = 9) a) All f orbitals consist of sets of seven (l = 3 and m l = 3,,, 0, +, +, +3). b) All p orbitals consist of sets of tree (l = and m l =, 0, +). c) All d orbitals consist of sets of five (l = and m l =,, 0, +, +). d) If n =, ten tere is a s ( orbital) and a p set (3 orbitals) for a total of four orbitals ( + 3 = 4). 7.5 Plan: Magnetic quantum numbers (m l ) can ave integer values from l to + l. Te l quantum number can ave integer values from 0 to n. a) l = so m l =,, 0, +, + b) n = so l = = 0 and m l = 0 c) l = 3 so m l = 3,,, 0, +, +, Magnetic quantum numbers can ave integer values from l to +l. Te l quantum number can ave integer values from 0 to n. a) l = 3 so m l = 3,,, 0, +, +, +3 b) n = so l = 0 or ; for l = 0, m l = 0; for l = l, m l =,0,+ c) l = so m l =, 0, + 7-0

11 7.53 Plan: Te s orbital is sperical; p orbitals ave two lobes; te subscript x indicates tat tis orbital lies along te x-axis. a) s: sperical b) p x : lobes along te x-axis z z x x y Te variations in coloring of te p orbital are a consequence of te quantum mecanical derivation of atomic orbitals tat are beyond te scope of tis course a) p z : lobes along te z-axis b) d xy : 4 lobes y z y y x x Te variations in coloring of te p and d orbitals are a consequence of te quantum mecanical derivation of atomic orbitals tat are beyond te scope of tis course Plan: Te following letter designations for te various sublevels (orbitals) correlate wit te following l quantum numbers: l = 0 = s orbital; l = = p orbital; l = = d orbital; l = 3 = f orbital. Remember tat allowed m l values are l to + l. Te number of orbitals of a particular type is given by te number of possible m l values. sublevel allowable m l # of possible orbitals a) d (l = ),, 0, +, + 5 b) p (l = ), 0, + 3 c) f (l = 3) 3,,, 0, +, +, sublevel allowable m l # of possible orbitals a) s (l = 0) 0 b) d (l = ),, 0, +, + 5 c) p (l = ), 0, Plan: Te integer in front of te letter represents te n value. Te letter designates te l value: l = 0 = s orbital; l = = p orbital; l = = d orbital; l = 3 = f orbital. Remember tat allowed m l values are l to + l. a) For te 5s subsell, n = 5 and l = 0. Since m l = 0, tere is one orbital. 7-

12 b) For te 3p subsell, n = 3 and l =. Since m l =, 0, +, tere are tree orbitals. 7-

13 c) For te 4f subsell, n = 4 and l = 3. Since ml = 3,,, 0, +, +, +3, tere are seven orbitals a) n = 6; l = 4; 9 orbitals (m l = 4, 3,,, 0, +, +, +3, +4) b) n = 4; l = 0; orbital (ml = 0) c) n = 3; l = ; 5 orbitals ( m l =,, 0, +, +) 7.59 Plan: Allowed values of quantum numbers: n = po sitive integers; l = integers from 0 to n ; m l = integers from l troug 0 to + l. a) n = ; l = 0; m l = : Wit n =, l can be 0 or ; wit l = 0, te only allowable m l value is 0. Tis combination is not allowed. To correct, eiter cange te l or ml value. Correct: n = ; l = ; m l = or n = ; l = 0; m l = 0. b) n = 4; l = 3; m l = : Wit n = 4, l can be 0,,, or 3; wit l = 3, te allowable m l values are 3,,, 0, +, +, +3. Combination is allowed. c) n = 3; l = ; m l = 0: Wit n = 3, l can be 0,, or ; wit l =, te allowable m l values are, 0, +. Combination is allowed. d) n = 5; l = ; m l = +3: Wit n = 5, l can be 0,,, 3, or 4; wit l =, te allowable m l values are,, 0, +, is not an allowable m l value. To correct, eiter cange l or m l value. Correct: n = 5; l = 3; m l = +3 or n = 5; l = ; m l = a) Combination is allowed. b) No; n = ; l = ; m l = + or n = ; l = ; m l = 0 c) No; n = 7; l = ; m l = + or n = 7; l = 3; m l = 0 d) No; n = 3; l = ; m l = or n = 3; l = ; m l = 7.6 Determine te max for -carotene by measuring its absorbance in te nm region of te visible spectrum. Prepare a series of solutions of -carotene of accurately known concentration (using benzene or cloroform as solvent), and measure te absorbance for eac solution. Prepare a grap of absorbance versus concentration for tese solutions and determine its slope (assuming tat tis material obeys Beer s law). Measure te absorbance of te oil expressed from orange peel (diluting wit solvent if necessary). Te -carotene concentration can ten eiter be read directly from te calibration curve or calculated from te slope (A = kc, were k = slope of te line and C = concentration). 7.6 Plan: For Ppart a, use te values of te constants, π, m e, and a 0 to find te overall constant in te equation. Use te resulting equation to calculate E in part b). Use te relationsip E = c to calculate te wavelengt in part c). Remember tat a joule is equivalent to kg m /s. a) = 6.66x0 34 J s; m = 9.094x0 3 kg; a 0 = 5.9x0 m E = = 8 man e 0 8 ma n e e x0 J s kg m /s E = x0 kg 5.9x0 m J n = (.7963x0 8 J) = (. 80x0 8 J) n n Tis is identical wit te result from Bor s teory. F or te H atom, Z = and Bor s constant =.8x0 8 J. For te ydrogen atom, derivation using classical principles or quantum-mecanical principles yields te same constant. b) Te n = 3 energy level is iger in energy tan t e n = level. Because te zero point of te atom s energy is defined as an electron s infinite distance from te nucleus, a larger negative number describes a lower energy level. Altoug tis may be confusing, it makes sense tat an energy cange would be a positive number. 7-3

14 E = (.80x0 8 J) = x0 9 = 3.08x0 9 J 3 c) E = c (m) = c = E x0 J s.9979x0 m/s x0 J (nm) = x0 m 9 = x0 7 = 6.56x0 7 m 7 nm = = 656. nm 0 m Tis is te wavelengt for te observed red line in te ydrogen spectrum Plan: Wen ligt of sufficient frequency (energy) sines on metal, electrons in te metal break free and a current flows. Solu tion: a) Te lines do not begin at te origin because an electron must absorb a minimum amount of energy before it as enoug energy to overcome te attraction of te nucleus and leave te atom. Tis minimum energy is te energy of potons of ligt at te tresold frequency. b) Te lines for K a nd Ag do not begin at t e same point. Te amount of energy tat an electron must absorb to leave te K atom is less tan te amount of energy tat an electron must absorb to leave te Ag atom, were te attraction between te nucleus and outer electron is stronger tan in a K atom. c) Wavelengt is inversely proportional to energy. Tus, te metal tat requires a larger amount of energy to be absorbed before electrons are emitted will require a sorter wavelengt of ligt. Electrons in Ag atoms require more energy to leave, so Ag requires a sorter wavelengt of ligt tan K to eject an electron. d) Te slopes of te line sow an increase in kinetic energy as te frequency (or en ergy) of ligt is increased. Since te slopes are te same, tis means tat for an increase of one unit of frequency (or energy) of ligt, te increase in kinetic energy of an electron ejected from K is te same as te increase in te kinetic energy of an electron ejected from Ag. After an electron is ejected, te energy tat it absorbs above te tresold energy becomes te kinetic energy of te electron. For te same increase in energy above te tresold energy, for eiter K or Ag, te kinetic energy of te ejected electron will be te same E = c 8 7 a) = 6.66x0 J s)(3.00x0 m/s nm 9 =.8397x0 9 J 700. nm 0 m Tis is te value for eac poton, tat is, J/poton. = 7 poton Number of potons.0x0 J = = potons.8397x0 J 34 8 c b) E = = 6.66x0 J s)(3.00x0 m/s nm 475. nm 0 9 m = x 0 9 J Tis is te value for eac poton, tat is, J/poton. Number of po ton s = 7 poton.0x0 J x0 J = = 48 potons 7.65 Determine te wavelengt: = /(953 cm ) = x0 4 cm 4 (nm) = x0 cm 0 m nm 3 cm 0 9 = = 5.0x0 nm m 4 (Å) = x0 cm 0 m Å cm 0 0 = = 5.0x0 4 Å m 7-4

15 8.9979x0 m/s cm Hz = c/= x0 cm 0 m s = x0 3 = 5.855x0 3 Hz 7.66 Plan: Te Bor model as been successfully applied to predict te spectral lines for one-electron species oter tan H. Common one-electron species are small cations wit all but one electron removed. Since te problem specifies a metal ion, assume tat te possible coices are Li + or Be 3+. Use te relationsip E = to convert te 8 Z f requency to energy and ten solve Bor s equation E =.8x0 J to verify if a wole number for Z n can be calculated. Recall tat te negative sign is a convention based on te zero point of te atom s energy; it is deleted in tis calculation to avoid taking te square root of a negative number. Te igest energy line corresponds to te transition from n = to n =. E = = (6.66x0 34 J s) (.96x0 6 Hz) (s /Hz) = x0 7 J E =.8x0 J 8 Z n Z = carge of te nucleus 7 Z En x0 ( ) = = = x0 J.8x0 J Ten Z = 9 and Z = 3. Terefore, te ion is Li + wit an atomic number of a) Electron: = 34 mu = 6.66x0 J s kg m /s =.394x0 3 6 m J 0 =.x0 0 m 9.x0 kg3.4x0 s x0 J s kg m /s Proton: = = =.6696x0 mu 7 6 m J 3 =.x0 3 m.67x0 kg3.4x0 s b) E = /mu terefore u = E/m u = E m Electron: u = Proton: u = Electron: = Proton: = 5.7x0 J kg m /s 3 9.x0 kg J 5.7x0 J kg m /s 7.67x0 kg mu = J = 7.699x0 7 m/s =.798x0 6 m/s x0 J s kg m /s = x0 3 7 m J = 9.4x0 m 9.x0 kg7.699x0 s mu = 7.67x0 kg x0 J s kg m /s =.0646 x 0 6 m J 3 =. x 0 3 m.798x0 s 7-5

16 7.68 Plan: Te electromagnetic spec trum sows tat te visible region goes from 400 to 750 nm (4000 Å to 7500 Å). Tus, wavelengts b, c, and d are for te tree transitions in te visible series wit n final =. Wavelengt a is in te ultraviolet region of te spectrum and te ultraviolet series as n final =. Wavelengt e is in te infrared region of te spectrum and te infrared series as n final = 3. Use te R ydberg equation to find te n initial for eac line. Convert te wavelengts from Å to units of m. n =? n = ; =.7 Å (sortest corresponds to te largest E) 0 0 m (m) =.7 Å =.7x0 Å 7 m 7 = x0 m n n 7 = x0 m.7x0 m n = n = n = n n = for line (a) (n = n = ) n =? n = 3; = 0,938 Å (longest corresponds to te smallest E) 0 0 m (m) = 0,938 Å =.0938x0 6 m Å = x0 m n n 6 = x0 m.0938x0 m 3 n = 3 n = 0. n = n = n n =36.03 n = 6 for line (e) (n = 6 n = 3) For te oter tree lines, n =. For line (d), n = 3 (largest smallest E). For line (b), n = 5 (smallest largest E). For line (c), n =

17 7.69 E = c tus = c E 34 8 = c 6.66x0 J s 3.00x0 m/s nm a) (nm) = 9 9 = = 43 nm E 4.60x0 J 0 m 34 8 = c 6.66x0 J s 3.00x0 m/s nm b) (nm) = 9 9 = = 86 nm E 6.94x0 J 0 m 34 8 c ) = = 6.66x0 J s 3.00x0 m/s nm c) (nm 9 9 = = 45 nm E 4.4x0 J 0 m 7.70 Index of refraction = c/v; v = c/(index of refraction) a) Water v = c/(index of refraction) = (3.00x0 8 m/s)/(.33) =.556x0 8 =.6x0 8 m/s b) Diamond v = c/(index of refraction) = (3.00x0 8 m/s)/(.4) =.39669x0 8 =.4x0 8 m/s 7.7 Extra significant figures are necessary because of te data presented in te problem. He Ne = 63.8 nm Ar = 6.48x0 4 s Ar Kr E = 3.499x0 9 J Dye = nm Calculating missing values: Ar = c/ = (.9979x0 8 m/s)/(6.48x0 s ) = x 0 7 = 4.876x0 7 m Ar Kr = c/e = (6.66x0 34 J s) (.9979x0 8 m/s)/(3.499x0 9 J) = x0 7 = 5.677x0 7 m Calculating missing values: He Ne = c/ = (.9979x0 m/s)/[63.8 nm (0 m/nm)] = x 0 = x 0 s Ar Kr = E/ = (3.499x0 9 J)/(6.66x0 34 J s) = 5.807x0 4 = 5.8x0 4 s Dye = c/ = (.9979x0 8 m/s)/[663.7 nm (0 9 m/nm)] = x0 4 = 4.57x0 4 s Calculating missing E values: He Ne E = c/ = [(6.66x0 34 J s)(.9979x0 8 m/s)]/[63.8 nm (0 9 m/nm)] 9 9 = x0 = 3.39x0 J Ar E = = (6.66x0 34 J s)(6.48x0 4 s ) = x0 9 = 4.074x0 9 J Dye E = c/ = [(6.66x0 34 J s)(.9979x0 8 m/s)]/[663.7 nm (0 9 m/nm)] =.9993x0 9 =.993x0 9 J Te colors may be predicted from Figure 7.3 and te frequencies. He Ne = 4.738x0 4 s Orange Ar = 6.48x0 4 s Green Ar Kr = 5.8x0 4 s Yellow Dye = 4.57x0 4 s Red 7.7 Plan: Allowed values of quantum numbers: n = positive integers; l = integers from 0 to n ; m l = integers from l troug 0 to + l. a) Te l value must be at least for m l to be, but cannot be greater tan n = 3 =. Increase te l value to or to create an allowable combination. b) Te l value must be at least for m l to be +, but cannot be greater tan n = 3 =. Decrease te l value to or to create an allowable combination. c) Te l value must be at least 3 for m l to be +3, but cannot be greater tan n = 7 = 6. Increase te l value to 3, 4, 5, or 6 to create an allowable combination. d) Te l value must be at least for m l to be, but cannot be greater tan n = 4 = 3. Increase te l value to or 3 to create an allowable combination. 7-7

18 = x0 m n n a) nm = nm 0 m x0 7 m n = n n = 5 nm b) 9 8 nm = 0 m x0 m n = 5 n n = 3 c) = x0 m 3 = 9.749x06 m 7 nm =.0573x0 m 9 = = 0.6 nm 0 m 7.74 Plan: Ionization occurs wen te electron is completely removed from te atom, or wen n final =. We can use te equation for te energy of an electron transition to find te quantity of energy needed to remove completely te electron, called te ionization energy (IE). To obtain te ionization energy per mole of species, multiply by Avogadro s number. Te carge on te nucleus must affect te IE because a larger nucleus would exert a greater pull on te escaping electron. Te Bor equation applies to H and oter one-electron species. Use te expression to determine te ionization energy of B 4+. Ten use te expression to find te energies of te transi tions listed and use E = c to convert energy to wavelengt. 8 Z a) E =.8x0 J Z = atomic number n E =.8x0 J Z nfinal n initial x0 E =.8x0 J Z n initial mol 6 = (.3796x0 ) Z for n = b) In te ground state n =, te initial energy level for te single electron in B 4+. Once ionized, n = is te final energy level. Z = 5 for B 4+. E = IE = (.3796x0 6 ) Z = (.3796x0 6 J/mol)(5 ) = 3.899x0 7 = 3.8x0 7 J/mol c) n and Z = for He + final =, n initial = 3,. E =.8x0 J Z 8 nfinal n initial = 8.8x0 J = x0 3 9 J 7-8

19 E = c (m) = c 34 8 E = 6.66x0 J s 3.00x0 m/s x0 J 7 nm.0568x0 m 9 =.0568x0 7 m (nm) = = = 05 nm 0 m d) n final =, n initial =, and Z = 4 for Be E =.8x0 J Z = n.8x0 J 4 initial (m) = c 6.66x0 34 J s 3.00x0 8 m/s = E 8 8.7x0 J 8 =.79587x0 8 m (nm) =.79587x0 8 m nm 0 9 m = =.8 nm = 8.7x0 8 J 7.75 a) Orbital D as te largest value of n, given tat it is te largest orbital. b) l = indicates a p orbital. Orbitals A and C are p orbitals. l = indicates a d orbital. Orbitals B and D are d orbitals. c) In an atom, tere would be four oter orbitals wit te same value of n and te same sape as orbital B. Tere would be two oter orbitals wit te same value of n and te same sape as orbital C. d) Orbital D as te igest energy a nd orbital C as te lowest energy Plan: Use te values and te equation given in te problem to calculate te appropriate values. a) r n = r = 7.77 a) r n = 0 e n me x0 J s 8.854x0 C x0 kg.60x0 9 C x0 kg.60x0 C J m kg m /s 0 J = 5.93x0 m = x 34 C x0 J s 8.854x0 J m kg m /s b) r 0 = = x0 J 9 = 5.93x0 9 m r 3 = n0 me e 34 J m kg m /s C x0 kg.60x0 C x0 J s 8.854x0 b) Z = for an H atom 8 Z E n =.8x0 J n c) Z = 3 for a Li atom 8 =.8x0 J J = x0 0 = 4.764x0 0 m =.4x0 3 9 =.4x0 9 J 7-9

20 Z 8 3.8x0 J n =.8x0 3 d) Te greater number of protons in te Li nucleus results in a greater interaction between te Li nucleus and its electrons. Tus, te energy of an electron in a particular orbital becomes more negative wit increasing atomic number. 8 E n =.8x0 J = 7.78 Plan: Refer to Capter 6 for te calculation of te amount of eat energy absorbed by a substance from its specific eat capacity and temperature cange (q = c x ma ss x T). Using tis equation, calculate te energy absorbed by te water. Tis energy equals te energy from te microwave potons. Te energy of eac poton can be calculated from its wavelengt: E = c/. Dividing te total energy by te energy of eac poton gives te number of potons absorbed by te water. q = c x mass x T q = (4.84 J/g C)(5 g)(98 0.) C = 8.407x0 4 J c 6.66x0 34 J s 3.00x0 8 m/s 3 E = = =.845x0 J/poton.55x0 m 4 poton Number of potons = 8.407x0 J 3.845x0 J = 6.478x07 = 6.4x0 7 potons 7.79 One sample calculation will be done using te equation in te book: r = e a 0 For r = 50 pm: 3 3 a0 = a e r 0 r 50 a pm 0 =.46553x0 3 e a.6974x0 4 =.46553x0 e =.46553x0 e 5.9 = 5 = (5.6974x0 4 ) = x0 7 4r = 4(50) (3.4585x0 7 ) =.097x0 r (pm) (pm 3/ ) 3 (pm ) 4r (pm ) x0.5x x x0 6. 0x x x x x0 0.00x x0 r Te plots are similar to Figure 7.7A in te text. 7-0

21 7.80 Plan: In general, to test for overlap of te two series, compare te longest wavelengt in te n series wit te sortest wavelengt in te n+ series. Te longest wavelengt in any series corresponds to te transition between te n level and te next level above it; te sortest wavelengt corresponds to te transition between te n level and te n = level. Use te relationsip = R to calculate te wavelengts. n n Solu tion: = R 7 = n n x0 m n n a) Te overlap between te n = series and te n = series would occur between te longest wavelengts for n = and te sortest wavelengts for n =. Lo ngest wavelengt in n = series as n equal to. 7 = x0 m = 8,5,80 m 7 7 = =.56847x0 =.5684x0 m 8,5,80 m Sortest wavelengt in te n = series: 7 = x0 m =,74,940 m 7 = = x0 = x0 7 m,74,940 m Since te longest wavelengt for n = series is sorter tan sortest wavelengt for n = series, tere is no overlap between te two series. b) Te overlap between te n = 3 series and te n = 4 series would occur between te longest wavelengts for n = 3 and te sortest wavelengts for n = 4. Longest wavelengt in n = 3 series as n equal to 4. = x0 m 3 4 = 533,55 m = = x0 6 =.87567x0 6 m 533,55 m Sortest wavelengt in n = 4 series as n =. 7 = x0 m = 685,485 m 4 = =.45887x0 6 =.4588x0 6 m 685,485 m Since te n = 4 series sortest wavelengt is sorter tan te n = 3 series longest wavelengt, te series do overlap. c) Sortest wavelengt in n = 5 series as n =. 7 = ,70.4 m 6x0 m 5 = 43 = = x0 6 =.79408x0 6 m 438,70.4 m Calculate te first few longest lines in te n = 4 series to determine if any overlap wit te sortest wavelengt in te n = 5 series: For n = 4, n = 5: 7 = x0 m = 46,774.6 m

22 = = 4.058x0 6 m 46,774.6 m For n = 4, n = 6: 7 =. 096,85 m 776x0 m 4 6 = 380 = = =.65878x0 6 m 380,85 m For n = 4, n = 7: = x0 m 4 7 = 46,653. m = =.668x0 6 m 46,653. m Te wavelengts of te first two lines of te n = 4 series are longer tan te sortest wavelengt in te n = 5 series. Terefore, only te first two lines of te n = 4 series overlap te n = 5 series. d) At longer wavelengts (i.e., lower energies), tere is increasing overlap between te lines from different series (i.e., wit different n values). Te ydrogen spectrum becomes more complex, since te lines begin to merge into a more-or-less continuous band, and muc more care is needed to interpret te information. 7.8 a) Te igest frequency would correspond to te greatest energy difference. In tis case, te greatest energy difference would be between E 3 and E. E = E 3 E = = ( 5x0 9 J) ( 0x0 9 J) = 5x0 9 J = E/ = (5x0 9 J)/(6.66x0 34 J s) = x0 4 = 8x0 4 s = c/ = (3.00x0 8 m/s)/( x0 4 s ) = x0 7 = 4x0 7 m 9 b) Te ionization energy (IE) is te same as te reverse of E. Tus, te value of te IE is 0x0 J/atom IE = (0x0 J/atom)(kJ/0 J)(6.0x0 atoms/mol) = 04.4 =.x0 kj/mol c) Te sortest wavelengt wo uld correspond to an electron moving from te n = 4 level to te igest level available in te problem (n = 6). E = E 6 E = c/ = ( x0 9 J) ( x0 9 J) = 9x0 9 J = c/e = x0 J 6.66x0 J s 3.00x0 m/s nm 9 = = x0 nm 0 m 7.8 Plan: Te energy differences sougt may be determined by looking at te energy canges in steps. Te wavelengt is calculated from te relationsip = c E. a) Te difference between levels 3 and (E 3 ) may be found by taking te difference in te energies for te 3 transition (E 3 ) and te transition (E ). = (4.854x0 7 J) (4.098x0 7 8 E 3 = E 3 E J) = 7.56x0 J = = 34 8 c E = 6.66x0 J s 3.00x0 m/s x0 J =.69365x0 8 =.63x0 8 m b) Te difference between levels 4 and (E 4 ) may be found by adding te energies for te 4 transition (E 4 ) and te transition (E ). E 4 = E 4 + E = (.04x0 7 J) + (4.098x0 7 J) = 5.x0 7 J c E = x0 J s 3.00x0 m/s 7 5.x0 J 9 9 = x0 = 3.88x0 m 7-

23 c) Te difference between levels 5 and 4 (E 54) may be found by taking te difference in te energies for te 5 transition (E 5 ) and te 4 transition (see part b)). E 0 7 J) (5.x0 7 J) =.x = E 5 E 4 = (5.4x J = c 34 8 E = 6.66x0 J s 3.00x0 m/s 8.x0 J =.6565x0 7 =.66x0 7 m 7.83 a ) A dark green color implies tat relatively few potons are being reflected from te leaf. A large fraction of te potons is being absorbed, par ticularly in te red region of te spectrum. A plant migt adapt in tis way wen potons are in sort supply i.e., in conditions of low ligt intensity. b) An increase in te concentration of cloropyll, te ligt-absorbing pigment, would lead to a darker green color (and vice versa) Plan: For part a), use te equation for kinetic energy, E k = ½mu. For part b), use te relationsip E = c/to find te energy of te poton absorbed. From tat energy subtract te kinetic energy of te dislodged electron to obtai n te work function. Solutio n: a) Te energy of te electron is a function of its speed leaving te surface of te metal. Te mass of te electron is 9.09x0 3 kg. E 3 5 J k = 9.09x0 kg 6.40x0 m/s mu = =.8655x0 9 =.87x0 9 J kg m /s b) Te minimum energy required to dislodge te electron () is a function of te incident ligt. In tis example, te incident ligt is iger tan te tresold frequency, so te kinetic energy of te electron, E k, must be subtracte d from te total energy of te incident ligt,, to yield te work function,. (Te number of significant figures given in te wavelengt requires more significant figures in te speed of ligt.) 9 0 m (m) = 358. nm = 3.58x0 nm 7 m x0 J s.9979x0 m/s E = c/ = x0 m = x0 = E k = ( x0 9 J) (.8655x0 9 J) = x0 9 = 3.58x0 9 J x0 J s kg m /s 7.85 a) = /mu = =.3568x0 3 4 m J 8 m 9.09x0 kg 5.5x0 s Smallest object = / = (.3568x0 8 m)/ = 6.684x0 9 = 6.6x0 9 m x0 J s kg m /s b) = /mu = =.44708x0 m 3 7 m J 9.09x0 kg3.0x0 s Smallest object = / = (.44708x0 m)/ =.354x0 =.x0 m 7.86 Plan: Examine Figure 7.3 and matc te given wavelengts to teir colors. For eac salt, convert te mass of salt to moles and multiply by Avogadro s number to find te number of potons emitted by tat amount of salt c (assuming tat eac atom undergoes one-electron transition). Use te relationsip E = to find te energy of one poton and multiply by te total number of potons for te total energy of emission. 9 J 7-3

24 a) Figure 7.3 indicates tat te 64 nm wavelengt of Sr falls in te red region and te 493 nm wavelengt of Ba falls in te green region. b) SrCl 3 mol SrCl 6.0x0 potons Number of potons = 5.00 g SrCl = x0 potons 58.5 g SrCl mol SrCl (m) = 64 nm E = poton E total = BaCl 9 0 m = 6.4x0 7 m nm 34 8 c = 6.66x0 J s 3.00x0 m/s kj = 3.009x0 kj/poton x0 m 0 J = x0 kj x0 potons p oton Number of potons = 5.00 g BaCl (m) = 9 0 m 493 nm = 4.93x0 nm 7 m c 6.66x0 34 J s3.00x 8 E poton = = = 5.89 kj 3 mol BaCl 6.0x0 potons = x0 08. g BaCl mol BaCl potons 0 m/s kj = x0 kj/poton 4.93x0 m 0 J x0 kj E total = x0 potons poton = = 5.83 kj 7.87 a) Te igest energy line corresponds to te sortest waveleng t. Te sortest wavelengt line is given by R = n n x0 m n n nm nm 0 m = x0 m n 7 304,69 m = ( x0 m ) (/n ) /n = n = 6 b) Te lowest energy line corresponds to te longest wavelengt. Te longest wavelengt line is given by 7 = x0 m n n nm nm 0 m = x0 m 7 34,048 m = x0 m 0.00 = n n n n n n 7-4

25 Rearranging and solving tis equation for n yields n = 5. (You and your students may well need to resort to trialand-error solution of tis equation!) 7.88 Plan: Examine Figure 7.3 to find te region of te electromagnetic spectrum in wic te wavelengt lies. Compare te absorbance of te given concentration of Vitamin A to te absorbance of te given amount of fis- liver oil to find te concentration of Vitamin A in te oil. a) At tis wavelengt te sensitivity to absorbance of ligt by Vitamin A is maximized wile minimizing interference due to te absor bance of ligt by oter substances in te fis-liver oil. b) Te wavelengt 39 nm lies in te ultraviolet region of te electromagnetic spectrum. c) A known quantity of vitamin A (.67x0 3 g) is dissolved in a known volume of solvent (50. ml) to give a standard concentration wit a known response (.08 units). Tis can be used to find te unknown quantity of Vitam in A tat gives a response of 0.74 units. An equality can be made between te two concentration-toabsorbance ratios. 3.67x0 g Concentration (C, g/ml) of Vitamin A = = 6.68x0 50. ml 6 g/ml Vitamin A Absorbance (A ) of Vitamin A =.08 units. Absorbance (A ) of fis-liver oil = 0.74 units Concentration (g/ml) of Vitamin A in fis-liver oil sample = C A A = C C C = A C A = x0 g/ml.08 Mass (g) of Vitamin A in oil sample = 500. ml oil Concentration of Vitamin A in oil sample = = x0 6 g/ml Vitamin A x0 g Vitamin A x0 g 0.3 g Oil ml oil =.3754x0 g Vitamin A =.9808x0 =.93x0 g Vitamin A/g oil = c/e = x0 J 6.66x0 J s 3.00x0 m/s nm = = 6 nm 9 0 m Silver is not a good coice for a potocell tat uses visible ligt because 6 nm is in te ultraviolet region Mr. Green must be in t e dining room were green ligt (50 nm) is reflected. Lower frequency, longer wavelengt ligt is reflected in te lounge and study. Bot yellow and red lig t ave longer wavelengts tan green ligt. Terefore, Col. Mustard and Ms. Scarlet must be in eiter te lounge or study. Te sortest wavelengts are violet. Prof. Plu m must be in te library. Ms. Peacock must be te murderer. 7.9 E k = mv v = E k = /mv = 5 4.7x0 J kg m /s = x0 kg J 8 m/s m = x x0 J s kg m /s = x0 3 8 m J = 7.5x0 m 9.09x0 kg x0 s 7-5

26 7.9 Plan: First find te energy in joules from te ligt tat sines on te text. Eac watt is one joule/s for a total c of 75 J; t ake 5% of tat amount of joules and ten 0% of tat amount. Use E = to find te energy of one poton of ligt wit a wavelengt of 550 nm. Divide te energy tat sines on te text by te energy of one poton to obtain te number of potons. Te amount of energy is calculated from te wavelengt of ligt: 9 0 m m) = 550 nm = 5.50x0 nm 7 m 6.66x0 34 J s 3.00x0 8 m/s E = = c 7 = 3.648x0 9 J/poton 5.50x0 m Amount of power from te bulb = 75 W J/s W = 75 J/s Amount of power converted to ligt = 75 J/s 5% 00% = 3.75 Js 0% Amount of ligt sining on book = 3.75 J/s 00% = J/s J poton Number of potons: 9 s 3.648x0 J =.0376x08 =.0x0 8 potons/s 7.93 a) Sodium ions emit yellow-orange ligt, and potassium ions emit violet ligt. b) Te cobalt glass filter absorbs te yellow-orange ligt, wile te violet ligt passes troug. c) Sodium salts used as te oxidizing agents would emit intense yellow-orange ligt wic would obscure te ligt emitted by oter salts in te fireworks a) 6CO (g) + 6HO(l) C 6 H O 6 (s) + 6O (g) H rxn = {( mol) Hf C 6 H O 6 ] + (6 mol) Hf O ]} {(6 mol) Hf CO ] + (6 mol) Hf H O]} H rxn = [ 73.3 kj + 6(0.0 kj)] [6( kj) + 6( kj)] = = 80.7 kj 6CO (g) + 6HO(l) C6HO 6(s) + 6O (g) H rxn = 80.7 kj (for.00 mol C 6 H O 6 ) x0 J s3.00x0 m/s nm b) E = c/= 9 =.93353x0 9 J/poton 680. nm 0 m 3 0 J poton 9 kj.93353x0 J Number of potons = 80.7 kj = x0 4 = 9.59x0 4 potons 7.95 Plan: In te visible series wit n final =, te transitions will end in eiter te s or p orbitals since tose are te only two types of orbitals in te second main energy level. Wit te restriction tat te angular momentum quantum number can cange by only ±, te allowable transitions are from a p orbital to s (l = to l = 0), from an s orbital to p (l = 0 to l = ), and from a d orbital to p (l = to l = ). Te problem specifies a cange in energy level, so n init must be 3, 4, 5, etc. (Altoug a cange from p to s would result in a + cange in l, tis is not a cange in energy level.) Te first four transitions are as follows: 3s p 3d p 4s p 3p s 7-6

27 P MP P = P MP P = 7.96 a) Absorbance A = kc 3 k = 4.5x0P slope b) A/k = C = (0.55)/(4.5x0P x 0 5 Concentration, mol/l 3 P) =.x0p 4 4.x0P P M 7-7