Physics Solution Manual for 1100 and 2101

# 6 p h where h is plancks constant using equation 3014

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online