Chapter6WorksheetKEY

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