Physics Solution Manual for 1100 and 2101

6 p h where h is plancks constant using equation 3014

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Unformatted text preview: ) = 2.11 × 10−34 J ⋅ s 2π 2π Chapter 30 Problems 1561 ______________________________________________________________________________ 26. REASONING The maximum value for the magnetic quantum number is ml = l ; thus, in state A, l = 2 , while in state B, l = 1 . According to the quantum mechanical theory of angular momentum, the magnitude of the orbital angular momentum for a state of given l is L = l(l + 1) ( h / 2π ) (Equation 30.15). This expression can be used to form the ratio LA / LB of the magnitudes of the orbital angular momenta for the two states. SOLUTION Using Equation 30.15, we find that LA LB = h 2π = 6 = 3 = 1.732 h 2 1(1 + 1) 2π 2(2 + 1) _____________________________________________________________________________________________ 27. SSM WWW REASONING The values that l can have depend on the value of n, and only the following integers are allowed: l = 0, 1, 2, ... ( n − 1) . The values that ml can have depend on the value of l , with only the following positive and negative integers being permitted: ml = – l, . . . –2, –1, 0, +1, +2, . . .+l . SOLUTION Thus, when n = 6, the possible values of l are 0, 1, 2, 3, 4, 5 . Now when ml = 2 , the possible values of l are 2, 3, 4, 5, . . . These two series of integers overlap for the integers 2, 3, 4, and 5. Therefore, the possible values for the orbital quantum number l that this electron could have are l = 2, 3, 4, 5 . ______________________________________________________________________________ 28. REASONING The greater the magnitude of the magnetic quantum number ml , the greater the magnitude of the z component Lz of the electron’s angular momentum, as we see from h Lz = ml (Equation 30.16), where h = 6.626×10−34 J·s is Planck’s constant. The 2π maximum value of the magnetic quantum number ml can have is the value of the orbital quantum number l , so we know that ml = l . The angular momentum L of the electron can be found from the orbital quantum number l via L = l ( l + 1) h (Equation 30.15). 2π SOLUTION Squaring both sides of Equation 30.15 and solving for the quantity l ( l + 1) , we obtain 1562 THE NATURE OF THE ATOM h L = l ( l + 1) 2π 2 2 or 2π L l ( l + 1) = h 2 (1) The orbital quantum number l can only take on nonnegative integral values: l = 0, 1, 2, 3, . . . . The quantity l ( l + 1) in Equation (1), therefore, is a product of two successive integers, and must itself be an integer. Substituting L = 8.948×10−34 J·s into Equation (1) yields ( 2π 8.948 ×10−34 J ⋅ s l ( l + 1) = 6.626 ×10−34 J ⋅ s ) 2 = 72 (2) The only positive integer l that satisfies Equation (2) is 8: l ( l + 1) = 8(8 + 1) = 8(9) = 72 . h (Equation 30.16), we obtain the maximum possible 2π magnitude for the z component of the electron’s angular momentum: Substituting ml = l = 8 into Lz = ml Lz = 8 ( 6.626 ×10−34 J ⋅ s ) = 8.436 ×10−34 J ⋅ s 2π ______________________________________________________________________________ 29. SSM WWW REASONING AND SOLUTION a. For the angular momentum, Bohr's value is given by Equation 30.8, with n = 1, Ln = nh h = 2π 2π According to quantum theory, the angular momentum is given by Equation 30.15. For n = 1, l = 0 h h L = l ( l + 1) = 0 ( 0 + 1) = 0 J ⋅s 2π 2π b. For n = 3; Bohr theory gives Ln = nh 3h = 2π 2π while quantum mechanics gives [n = 3, l = 0] h L = l ( l + 1) 2π h = 0 ( 0 + 1) 2π = 0 J ⋅s Chapter 30 Problems [n = 3, l = 1] h L = l ( l + 1) 2π h = 1(1 + 1) 2π = 1563 2h 2π 6h h h L = l ( l + 1) = 2 ( 2 + 1) = 2π 2π 2π ______________________________________________________________________________ [n = 3, l = 2] 30. REASONING From the drawing, we see that the angle θ is related to the magnitude L of the orbital angular momentum and the magnitude Lz of the z component of the orbital angular momentum by Lz L θ = cos −1 z L (1.5) Lz θ The magnitude L of the orbital angular momentum is given by h (Equation 30.15), where l is the orbital quantum number L = l ( l + 1) 2π and h = 6.63×10−34 J·s is Planck’s constant. The orbital quantum number can have any integer value from l = 0 to l = n − 1 , where n = 5 is the principal quantum number. The h magnitude Lz of the z component of the orbital angular momentum is given by Lz = ml 2π (Equation 30.16), where ml is the magnetic quantum number, an integer ranging in value from −l to l . For a given magnitude L of the orbital angular momentum (and, hence, a given value of the orbital quantum number l ), the smaller the angle θ becomes, the greater the magnitude Lz of the z component of the orbital angular momentum. Therefore, we will determine the smallest value of θ by choosing the greatest possible value of the magnetic quantum number: ml = l . SOLUTION Substituting ml = l into Lz = ml Lz = ml Substituting L = l ( l + 1) h (Equation 30.16), we obtain 2π h h =l 2π 2π h and Equation (1) into Equation (1.5) yields 2π (1) 1564 THE NATURE OF THE ATOM h l L l l 2π −1 = cos −1 θ = cos −1 z = cos −1 = cos l + 1 l ( l + 1) L h l ( l + 1) 2π (2) We must choose the value of the orbital quantum number l such that th...
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