Physics Solution Manual for 1100 and 2101

5 lz the magnitude l of the orbital angular momentum

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Unformatted text preview: ive direction. The conservation of linear momentum gives rise to Equation 29.7, which relates the difference λ ′ − λ between the scattered and incident X-ray photon wavelengths to the scattering angle θ of the electron as λ′ − λ = b h 1 − cos θ mc h 1 g= mc b− cos 180.0° g= 2h mc (1) The conservation of total energy is written as hc λ 4 Energy of incident photon + 0 { Initial kinetic energy of electron = hc λ′ 4 Energy of scattered photon + mv 2 123 1 2 Final kinetic energy of electron (2) Chapter 29 Problems 1543 Equations (1) and (2) will permit us to find the wavelength λ of the incident X-ray photon. SOLUTION Solving Equation (1) for λ ′ and substituting the result into Equation (2) gives hc λ = hc + 2h +λ mc 1 2 mv 2 Algebraically rearranging this result, we obtain a quadratic equation for λ: λ2 + 2 Fh I GJ H3 mc K 1 2 4 .85 × 10 −12 m λ− 2h2 c h=0 1 m 2 mv 2 14243 9 .70 × 10 −20 m 2 where we have used h = 6.63 × 10–34 J⋅s, m = 9.11 × 10–31 kg, c = 3.00 × 108 m/s, and v = 4.67 × 106 m/s. Solving this quadratic equation for λ, we obtain λ = 3.09 × 10 −10 m CHAPTER 30 THE NATURE OF THE ATOM ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (d) The Bohr model deals with a single negatively charged electron in orbit about a positively charged nucleus. Since only one electron is assumed to be present, the model does not take into account the electrostatic repulsion between electrons that would exist in a multipleelectron atom, such as helium (two electrons). 2. (c) According to Equation 30.10, the radius rn of an orbit is proportional to the square of the quantum number n. An increase in the radius by a factor of four, therefore, means that n has doubled. According to Equation 30.8, the magnitude Ln of the orbital angular momentum is proportional to n. This means that Ln doubles when rn doubles. 3. (b) The shortest wavelength λshortest in a given series can be calculated from Equation 30.14 1 1 RZ 2 = 0 ) as = 2 , where R is the Rydberg constant and Z is the number of λshortest ni2 nf protons in the nucleus. (with 4. 1282 nm 5. (e) To remove an electron completely from an atom and place it at rest infinitely far from the nucleus of the atom, energy must be supplied in the amount of the ionization energy. 6. (e) In the quantum mechanical picture, the principal quantum number n can have the values 1, 2, 3, …. The orbital quantum number l can have the values l = 0, 1, 2, ..., ( n − 1) . Thus, for n = 3, the possible values for l are 0, 1, and 2. 7. (a) According to Equation 30.15, the magnitude L of the orbital angular momentum is h L = l ( l + 1) . This expression can be solved for l, the orbital angular momentum 2π quantum number, with the result that l = 4. Since l can be at most n − 1, the principal quantum number n must be 5 or larger. Of the options given in this case, 5 is the only feasible answer. 8. (c) The spin angular momentum of the electron plays no role in the Bohr model, but it does play a role in the quantum mechanical picture. 9. 6.33 × 10−34 J·s 10. (e) The Pauli exclusion principle states that no two electrons in an atom can have the same set of the four quantum numbers n, l, ml, and ms. Chapter 30 Answers to Focus on Concepts Questions 1545 11. (a) A subshell can contain up to a maximum of 2(2l + 1) electrons, where l is the orbital quantum number. For the 5f subshell, l = 3, so the maximum number of electrons is 2(2l + 1) = 14. For the 6h subshell, l = 5, so the maximum number of electrons is 2(2l + 1) = 22. Therefore, 19 electrons can fit into the 6h subshell but not the 5f subshell. 12. (c) The maximum number is obtained by adding the maximum number for each subshell within the n = 5 shell. The value of the orbital quantum number l defines a subshell, and for n = 5, the possible values for l are 0, 1, 2, 3, and 4. Each subshell can contain up to a maximum of 2(2l + 1) electrons. Thus, the l = 0 subshell can hold 2 electrons, the l = 1 subshell can hold 6 electrons, the l = 2 subshell can hold 10 electrons, the l = 3 subshell can hold 14 electrons, and the l = 4 subshell can hold 18 electrons, for a total of 50. 13. (d) For the s, p, and d subshells the values for l are 0, 1, and 2, respectively. Each subshell can contain up to a maximum of 2(2l + 1) electrons. Thus, the s subshell can hold 2 electrons, the p subshell can hold 6 electrons, and the d subshell can hold 10 electrons. In addition, the order in which the subshells fill is important. The subshells fill in order of increasing principal quantum number n and, for a given n, in order of increasing l. There are exceptions, however. For example, the 4s subshell (l = 0) fills before the 3d subshell (l = 2). This answer shows the reverse, namely, that the 3d subshell fills before the 4s subshell. 14. (c) More energy is required to knock a K-shell electron completely out of an atom, when the number Z of protons in the nucleus is greater. This is because the electrostatic force of attraction that the nucleus exerts on an electro...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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