ch37 - CHAPTER 37 Atoms 1 As n increases does the spacing...

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Unformatted text preview: 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 = 13.6 eV. ( a ) 9 E , ( b ) 3 E , ( c ) E /3, ( d ) E /9. ( a ) 3 Bohrs 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 , where E 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 . The kinetic energy of the electron in the state n = 2 is _____. ( a ) 4 E , ( b ) 2 E , ( c ) E /2, ( d ) E /4. ( d ) 6 The radius of the n = 1 orbit in the hydrogen atom is a = 0.053 nm. What is the radius of the n = 5 orbit? ( a ) 5 a , ( b ) 25 a , ( c ) a , ( d ) (1/5) a , ( e ) (1/25) a . ( b ) 7 Use the known values of the constants in Equation 37-11 to show that a is approximately 0.0529 nm. Evaluate Equ. 37-12, a = h _ 2 / mke 2 a = 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. Chapter 37 Atoms ? = (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....
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This note was uploaded on 01/23/2011 for the course PHYS 122 taught by Professor Cramer during the Winter '08 term at University of Washington.

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ch37 - CHAPTER 37 Atoms 1 As n increases does the spacing...

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