Physics Solution Manual for 1100 and 2101

1 3 136 ev the closest integer to this result is z 32

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Unformatted text preview: n the proton by the electric force in moving the electron from infinity (position A) to the “surface” of the nucleus (position B) is given by WAB = − q0 (VB − VA ) (Equation 19.4), where q0 = e = 1.6×10−19 C is the charge of a proton, VB is the electric potential at position B, and VA is the electric potential at position A. The electric potentials VA, VB are due to the net charge q = +Ze of the Z = 29 protons in the copper nucleus. The electric potential V a distance r from a point charge q = +Ze is given by kq kZe (Equation 19.6), where k = 8.99×109 N·m2/C2. Therefore, when the proton is V= = r r infinitely far from the copper nucleus (position A), the electric potential is VA = 0 V, and when the proton is at the “surface” of the copper nucleus (position B), the electric potential is kZe (1) VB = rB where rB = 4.8×10−15 m is the radius of the copper nucleus. We note that the work WAB given by Equation 19.4 will be in joules (J), which we will convert to electron volts (eV) via the equivalence 1 eV = 1.6×10−19 J. Chapter 30 Problems 1549 SOLUTION Substituting VA = 0 V and q0 = −e into WAB = − q0 (VB − VA ) (Equation 19.4) yields WAB = − q0 (VB − VA ) = −e (VB − 0 V ) = −eVB (2) Substituting Equation (1) into Equation (2), we obtain WAB kZe2 29 ( 8.99 ×109 N ⋅ m 2 /C2 )(1.6 ×10−19 C ) = −eVB = − =− rB 4.8 ×10−15 m 2 Converting this result to electron volts with the equivalence 1 eV = 1.6×10−19 J, we find that WAB 7. 2 29 ( 8.99 × 109 N ⋅ m 2 /C2 )(1.6 × 10−19 C ) 1 eV = − = −8.7 × 106 eV −15 −19 1.6 × 10 4.8 × 10 m J SSM WWW REASONING AND SOLUTION a. The longest wavelength in the Pfund series occurs for the transition n = 6 to n = 5, so that according to Equation 30.14 with Z = 1, we have ( ) 1 1 1 1 = R 2 – 2 = 1.097 × 107 m –1 2 – 2 λ n 6 5 5 1 or λ = 7458 nm b. The shortest wavelength occurs when 1/n2 = 0, so that ( ) 1 1 1 = R 2 – 2 = 1.097 × 107 m –1 2 λ n 5 5 1 or λ = 2279 nm c. The lines in the Pfund series occur in the infrared region . ______________________________________________________________________________ 8. REASONING The atomic number for helium is Z = 2. The ground state is the n = 1 state, the first excited state is the n = 2 state, and the second excited state is the n = 3 state. With Z = 2 and n = 3, we can use Equation 30.10 to find the radius of the ion. SOLUTION The radius of the second excited state is ( ) ( ) n2 32 (30.10) = 5.29 × 10−11 m = 2.38 × 10−10 m Z 2 ______________________________________________________________________________ r3 = 5.29 × 10−11 m 9. REASONING According to the Bohr model, the energy (in joules) of the nth orbit of an atom containing a single electron is 1550 THE NATURE OF THE ATOM En = − ( 2.18 ×10−18 J ) Z2 n2 (30.12) where Z is the atomic number of the atom. The ratio of the energies of the two atoms can be obtained directly by using this relation. SOLUTION Taking the ratio of the energy En, Be3+ of the nth orbit of a beryllium atom ( Z Be 3+ = 4 ) to the energy En, H of the nth orbit of a hydrogen (ZH = 1) atom gives 2 Z − ( 2.18 × 10−18 J ) Be 2 3+ 2 2 n = Z Be3+ = ( 4 ) = 16 2 2 En, H ZH (1)2 ( 2.18 ×10−18 J ) Z H − n2 ______________________________________________________________________________ En, Be3+ = 10. REASONING a. The total energy En for a single electron in the nth state is given by Z2 n2 En = − (13.6 eV ) (30.13) where Z = 2 for helium. The minimum amount of energy required to remove the electron from the ground state (n = 1) is that needed to move the electron into the state for which n = 2. This amount equals the difference between the two energy levels. b. The ionization energy defined as the minimum amount of energy required to remove the electron from the n = 1 orbit to the highest possible excited state ( n = ∞ ) . SOLUTION a. The minimum amount of energy required to remove the electron from the ground state (n = 1) and move it into the state for which n = 2 is Minimum energy = E2 − E1 = − (13.6 eV ) ( 2 ) 2 22 − (13.6 eV ) ( 2 )2 = 40.8 eV − 12 b. The ionization energy is the difference between the ground-state energy (n = 1) and the energy in the highest possible excited state ( n = ∞ ) . Thus, Ionization energy = E∞ − E1 = − (13.6 eV ) ( 2 ) ( ∞ )2 2 − (13.6 eV ) ( 2 )2 = 54.4 eV − 12 ____________________________________________________________________________________________ Chapter 30 Problems 11. 1551 SSM REASONING According to Equation 30.14, the wavelength λ emitted by the hydrogen atom when it makes a transition from the level with ni to the level with nf is given by 1 1 1 2π 2 mk 2e4 (Z 2 ) 2 = with ni , nf = 1, 2, 3,. . . and ni > nf nf ni2 λ h 3c where 2π 2 mk 2 e4 /(h3c) = 1.097 ×107 m −1 and Z = 1 for hydrogen. Once the wavelength for the particular transition in question is determined, Equation 29.2 ( E = hf = hc / λ ) can be used to find the energy of the emitted photon. SOLUTION In the Paschen series, nf = 3. Using the above expression with Z = 1, ni = 7 a...
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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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