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Physics Solution Manual for 1100 and 2101

# From the periodic table we see that z 78 corresponds

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Unformatted text preview: total number of electrons that can fit into the first and second (n = 2) energy levels. The first energy level has only an s subshell, while the second energy level has both an s subshell and a p subshell. The number of electrons that will fit into either kind of subshell is equal to the number of unique configurations of the quantum numbers l , ml , and ms. Using the “new” rule for the spin quantum number ms and the “old” rules for l and ml , we will count up all of the possible electron configurations for both subshells. Chapter 30 Problems 1567 SOLUTION a. In an s subshell, both the orbital quantum number l and the magnetic quantum number ml must be zero: l = 0 , ml = 0 . This means that there is only one (1) combination of these two quantum numbers to be matched with the three (3) values of the spin quantum number ms. Therefore, there are three (1× 3 = 3) unique configurations of the quantum numbers in the s subshell. We conclude that the first “noble gas” under these quantum rules would be the element with atomic number Z = 3, which is lithium (Li) . b. In a p subshell, the orbital quantum number is l = 1 , and the magnetic quantum number can have three values: ml = −1, 0, 1 . This makes for three (3) combinations of l and ml , each of which may be combined with one of the three (3) unique values of ms, so there are a total of nine (3 × 3 = 9) unique configurations in the p subshell. For a ground-state configuration to fill the n = 1, l = 0 subshell (3 electrons), the n = 2, l = 0 subshell (3 electrons), and the n = 2, l = 1 subshell (9 electrons), the element must have 3 + 3 + 9 = 15 electrons. The element phosphorus (P) has an atomic number Z = 15 and, therefore, would be the second “noble gas” under these quantum rules. ______________________________________________________________________________ 37. SSM REASONING This problem is similar to Example 11 in the text. We use Equation 30.14 with the initial value of n being ni = 2, and the final value being nf = 1. As in Example 11, we use a value of Z that is one less than the atomic number of the atom in question (in this case, a value of Z = 41 rather than 42); this accounts approximately for the shielding effect of the single K-shell electron in canceling out the attraction of one nuclear proton. SOLUTION Using Equation 30.14, we obtain 1 1 = (1.097 × 107 m −1 )(41) 2 2 or λ = 7.230 ×10−11 m 1 λ 22 ______________________________________________________________________________ 1 38. REASONING As discussed in text Example 11, the wavelengths of the Kα electrons depend upon the quantity Z − 1, where Z is the atomic number of the target. The Bohr model Z2 predicts that the energies of the electrons are given by En = − (13.6 eV ) 2 n (Equation 30.13), where n can take on any integer value greater than zero. However, the Kα photons correspond to a transition to the n = 1 shell from the n = 2 shell, and during this transition the nuclear charge is partly screened by the electron that remains in the n = 1 shell, as discussed in text Example 10. We will replace Z with Z − 1 in Equation 30.13 to account for this screening. Therefore, the energy of the electron’s initial and final states are calculated from 1568 THE NATURE OF THE ATOM En = − (13.6 eV ) ( Z −1)2 (1) n2 The energy E of the emitted X-ray is equal to the higher initial energy E2 of the electron minus the lower final energy E1: E = E2 − E1 (2) SOLUTION Substituting Equation (1) into Equation (2), we obtain ( Z −1)2 − − 13.6 eV ( Z − 1)2 = 13.6 eV Z −1 2 1 − E = − (13.6 eV ) ( )2 ( )( ) 2 2 2 = 1 1 1 22 (3) 3 (13.6 eV )( Z −1)2 4 Solving Equation (3) for ( Z − 1) and taking the square root of both sides yields 2 ( Z − 1)2 = 4E 3 (13.6 eV ) or Z= 4E +1 3 (13.6 eV ) The prediction of the Bohr model is, then, Z= 4 ( 9890 eV ) + 1 = 32.1 3 (13.6 eV ) The closest integer to this result is Z = 32 ; the atomic number of germanium (Ge) . 39. REASONING The wavelength λ of a photon is λ = c/f, according to Equation 16.1, where f is the frequency and c is the speed of light in a vacuum. The frequency is given by Equation 29.2 as f = E/h, where E is the energy of the Bremsstrahlung photon and h is Planck’s constant. Substituting this expression for f into the expression for the wavelength gives c c hc λ= = = f E/h E The energy of the photon is 35.0% of the kinetic energy KE of the electron that collides with the metal target. According to our discussions in Section 19.2, the electron acquires its kinetic energy by accelerating from rest through a potential difference V, the kinetic energy Chapter 30 Problems 1569 being KE = eV, where e is the magnitude of the charge on the electron. Thus, we have that E = 0.350 KE = 0.350 eV. Substituting this result into the wavelength expression shows that λ= hc hc = E 0.350 eV (1) SOLUTION Using Equation (1), we find that ( )( ) 6.63 × 10−34 J ⋅ s 3.00 × 108 m/s hc = = 6.83 × 10−11 m λ= 0.350 eV 0.350 1.60 × 10−19 C 52.0 × 103 V ( )( ) 40. REASONING The cutoff wavelength λ0 depends only on the voltage V across the X-ray tube, according to Equation 30.17; λ0 = hc / ( eV ) . Since the volt...
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