Physics Solution Manual for 1100 and 2101

# In order to determine the difference zd zp between

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: hen series occurs for ni = 4 and is given by 1 = 1.097 × 107 m –1 2 3 λ 1 ( 1 42 ) λ = 1.875 × 10 –6 3 m 144 2444 4 or Longest wavelength in Paschen series For the Brackett series, nf = 4. The range of wavelengths occurs for values of ni = 5 to ni = ∞. Using Equation 30.14, we find that the shortest wavelength occurs for ni = ∞ and is given by 1 = 1.097 × 107 m –1 (1)2 2 nf λ 1 ( ) 1 7 –1 1 –6 m 2 or λ = 1.459 × 10 3 2 = 1.097 × 10 m 144 2444 4 ni 4 Shortest wavelength in ( ) Brackett series The longest wavelength in the Brackett series occurs for ni = 5 and is given by 1 = 1.097 × 107 m –1 2 4 λ 1 ( ) 1 52 λ = 4.051 × 10 –6 m or 144 2444 4 3 Longest wavelength in Brackett series Since the longest wavelength in the Paschen series falls within the Brackett series, the wavelengths of the two series overlap. ______________________________________________________________________________ 60. REASONING AND SOLUTION From the diagram given with the problem statement we see that an integral number of half wavelengths fit into the "box". That is, nλ/2 = L, where n = 1, 2, 3, ... Using the de Broglie equation yields λ = h/(mv). Combining these two expressions yields, nh/(2mv) = L, and rearranging gives the velocity as v = nh/(2mL). Finally, the kinetic energy can be written KE = 1 mv 2 2 = 1 m nh 2 2mL 2 1582 THE NATURE OF THE ATOM n2 h2 where n = 1,2,3,. . . 8mL2 ______________________________________________________________________________ KE = 61. SSM WWW REASONING AND SOLUTION a. To find an expression for vn, we use Equation 30.8, Ln = mvnrn = nh/(2π), and substitute for rn from Equation 30.9: h 2 n 2 nh mvn 2 4π mke2 Z = 2π or vn = 2π ke2 Z nh b. For the hydrogen atom (Z = 1) in the n = 1 orbit, vn = 2π (8.99 × 109 N ⋅ m 2 /C2 )(1.602 × 10−19 C) 2 (1) = 2.19 × 106 m/s (1) ( 6.626 ×10−34 J ⋅ s ) c. For the n = 2 orbit of hydrogen, vn = 2π (8.99 × 109 N ⋅ m 2 /C2 )(1.602 ×10−19 C) 2 (1) = 1.09 × 106 m/s ( 2 ) ( 6.626 ×10−34 J ⋅ s ) d. The speeds found in parts (b) and (c) are well below the speed of light, 3.0 × 108 m/s, and are consistent with ignoring relativistic effects . ______________________________________________________________________________ CHAPTER 31 NUCLEAR PHYSICS AND RADIOACTIVITY ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. A (e) In the notation Z In, the symbol Z represents the number of protons, and the symbol A represents the number of protons plus the number of neutrons. 2. (c) Because the atomic number of the nucleus is 37, there are 37 protons in the nucleus. Thus, there must be 37 electrons in orbit about the electrically neutral atom. 3. r = 4.8 × 10−15 m 4. (e) According to the discussion in Section 31.1, the nuclear density of all nuclei is approximately the same. 5. (c) The graph in Section 31.2 displays the nucleon number versus the proton number for the elements, and it shows that the number of neutrons in the nucleus is about equal to the number of protons for elements whose proton number is less than or equal to Z = 8. 6. (d) The mass defect depends is equal to the mass of the separated nucleons (63 u) minus the mass (62.5 u) of the stable nucleus (see Section 31.3). 7. 8. (a) Both α and β − decays produce a daughter nucleus that has a different atomic number than the parent nucleus (see Section 31.4). Thus, each decay results in a new element. (b) When a nucleus decays, electric charge is conserved. The number of protons in the 240 92 U daughter nucleus and the α particle (92 + 2) is the same as the number (94) of protons in the 244 94 Pu parent nucleus. Also, when a nucleus decays, the nucleon number is conserved. The number of nucleons in the 240 92 U daughter nucleus and the α particle (240 + 4) is the same as the number (244) of protons in the 9. 244 94 Pu parent nucleus. (d) The radius of a nucleus depends on the number of nucleons it contains (see Section 31.1). In α decay the number of nucleons in the daughter nucleus is four less than that in the parent nucleus, while in β − and γ decay the number of nucleons is the same as that in the parent nucleus (see Section 31.4). 10. (b) When a nucleus decays, electric charge is conserved. The combined charge (80 − 1) of the − 198 198 80 Hg daughter nucleus and the β particle is the same as the charge (79) in the 79 Au parent nucleus. Also, when a nucleus decays, the nucleon number is conserved. The number (198 + 0) 1584 NUCLEAR PHYSICS AND RADIOACTIVITY of nucleons in the of nucleons in the 198 80 Hg 198 79 Au − daughter nucleus and the β particle is the same as the number (198) parent nucleus. 11. (c) The activity of a sample is directly proportional to the number of radioactive nuclei (see Equation 31.4). Therefore, if the mass of a substance increases, the number of nuclei increases. The decay constant λ depends on the specific type of radioactive nuclei in the 4 sample, e.g., 16 C , but not on the mass of the radioactive material (See Equations 31.4...
View Full Document

## 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.

Ask a homework question - tutors are online