Chapter6WorksheetKEY

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Dec 6, 2023

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CHEM 1211 Chapter 6: Electron Structure Worksheet 6 KEY Electromagnetic Energy [1] Determine the frequency of light (in Hz) corresponding to a wavelength of 750 nm. Determine the wavelength corresponding to a frequency of 104.7 MHz. Show your work ν = c λ = 2.998×10 17 nm s 750 nm = 3.997×10 14 s -1 ; λ = c ν = 2.998×10 10 cm s 104.7×10 6 s = 286 cm [2] Light with a wavelength of 614.5 nm looks orange. What is the energy, in joules, per photon of this orange light? What is the energy in eV (1 eV = 1.602 × 10 19 J)? Show your work E = hc λ = 6.626×10 -34 J s �� 2.998×10 17 nm s ⁄ � 614.5 nm = 3.233×10 -19 J ; × 1 eV 1.602×10 -19 J = 2.018 eV [3] The emission spectrum of cesium contains two lines whose frequencies are (a) 3.45 × 10 14 Hz and (b) 6.53 ×10 14 Hz. What are the wavelengths and energies per photon of the two lines? What color are the lines? Show your work (a) E = h ν = 4.14×10 -15 eV s � � 3.45×10 14 s = 1.4 3 eV ; λ = c ν = 2.998×10 17 nm s 3.45×10 14 s = 869 nm ; IR (b) E = h ν = 4.14×10 -15 eV s � � 6 .53×10 14 s = 2 .70 eV ; λ = c ν = 2.998×10 17 nm s 6.53 ×10 14 s = 45 9 nm ; Blue [4] The Paschen Series of the hydrogen spectrum involves transitions from n i to n=3, where ni > 3. Using the Rydberg equation, determine the longest and shortest wavelengths of emission lines in this series. Show your work ν = R 1 τ 2 2 - 1 τ 1 2 = 0.01097 nm 1 3 2 - 1 4 2 = 5.333×10 -4 nm = 0.01097 nm 1 3 2 - 1 2 = 1.219×10 -2 nm 𝜆𝜆 = 1 ν = nm 5.333×10 -4 = 1875 nm; = nm 1.219×10 -2 =820 nm; The shortest and longest wavelengths are respectively 820 nm and 1875 nm [5] Using the Rydberg Equation calculate the wavelength of a line in the H-atom spectrum corresponding to the n i =4 and nj=7 Show your work ν = R 1 τ 2 2 - 1 τ 1 2 = 0.01097 nm 1 4 2 - 1 7 2 = 4.617×10 -4 nm ; 𝜆𝜆 = 1 ν = nm 4.617×10 -4 = 2166 nm [6] Determine the wavelength of electromagnetic radiation that is needed to irradiate the surface of aluminum in order to eject electrons with a kinetic energy of 1.91 eV. (Aluminum has a work function of φ =4.3 eV). Show your work 𝜆𝜆 = hc T+ ϕ = 4.14×10 -15 eV s �� 2.998×10 17 nm s ⁄ � 1.91 eV+4.3 eV = 2.00×10 2 nm [7] What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 10 14 s 1 ejects an electron with 7.74 × 10 20 J kinetic energy? Will the photoelectric effect be observed if sodium is exposed to orange light (590 nm to 625 nm)? Show your work T = h ( ν - ν° ) ; ν° = ν - T h = 6.66×10 14 s - 7.74×10 -20 J 6.626×10 -34 J s = 5.49×10 14 s -1 λ° = 2.998×10 17 nm s 5.49×10 14 s =545 nm ; Since wavelength range for orange light is higher, the photoelectric effect is not observed [8] Determine the DE Broglie wavelength of an electron traveling with a speed of 2.65 × 10 6 m/s (Mass of an electron is 9.11 × 10 -31 kg Show your work λ = h mv = 6.626×10 -34 J s 9.11×10 -31 kg �� 2.65×10 6 m s ⁄ � kg m 2 s 2 J = 2.74×10 -10 m = 0.274 nm
[9] Using the Bohr model, determine for an electron in the n = 8 stationary state orbital in Be 3+ (z=4) Show your work (a) the energy in eV E = - 13.6 eV 4 8 2 = - 3.4 eV (b) the radius in pm. r = 52.92 pm× 8 2 4 = 846.7 pm [10] Consider the n=4 and n=7 levels in the hydrogen atom (Z=1) Show your work (a) What is the energy (in eV) at the n=4 and n=7 level? E(n=4) = - 13.6 eV 1 4 2 = - 0.850 eV ; E(n= 7 ) = - 13.6 eV 1 7 2 = - 0.278 eV (b) What is the energy of a photon absorbed or emitted (in eV) in going from the n=4 to the n=7 level? E rad = | E(n=4) E(n=7) | = | (- 0.850 eV) E(- 0.278 eV) | = 0.572 eV (c) What is the frequency and wavelength of this photon? ν = E rad h = 0.572 eV 4.14×10 -15 eV s = 1.38×10 14 s -1 ; λ = c ν = 2.998×10 17 nm s 1.38×10 14 s = 2170 nm [11] A sheet of copper metal ( φ = 4.7 eV) is irradiated with electromagnetic radiation with a wavelength of 150 nm, what is the DE Broglie wavelength (in m/s) of the ejected electrons? Show your work, mass of electron is 9.11 × 10 -31 kg 𝐸𝐸 𝑟𝑟𝑟𝑟𝑟𝑟 = h c λ = 4.14×10 -15 eV s �� 2.998×10 17 nm s ⁄ � 150 𝑛𝑛𝑛𝑛 = 8.27 𝑒𝑒𝑒𝑒 ; 𝑇𝑇 = 𝐸𝐸 𝑟𝑟𝑟𝑟𝑟𝑟 − 𝜙𝜙 = 8.27 𝑒𝑒𝑒𝑒 − 4.7 𝑒𝑒𝑒𝑒 = 3.57 𝑒𝑒𝑒𝑒 E rad (J) =3.57 eV× 1.602×10 -19 J eV =5.72× 10 -19 J λ = h mv = h 2mT = 6.626×10 -34 J s 2 9.11×10 -31 kg �� 5.72×10 -19 J = 6.49×10 -10 m = 0.649 nm Note: mv = ( mv ) 2 = m ( m v 2 ) = 2m 1 2 m v 2 = 2mT [12] Order the orbitals for a multi-electron atom in each of the following lists according to increasing energy (refer to the aufbau table). a) 4d, 3p, 2p, 5s ( 2p < 3p < 5s < 4d) b) 2s, 4s, 3d, 4p (2s < 4s < 3d < 4p) c) 3d, 3p, 4p, 4s, 4d (3p < 4s < 3d < 4p < 4d) [13] Which electron is, on average closer to the nucleus (a) an electron in a 2p or an electron in a 3p orbital 2p (b) an electron in a 3s orbital or an electron in a 3p orbital 3s [14] Write electron configurations (condensed notation) for the following atoms: (a) P (Z=15), [Ne]3s 2 3p 3 (b) Cr (z = 24), [Ar] 4s 1 3 d 5 (c) Ni (z = 28), [Ar] 4s 2 3 d 8 (d) Cu (z = 29), [Ar] 4s 1 3 d 10 (e) Br (z = 35) [Ar] 4s 2 3 d 10 5p 5 [15] Which atom has the electron configuration 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 2 ? Zr
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