Physics Solution Manual for 1100 and 2101

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Unformatted text preview: carbon (Z = 6). ______________________________________________________________________________ 55. SSM REASONING According to the Bohr model, the energy En (in eV) of the electron ( ) in an orbit is given by Equation 30.13: En = –13.6 Z 2 / n 2 . In order to find the principal quantum number of the state in which the electron in a doubly ionized lithium atom Li 2+ 1578 THE NATURE OF THE ATOM has the same total energy as a ground state electron in a hydrogen atom, we equate the right hand sides of Equation 30.13 for the hydrogen atom and the lithium ion. This gives Z2 Z2 – (13.6 ) 2 = – (13.6 ) 2 n H n Li or n2 2 2 nLi = 2 Z Li Z H This expression can be evaluated to find the desired principal quantum number. SOLUTION For hydrogen, Z = 1, and n = 1 for the ground state. For lithium Li 2+ , Z = 3 . Therefore, n2 12 2 2 nLi = 2 Z Li = 2 32 or nLi = 3 Z 1 H ______________________________________________________________________________ () 56. REASONING If the electron undergoing the Kα transition to the nf = 1 shell from the ni = 2 shell were the only electron in the niobium atom, the Bohr model would predict the 1 1 1 = 1.097 ×107 m−1 ( Z 2 ) 2 − 2 wavelength λpredicted of the emitted X-ray via n n λpredicted f i (Equation 30.14), where Z = 41 is the atomic number of niobium. But in actuality, the charge +Ze of the niobium nucleus is partially screened by the single electron (charge = −e) remaining in the n = 1 shell “below” the electron in the ni = 2 shell. Therefore, we will use Z − 1 in Equation 30.14 to account for this partial screening of the nuclear charge. ( ) SOLUTION a. Taking the reciprocal of Equation 30.14 and replacing Z with Z − 1 yields λpredicted = 1 (1.097 ×107 m−1 ) ( Z −1)2 (1 nf2 ) − (1 ni2 ) Substituting nf = 1, ni = 2, and Z = 41 into Equation (1), we find that λpredicted = (1.097 ×10 7 m −1 ) 1 ( )( ) ( 41 − 1) 1 1 − 1 2 2 2 2 = 7.596 ×10−11 m The difference ∆λ between the predicted wavelength and the observed wavelength is ∆λ = λpredicted − λobserved = 7.596 × 10 −11 m − 7.462 × 10 −11 m = 1.34 × 10 −12 m (1) Chapter 30 Problems 1579 b. Expressed as a percentage of the observed wavelength, the difference is ∆λ λobserved × 100% = 1.34 × 10−12 m × 100% = 1.80% 7.462 × 10 −11 m 57. REASONING The total energy En for a hydrogen atom in the quantum mechanical picture is the same as in the Bohr model and is given by Equation 30.13: En = − (13.6 eV ) 1 n2 (30.13) Thus, we need to determine values for the principal quantum number n if we are to calculate the three smallest possible values for E. Since the maximum value of the orbital quantum number l is n − 1, we can obtain a minimum value for n as nmin = l + 1. But how to obtain l? It can be obtained, because the problem statement gives the maximum value of Lz, the z component of the angular momentum. According to Equation 30.16, Lz is Lz = ml h 2π (30.16) where ml is the magnetic quantum number and h is Planck’s constant. For a given value of l the allowed values for ml are as follows: −l, …, −2, −1, 0, +1, +2, …, +l. Thus, the maximum value of ml is l, and we can use Equation 30.16 to calculate the maximum value of ml from the maximum value given for Lz. SOLUTION Solving Equation 30.16 for ml gives ml = 2π Lz h = ( 2π 4.22 × 10−34 J ⋅ s 6.63 × 10−34 J ⋅ s )=4 As explained in the REASONING, this maximum value for ml indicates that l = 4. Therefore, a minimum value for n is nmin = l + 1 = 4 + 1 = 5 or n≥5 This means that the three energies we seek correspond to n = 5, n = 6, and n = 7. Using Equation 30.13, we find them to be 1580 THE NATURE OF THE ATOM [n = 5] E5 = − (13.6 eV ) 1 = −0.544 eV 52 [n = 6] E6 = − (13.6 eV ) 1 = −0.378 eV 62 [n = 7] E7 = − (13.6 eV ) 1 = −0.278 eV 72 58. REASONING To obtain the quantum number of the higher level from which the electron falls, we will use Equation 30.14 for the reciprocal of the wavelength λ of the photon: 1 1 = R 2 − 2 n λ f ni 1 where R is the Rydberg constant and nf and ni, respectively, are the quantum numbers of the final and initial levels. Although we are not directly given the wavelength, we do have a value for the magnitude p of the photon’s momentum, and the momentum and the wavelength are related according to Equation 29.6: p= h λ where h is Planck’s constant. Using Equation 30.14 to substitute for p= 1 λ , we obtain 1 1 = hR 2 − 2 n λ f ni h (1) SOLUTION Rearranging Equation (1) gives 1 1 p − 2= hR nf2 ni or 1 1 p = 2− ni2 nf hR Thus, we find 1 1 p 1 5.452 ×10−27 kg ⋅ m/s = 2− = 2− = 0.2499 ni2 nf hR 1 6.626 ×10−34 J ⋅ s 1.097 ×107 m −1 ( )( ) or ni = 2 Chapter 30 Problems 59. 1581 SSM REASONING AND SOLUTION For the Paschen series, nf = 3. The range of wavelengths occurs for values of ni = 4 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 –7 m 2 or λ = 8.204 × 10 3 2 = 1.097 × 10 m 144 2444 4 ni 3 Shortest wavelength in ( ) Paschen series The longest wavelength in the Pasc...
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