Session #3: Homework Solutions

Session #3: Homework Solutions Problem #1 From a standard radio dial, determine the maximum and minimum wavelengths ( λmax and λmin ) for broadcasts on the (a) AM band (b) FM band Solution c = νλ, ∴ λmin = AM λmin = FM λmin = c c ; λmax = νmax νmin 3 x 108m / s 1600 x 103Hz 3 x 108 108 x 106 = 188 m = 2.78 m λmax = λmax = 3 x 108 530 x 103 3 x 108 88 x 106 Problem #2 For light with a wavelength ( λ ) of 408 nm determine: (a) (b) (c) (d) (e) the the the the the frequency wave number wavelength in Å total energy (in Joules) associated with 1 mole of photons “color” Solution To solve this problem we must know the following relationships: νλ = c ; 1/λ = ν ; 1 nm = 109 m = 10 Å E = hν ; Emolar = hν x NA (NA = 6.02 x 1023 ) (a) ν (frequency) = c 3 x 108 m/s = = 7.353 x 1014s−1 λ 408 x 10-9 m = 566 m = 3.41 m (b) ν (wavenumber) = (c) λ = 408 x 10-9m x 1 1 = = 2.45 x 106m−1 λ 408 x 10-9m 1010 Å = 4080Å m (d) E = hν x NA = 6.63 x 10-34 x 7.353 x 1014 x 6.02 x 1023 J/mole = 2.93 x 105 J/mole = 293 kJ/mole (e) visible spectrum: violet (500 nm) red (800 nm) 408 nm = UV Problem #3 For “yellow radiation” (frequency, ν , = 5.09 x 1014 s–1) emitted by activated sodium, determine: (a) the wavelength ( λ ) in [m] (b) the wave number ( ν ) in [cm–1] (c) the total energy (in kJ) associated with 1 mole of photons Solution (a) The equation relating ν and λ is c = νλ where c is the speed of light = 3.00 x 108 m. λ = c 3.00 x 108 m/s = = 5.89 x 10-7m 14 -1 ν 5.09 x 10 s (b) The wave number is 1/wavelength, but since the wavelength is in m, and the wave number should be in cm–1, we first change the wavelength into cm: λ = 5.89 x 10–7 m x 100cm/m = 5.89 x 10–5 cm Now we take the reciprocal of the wavelength to obtain the wave number: ν = 1 1 = = 1.70 x 104 cm-1 -5 λ 5.89 x 10 cm (c) The Einstein equation, E =h ν , will give the energy associated with one photon since we know h, Planck’s constant, and ν . We need to multiply the energy obtained by Avogadro’s number to get the energy per mole of photons. h = 6.62 x 10–34 J.s ν = 5.09 x 1014 s-1 E = hν = (6.62 x 10-34 J.s) x (5.09 x 1014 s-1 ) = 3.37 x 10-19 J per photon This is the energy in one photon. Multiplying by Avogadro’s number: ⎛ 6.023 x 1023 photons ⎞ E ⋅ NAv = (3.37 x 10-19 J per photon) ⎜ ⎟ ⎜ ⎟ mole ⎝ ⎠ = 2.03 x 105 J per mole of photons we get the energy per mole of photons. For the final step, the energy is converted into kJ: 1 kJ = 2.03 x 102 kJ 1000 J 2.03 x 105 J x Problem #4 Potassium metal can be used as the active surface in a photodiode because electrons are relatively easily removed from a potassium surface. The energy needed is 2.15 x 105 J per mole of electrons removed (1 mole =6.02 x 1023 electrons). What is the longest wavelength light (in nm) with quanta of sufficient energy to eject electrons from a potassium photodiode surface? Solution anode + e– Ip, the photocurrent, is proportional to the intensity of incident radiation, i.e. the number of incident photons capable of generating a photoelectron. Ip hν This device should be called a phototube rather than a photodiode – a solar cell is a photodiode. K layer Required: 1eV = 1.6 x 10-19 J Erad = hν = (hc)/λ The question is: below what threshold energy (h ν ) will a photon no longer be able to generate a photoelectron? 2.15 x 105 J/mole photoelectrons x 1 mole 23 6.02 x 10 photoelectrons = 3.57 x 10-19 J/photoelectron λ threshold = hc 3.57 x 10-19 = 6.62 x 10-34 x 3 x 108 3.57 x 10-19 = 5.6 x 10-7m = 560 nm Problem #5 For red light of wavelength ( λ ) 6.7102 x 10–5 cm, emitted by excited lithium atoms, calculate: (a) the frequency ( ν ) in s–1; (b) the wave number ( ν ) in cm–1; (c) the wavelength ( λ ) in nm; (d) the total energy (in Joules) associated with 1 mole photons of the indicated wavelength. Solution (a) c = λν and ν = c/ λ where ν is the frequency of radiation (number of waves/s). For: λ = 6.7102 x 10-5cm = 6.7102 x 10-7m ν = (b) 2.9979 x 108 ms-1 6.7102 x 10-7 m 1 1 = = 1.4903 x 106 m−1 = 1.4903 x 104 cm−1 -7 λ 6.7102 x 10 m ν = (c) λ = 6.7102 x 10-5 cm x (d) E = = 4.4677 x 1014 s−1 = 4.4677 Hz 1 nm 10-7 cm = 671.02 cm hc 6.62 x 10-34 Js x 2.9979 x 108 ms-1 = λ 6.7102 x 10-7 m = 2.96 x 10-19 J/photon = 1.78 x 105 J/mole photons Problem #6 Calculate the “Bohr radius” for He+. Solution In its most general form, the Bohr theory considers the attractive force (Coulombic) between the nucleus and an electron being given by: Fc = Ze2 4πε or2 where Z is the charge of the nucleus (1 for H, 2 for He, etc.). Correspondingly, the electron energy (Eel) is given as: Eel = - Z2 me4 n2 8h2 ε o2 and the electronic orbit (rn): h2εo rn = n2 Z rn = n2 a Z o πme2 For He+ (Z=2), r1 = 1 0.529 ao = x 10-10 m = 0.264Å 2 2 Problem #7 (a) Determine the atomic weight of He++ from the values of its constituents. (b) Compare the value obtained in (a) with the value listed in your Periodic Table and explain any discrepancy if such is observed. (There is only one natural 42He isotope.) Solution (All relevant data are in the P/T and T/C.) (a) The mass of the constituents (2p + 2n) is given as: 2p = 2 x 1.6726485 x 10-24 g 2n = 2 x 16749543 x 10-24 g (2p + 2n) = 6.6952056 x 10-24 g The atomic weight (calculated) in amu is given as: 6.6952056 x 10−24 g 1.660565 x 10-24 g / amu He = 4.03188 amu (b) The listed atomic weight for He is 4.00260 (amu). The data indicate a mass defect of 2.92841 x 10–2 amu, corresponding to 4.8628 x 10–26 g/atom. This mass defect appears as nuclear bond energy: ΔE = 4.8628 x 10-29 kg x 9 x 1016 m2 / s2 = 4.3765 x 10-12 J/atom = 2.6356 x 1012 J/mole Δm = ΔE 2 c = 2.928 x 10-5 kg/mole = 0.02928 g/mole