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Unformatted text preview: c = ν λ, so 1 1 2 − n2 n2 1 c 1 1 − =R λ n2 n2 2 1 1 1 c=R λ 2 − n2 n2 1 c λ= 1 1 R 2 − n2 n2 1 3 × 108 m/s = 1 1 − (3.29 × 1015 1/s) 4 16 −7 = 4.86322 × 10 m = 486.322 nm ν=R 004 (part 2 of 3) 10 points What is the name given to the spectroscope series to which this transition belongs? 1. Pfund series 2. Balmer series correct Explanation: This absorption lies in the visible, blue region. 006 (part 1 of 1) 10 points The ﬁrst line (lowest energy) in the Balmer series appears at 15,233 cm−1 . Where does the second line appear? 1. 109,678 cm−1 2. 20,311 cm−1 3. 45,699 cm−1 4. 30,466 cm−1 5. 20,563 cm−1 correct Explanation: The Balmer series represents emission due to electrons falling from higher levels down to n = 2. The ﬁrst line is from 3, the second from 4, and so on. So the second line is from n2 = 4 to n1 = 2. Using the Ryberg formula, the frequency is ν=R 1 1 2 − n2 n1 2 1 1 − 4 16 3. Paschen series 4. Brackett series 5. Lyman series Explanation: 005 (part 3 of 3) 10 points In what region will the light lie? 1. xray, gamma ray 2. visible, blue correct 3. visible, red 4. visible, yellow 5. infrared 6. ultraviolet = (3.29 × 1015 Hz) = 6.16875 × 1014 Hz so the wavelength is λ= = c ν 3 × 108 m/s 6.16875 × 1014 Hz = 4.86322 × 10−7 m . Wave numbers (cm−1 ) are the reciprocal of wavelengths (in cm): 1 = 20562.5 cm−1 . −5 cm 4.86322 × 10 007 (part 1 of 1) 10 points In a onedimensional particle in a box, for Ψ4 , how many nodes are predicted? ...
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This note was uploaded on 01/23/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Fakhreddine/Lyon
 Chemistry

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