Chapter 37 - CHAPTER Atoms 37 As n increases, does the...

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CHAPTER 37 Atoms 1* As n increases, does the spacing of adjacent energy levels increase or decrease? The spacing decreases. 2 The energy of the ground state of doubly ionized lithium ( Z = 3) is ______ ,where E 0 = 13.6 eV. ( a ) –9 E 0 , ( b ) –3 E 0 , ( c ) E 0 /3, ( d )– E 0 /9. ( a ) 3 Bohr’s quantum condition on electron orbits requires ( a ) that the angular momentum of the electron about the hydrogen nucleus equal nh _ . ( b ) that no more than one electron occupy a given stationary state. ( c ) that the electrons spiral into the nucleus while radiating electromagnetic waves. ( d ) that the energies of an electron in a hydrogen atom be equal to nE 0 , where E 0 is a constant energy and n is an integer. ( e ) none of the above. ( a ) 4 ∙∙ If an electron moves to a larger orbit, does its total energy increase or decrease? Does its kinetic energy increase or decrease? Its total energy increases; its kinetic energy decreases. 5* ∙∙ The kinetic energy of the electron in the ground state of hydrogen is 13.6 eV = E 0 . The kinetic energy of the electron in the state n = 2 is _____. ( a ) 4 E 0 , ( b ) 2 E 0 , ( c ) E 0 /2, ( d ) E 0 /4. ( d ) 6 The radius of the n = 1 orbit in the hydrogen atom is a 0 = 0.053 nm. What is the radius of the n = 5 orbit? ( a ) 5 a 0 , ( b ) 25 a 0 , ( c ) a 0 , ( d ) (1/5) a 0 , ( e ) (1/25) a 0 . ( b ) 7 Use the known values of the constants in Equation 37-11 to show that a 0 is approximately 0.0529 nm. Evaluate Equ. 37-12, a 0 = h _ 2 / mke 2 a 0 = m ) 10 (1.6 10 8.99 10 9.11 ) 10 (1.05 2 19 9 31 2 34 × × × × × × = 5.26 × 10 –11 m = 0.0526 nm 8 The longest wavelength of the Lyman series was calculated in Example 37-2. Find the wavelengths for the transitions ( a ) n 1 = 3 to n 2 = 1 and ( b ) n 1 = 4 to n 2 = 1.
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Chapter 37 Atoms λ = (1240/ E ) nm, where E = E i E f and the energies are in eV (see Equs. 17-5 and 17-21); here E f = -13.6 eV. ( a ) E 3 = –13.6/9 eV = –1.51 eV ( b ) E 4 = –13.6/16 eV = –0.85 eV E = 12.09 eV; = 102.6 nm E = 12.75 eV; = 97.25 nm 9* Find the photon energy for the three longest wavelengths in the Balmer series and calculate the wavelengths. For the Balmer series, E f = E ( n = 2) = –3.40 eV. Use Equs. 17-5 and 17-21, i.e., = (1240 eV . nm)/( E eV). 1. E = E 3 E 2 ; E 3 = –13.6/9 eV = –1.51 eV 2. E = E 4 E 2 ; E 4 = –13.6/16 eV = –0.85 eV 3. E = E 5 E 2 ; E 5 = –13.6/25 eV = –0.544 eV E = 1.89 eV; 3,2 = 656.1 nm E = 2.55 eV; 4,2 = 486.3 nm E = 2.856 eV; 5,2 = 434.2 nm 10 ( a ) Find the photon energy and wavelength for the series limit (shortest wavelength) in the Paschen series ( n 2 = 3). ( b ) Calculate the wavelengths for the three longest wavelengths in this series and indicate their positions on a horizontal linear scale. ( a ) For min , n = and E i = 0 h _ f = (13.6/9) eV = 1.51 eV; min = 1240/1.51 nm = 820.6 nm ( b ) For the three longest , n i = 4, 5, and 6 E 4,3 = 0.66 eV; 4,3 = 1876 nm E 5,3 = 0.967 eV; 5,3 = 1282 nm E 6,3 = 1.133 eV; 6,3 = 1094 nm The positions of these lines on a horizontal linear scale are shown below with the wavelengths indicated. 11 Repeat Problem 10 for the Brackett series ( n 2 = 4). ( a ) For min , n = and E i = 0 ( b ) For the three longest , n i = 5, 6, and 7 h _ f = (13.6/16) eV = 0.85 eV; λ min = (1240/0.85) nm = 1459 nm E 5,4 = 0.306 eV; 4,3 = 4052 nm E 6,4 = 0.472 eV; 5,3 = 2627 nm E 7,4 = 0.572 eV; 6,3 = 2166 nm The positions of these lines on a horizontal linear scale are shown below with the wavelengths indicated.
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Chapter 37 Atoms 12
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This note was uploaded on 02/20/2011 for the course PHY 241 taught by Professor Virgil.e.barnes during the Spring '11 term at Purdue University-West Lafayette.

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Chapter 37 - CHAPTER Atoms 37 As n increases, does the...

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