Physics Solution Manual for 1100 and 2101

Using equation 3014 for 1 we find that v 2 hc h cr

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: nd nf = 3, we find that 1 λ ( )( ) 312 = 1.097 × 107 m −1 12 1 72 or λ = 1.005 ×10−6 m The photon energy is E= hc ( 6.63 ×10−34 J ⋅ s ) (3.00 ×108 m/s ) = = 1.98 × 10 –19 J λ 1.005 × 10−6 m ______________________________________________________________________________ 12. REASONING The ionization energy for a given state is the energy needed to remove the electron completely from the atom. The removed electron has no kinetic energy and no electric potential energy, so its total energy is zero. The ionization energy for a given excited state is less than the ionization energy for the ground state. In the excited state the electron already has part of the energy necessary to achieve ionization, so less energy is required to ionize the atom from the excited state than from the ground state. SOLUTION a. The energy of the nth state in the hydrogen atom is given by Equation 30.13 as En = − (13.6 eV ) Z 2 / n2 . When n = ∞ , E∞ = 0 J , and when n = 4, E4 = − (13.6 eV )(1) / 42 = − 0.850 eV . The difference in energies between these two states is the ionization energy: 2 Ionization energy = E∞ – E4 = 0.850 eV 1552 THE NATURE OF THE ATOM b. In the same manner, it can be shown that the ionization energy for the n = 1 state is 13.6 eV. The ratio of the ionization energies is 0.850 eV = 0.0625 13.6 eV ______________________________________________________________________________ 13. REASONING Since the atom emits two photons as it returns to the ground state, one is emitted when the electron falls from n = 3 to n = 2, and the other is emitted when it subsequently drops from n = 2 to n = 1. The wavelengths of the photons emitted during these transitions are given by Equation 30.14 with the appropriate values for the initial and final numbers, ni and nf. SOLUTION The wavelengths of the photons are n = 3 to n = 2 ( ) 1 2 1 = 1.097 × 107 m −1 (1) 2 − 2 = 1.524 × 106 m −1 λ 3 2 1 (30.14) λ = 6.56 × 10−7 m n = 2 to n = 1 ( ) 1 2 1 = 1.097 × 107 m −1 (1) 2 − 2 = 8.228 × 106 m −1 λ 2 1 1 (30.14) λ = 1.22 × 10−7 m ______________________________________________________________________________ 14. REASONING The Bohr expression as it applies to any one-electron species of atomic number Z, is given by Equation 30.13: En = –(13.6 eV)( Z 2 / n 2 ) . For certain values of the quantum number n, this expression predicts equal electron energies for singly ionized helium He + (Z = 2) and doubly ionized lithium Li + (Z = 3). As stated in the problem, the quantum number n is different for the equal energy states for each species. SOLUTION For equal energies, we can write ( En ) He = ( En ) Li or –(13.6 eV) 2 Z He 2 nHe = –(13.6 eV) Simplifying, this becomes 2 Z He 2 nHe = 2 Z Li 2 nLi or 4 2 nHe = 9 2 nLi 2 Z Li 2 nLi Chapter 30 Problems 1553 Thus, 4 2 = nLi 9 3 nHe = nLi Therefore, the value of the helium energy level is equal to the lithium energy level for any value of nHe that is two-thirds of nLi . For quantum numbers less than or equal to 9, an equality in energy levels will occur for nHe = 2, 4, 6 corresponding to nLi = 3, 6, 9. The results are summarized in the following table. nHe nLi Energy 2 3 –13.6 eV 4 6 –3.40 eV 6 9 –1.51 eV ______________________________________________________________________________ 15. REASONING In the Bohr model of the hydrogen atom the total energy En of the electron is given in electron volts by Equation 30.13 and the orbital radius rn is given in meters by Equation 30.10: −13.6 En = and rn = 5.29 × 10−11 n 2 2 n ( ) Solving the radius equation for n2 and substituting the result into the energy equation gives En = ( −13.6 rn / 5.29 × 10−11 ) = ( −13.6 ) ( 5.29 ×10−11 ) rn Thus, the energy is inversely proportional to the radius, and it is on this fact that we base our solution. SOLUTION We know that the radius of orbit B is sixteen times greater than the radius of orbit A. Since the total energy is inversely proportional to the radius, it follows that the total energy of the electron in orbit B is one-sixteenth of the total energy in orbit A: EB = EA −3.40 eV = = −0.213 eV 16.0 16.0 1554 THE NATURE OF THE ATOM 16. REASONING We expect Zeffective to be greater than one. Since the orbiting “inner” electrons spend part of the time on the side of the nucleus opposite to the outermost electron, they are not always between the nucleus and the outermost electron. Therefore, they are sometimes farther away from the outermost electron than the nuclear protons are. As a result, the repulsive force that they exert on the outermost electron sometimes has a magnitude less than the attractive force exerted by the nuclear protons. The net effect of this is that the 10 “inner” electrons only partially shield the outermost electron from the attractive force of 10 nuclear protons, and it follows that Zeffective is greater than one. In the Bohr model the orbital radius rn is ( rn = 5.29 ×10−11 m ) nZ 2 (30.10) where n is the principal quan...
View Full Document

Ask a homework question - tutors are online