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Chapter_42.pdf

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- South-carolina
- Clemson University
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- Physics 122
- Brown
- Chapter_42.pdf

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42.1. Solve: (a) A 4p state corresponds to n = 4 and l = 1. From Equation 42.3, the orbital angular momentum is 1(1 1) 2 .L = + == = (b) In the case of a 5f state, n = 5 and l = 3. So, ( )3 3 1 12 .L = + == = 42.2. Solve: (a) Excluding spin, a state is described by three quantum numbers: n, l, and m. 3p states correspond to n = 3 and l = 1. The quantum number m takes values from −l to l. The quantum numbers of the various 3p states are displayed in the table below. n 3 3 3 l 1 1 1 m −1 0 +1 (b) A 3d state is described by n = 3 and l = 2. Including the quantum number m, the quantum numbers of the various 3d states are displayed in the table below. n 3 3 3 3 3 l 2 2 2 2 2 m −2 −1 0 +1 +2 42.3. Solve: (a) The orbital angular momentum is ( )1 .L l l= + = Thus, ( ) 22 34 34 3.65 10 J s 1 12 1.05 10 J s L l l − − ⎛ ⎞×⎛ ⎞+ = = =⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠= ⇒ l = 3 This is an f electron. (b) The l quantum number is required to be less than n. Thus, the minimum possible value of n for an electron in the f state is nmin = 4. The corresponding minimum possible energy is min 4 2 13.60 eV 0.85 eV 4 E E= = − = − 42.4. Solve: From Equation 42.2, the hydrogen atom’s energy is 2 13.60 eV 0.544 eV 5nE nn = − = − ⇒ = The largest l value for an n = 5 state is 4. Thus, the magnitude of the maximum possible angular momentum L is ( ) ( )1 4 4 1 20 .L l l= + = + == = = 42.5. Solve: A 6f state for a hydrogen atom corresponds to n = 6 and l = 3. Using Equation 42.2, 6 2 13.6 eV 0.378 eV 6 E −= = − The magnitude of the angular momentum is ( ) ( )1 3 3 1 12 .L l l= + = + == = = 42.6. Model: No two electrons can have exactly the same set of quantum numbers (n, l, m, ms). Solve: For n = 1, there are a total of two states with the quantum numbers given by ( )121, 0, 0, ± . For n = 2, there are a total of eight states: ( ) ( ) ( ) ( )1 1 1 12 2 2 22, 0, 0, 2, 1, 1, 2, 1, 0, 2, 1, 1, ± − ± ± ± For n = 3, there are a total of 18 states: ( ) ( ) ( ) ( )1 1 1 12 2 2 23, 0, 0, 3, 1, 1, 3, 1, 0, 3, 1, 1, ± − ± ± ± ( ) ( ) ( ) ( ) ( )1 1 1 1 12 2 2 2 23, 2, 2, 3, 2, 1, 3, 2, 0, 3, 2, 1, 3, 2, 2, ± ± ± − ± − ± 42.7. Solve: (a) A lithium atom has three electrons, two are in the 1s shell and one is in the 2s shell. The electron in the 2s shell has the following quantum numbers: n = 2, l = 0, m = 0, and ms. ms could be either 12+ or 1 2− . Thus, lithium atoms should behave like hydrogen atoms because lithium atoms could exist in the following two states: ( )122, 0, 0, + and ( )122, 0, 0, − . Thus there are two lines. (b) For a beryllium atom, we have two electrons in the 1s shell and two electrons in the 2s shell. The electrons in both the 1s and 2s states are filled. Because the two electron magnetic moments point in opposite directions, beryllium has no net magnetic moment and is not deflected in a Stern-Gerlach experiment. Thus there is only one line. 42.8. Solve: Mg, Sr, and Ba are all in the second column of the periodic table. The electron configuration of Mg (Z = 12) is 1s22s22p63s2. The electron configuration of Sr (Z = 38) is 1s22s22p63s23p64s23d104p65s2 The electron configuration of Ba (Z = 56) is 1s22s22p63s23p64s23d104p65s24d105p66s2 Assess: It is necessary to recall that the 4s subshell fills before the 3d subshell, the 5s before the 4d, and the 6s before the 4f or 5d. 42.9. Solve: P, As, and Sb are all in the same column of the periodic table as nitrogen. The electron configuration of P (Z = 15) is 1s22s22p63s23p3. The electron configuration of As (Z = 33) is 1s22s22p63s23p64s23d104p3 The electron configuration of Sb (Z = 51) is 1s22s22p63s23p64s23d104p65s24d105p3 Assess: It is necessary to recall that the 4s subshell fills before the 3d subshell. 42.10. Solve: (a) Nine electrons (Z = 9) make the element fluorine (F). These are the nine lowest energy states, so this is the ground state of F. (b) Thirty-one electrons (Z = 31) make the element gallium (Ga). States fill in the order 4s – 3d – 4p. So 3d104s24p, with filled 3d and 4s shells, has the 31 electrons in the lowest possible energy states. This is the ground state of Ga. 42.11. Solve: (a) Ten electrons (Z = 10) make the element neon (Ne). These are not the ten lowest energy states because 1s22s22p6 would be lower in energy than 1s22s22p53d. This is an excited state of Ne. (b) Twenty-six electrons (Z = 26) make the element iron (Fe). These are not the 26 lowest energy states because the 3d shell is not filled. This is an excited state of Fe. 42.12. Solve: Accurate to three significant figures, the factor is ( )( )34 8 26 191 eV6.63 10 J s 3.00 10 m/s 19.89 10 J m 1243 eV nm 1240 eV nm1.60 10 Jhc − − −= × × = × × = ≅× 42.13. Visualize: Please refer to Figure 42.15. Solve: The diagram shows the energy levels for electrons in a multielectron atom. A lithium atom in the ground state has two electrons in the 1s shell and 1 electron in the 2s shell. In other words, the ground-state electron configuration of lithium is 1s22s1. The electron configuration for the first excited state is 1s22p1 and for the second excited state is 1s23s1. 42.14. Model: Assume the hydrogen atom starts in the ground state. Visualize: The electron gains 12 5 eV. of energy in the acceleration. This is enough to excite the hydrogen atom to 3n = (with the electron left with some energy) but no higher. (See Figure 42.4.) Solve: The atom could emit a photon in the two possible quantum-jump transitions: 3 2→ and 3 1→ . The corresponding energies of the emitted photons are 1 51 eV ( 3 40 eV) 1 89 eV− . − − . = . and 1 51 eV ( 13 60 eV) 12 09 eV.− . − − . = . The corre-sponding wavelengths are given by photon .hc Eλ = / photontransition (eV) (nm) 3 2 1.89 656 3 1 12.09 102 E λ → → Assess: The wavelength for the 3 2→ transition is in the visible Balmer series; while the one for the 3 1→ transition is in the ultraviolet region. 42.15. Solve: (a) A 4p → 4s transition is allowed because ∆l = 1. Using the sodium energy levels from Figure 42.25, the wavelength is 1240 eV nm 2210 nm 2.21 m 3.75 eV 3.19 eV hc E λ μ= = = =Δ − (b) A 3d → 4s transition is not allowed because ∆l = 2 violates the selection rule that requires ∆l = 1. 42.16. Solve: Because the probability is very small, we can write the decay rate 10.010 0.10 ns 0.10 ns P r t −= = =Δ The lifetime of the excited state is 1 1 1 10 ns 0.10 nsr τ −= = = 42.17. Solve: The interval ∆t = 0.50 ns is very small in comparison with the lifetime τ = 25 ns, so we can write Prob(decay in ∆t) = r∆t where r is the decay rate. The decay rate is related to the lifetime by r = 1/τ = 1/(25 ns) = 0.040 ns–1. Thus Prob(decay in ∆t) = (0.040 ns–1)(0.50 ns) = 0.020 = 2.0% 42.18. Solve: (a) If there are initially N0 atoms in a state with lifetime τ, then the number remaining at time t is Nexc = N0e−t/τ. From Table 42.3, the 3p state has a lifetime τ = 17 ns. Hence, Nexc = 1.0 × 106e−10/17 = 555,000 (b) Likewise at t = 30 ns, Nexc = 1.0 × 106e−30/17 = 171,000. (c) For t = 100 ns, Nexc = 1.0 × 106e−100/17 = 3000. 42.19. Solve: (a) The number of atoms that have undergone a quantum jump to the ground state is 0.90N0 = 0.90(1.0 × 106) = 9.0 × 105 Because each transition is accompanied by a photon, the number of emitted photons is also 9.0 × 105. (b) 10% of the atoms remain excited at t = 20 ns. Thus ( )/ 20 ns/exc 0 0 0 20 ns0.10 ln 0.10 8.7 nstN N e N N eτ τ ττ− − −= ⇒ = ⇒ = ⇒ = 42.20. Visualize: We’ll first use Equation 42.18 to find the time t when 80% of the excited sodium atoms have decayed (that is, when 20% of the sodium atoms are still excited). Then we’ll take that t and use it again in Equa-tion 42.18 to find how many potassium atoms are still excited; t is the same in both equations. We look up the two lifetimes in Table 42.3: Na 17 nsτ = and P 26 ns.τ = We are given 80 1 0 10N = . × for both kinds of atoms. Solve: Solve Equation 42.18 for t and plug in the values for sodium. exc 0 ln (17 ns)ln(0 20) 27 4 ns N t N τ ⎛ ⎞= − = − . = .⎜ ⎟⎝ ⎠ Now use this t to find excN for potassium. 8 27 4 ns 26 ns 7 exc 0 (1 0 10 ) 3 5 10 tN N e eτ− / − . /= = . × = . × Assess: 35% of the potassium atoms are still excited when only 20% of the sodium ones are; this seems reasonable. 42.21. Solve: Since 1 mW = 0.001 W, the laser emits Elight = 0.001 J of light energy per second. This energy consists of N photons. The energy of each photon is 19 photon 3.14 10 J hc E hf λ −= = = × Because Elight = NEphoton, the number of photons is light 15 19 photon 0.001 J 3.2 10 3.13 10 J E N E − = = = ×× So, photons are emitted at the rate 3.2 × 1015 s−1. 42.22. Solve: The energy of each photon is 20 photon 1.876 10 J hc E hf λ −= = = × Because Elight = NEphoton, 22 20 light (5.0 10 )(1.876 10 J) 940 JE −= × × = So, the energy output is 940 J every second, or 940 W. 42.23. Solve: (a) The wavelength is 1240 eV nm 1060 nm 1.06 m 1.17 eV 0 eV hc E λ μ= = = =Δ − (b) The energy per photon is Eph = 1.17 eV = 1.87 × 10−19 J. The power output of the laser is the number of photons per second times the energy per photon: P = (1.0 × 1019 s−1)(1.87 × 10−19 J) = 1.9 J/s = 1.9 W 42.24. Solve: (a) For l = 3, the magnitude of the angular momentum vector is ( ) ( )1 3 3 1 12L l l= + = + == = = The angular momentum vector has a z-component Lz = Lcosθ along the z-axis. Lz = m= where m is an integer between −l and l, that is, between −3 and 3. The seven possible orientations of the angular momentum vector for l = 3 are shown in figure below. (b) From Equation 42.5, the minimum angle between L G and the z-axis is 1 33 3 cos 30 12 θ − ⎛ ⎞= = °⎜ ⎟⎝ ⎠ = = 42.25. Solve: (a) For s = 1, ( ) 341 2 1.48 10 J s.S s s −= + = = ×= = (b) The spin quantum number is ms = −1, 0, or 1. (c) The figure below shows the three possible orientations of .S G 42.26. Solve: (a) Since 2 2 2 2x y zL L L L+ + = , 2 2 2 2x y zL L L L+ = − . For a hydrogen atom with l = 2, the magnitude of L2 is always l(l + 1)=2 or 6=2. The value of Lz is m=, where m is an integer between −l and l. Hence, the maximum value of Lz is 2= and the minimum value of ( )1/ 22 2x yL L+ is ( ) ( )22 2 21 2 6 4 2l l + − = − == = = = = (b) Because the minimum value of Lz is 0 J s, the maximum value of 2 2 x yL L+ is 2 2 26 0 J s 6zL L− = − == = 42.27. Solve: A hydrogen atom in its fourth excited state is in the n = 5 state. The energy of the wavelength of the emitted photon is 1240 eV nm 0.967 eV 1282 nm hc E λΔ = = = From Equation 42.2, the hydrogen atom’s energy must be one of the following values 2 13.60 eV nE n = − The transition 5 → n corresponds to the energy 2 2 1 1 13.60 eV 0.967 eV 3 5 E n n ⎛ ⎞Δ = − − = ⇒ =⎜ ⎟⎝ ⎠ This means the angular momentum quantum number is l = 2, 1, or 0. The atom’s maximum possible orbital angular momentum after the emission is ( ) ( )1 2 2 1 6L l l= + = + == = = 42.28. Solve: (a) From Equation 42.7, the radial wave function of hydrogen in the 1s state is ( ) ( ) 1B 211 1 B23 3 3 B B B 1 1 0.342r a s sR r e R a e a a aπ π −−= ⇒ = = (b) From Equation 42.10, the probability density is ( ) ( ) ( ) 122 222 B1r 1 B2 3 BB 1 0.368 4 4 2nl s a P r r R r P a e aa π π π −⎛ ⎞ ⎛ ⎞= ⇒ = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ 42.29. Solve: (a) From Equations 42.7 and 42.20, the radial wave function and radial probability density are ( ) ( )12 2 2B1 11 B B2 23 3 3 BB B B 1 0.607 0.607 0.368 4 2s r a R a e P a aa a a ππ π π − ⎛ ⎞ ⎛ ⎞= = ⇒ = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ From Equation 42.9, the probability is Prob(in δ r at r) ( ) ( ) ( ) ( ) 31 B B B2 B 0.368 0.01 0.01 3.7 10r rP r r P a a aa δ −⎛ ⎞= = = = ×⎡ ⎤ ⎜ ⎟⎣ ⎦ ⎝ ⎠ (b) Likewise, ( ) ( ) ( ) 2 1 2 1 B B B 33 3 B BB B 0.3681 0.368 0.541 4s rR a e P a a a aa a π ππ π −= = ⇒ = = The probability is ( ) ( )( ) 3B B0.01 5.4 10r rP r r P a aδ −= = × . (c) For r = 2aB, ( ) ( ) ( ) ( ) 2 22 1 B B B 33 3 B BB B 0.1351 0.135 0.293 2 2 4 2s rR a e P a a a aa a π ππ π −= = ⇒ = = The probability is ( ) ( )( ) 3B B2 0.01 2.9 10 .r rP r r P a aδ −= = × 42.30. Solve: The normalization condition for the three-dimensional hydrogen atom is ( ) ( ) 22 0 0 4 1r nlP r dr r R r drπ∞ ∞= =∫ ∫ . From Equation 42.7, the 1s radial wave function can be written ( ) B1 1 r as sR r A e−= . Hence, ( )B 22 2 2 3 2 1 1 B 1 13 30 B B 2! 1 1 4 4 2 r a s s s sA r e dr A a A A a a π π π π ∞ −= = = ⇒ =∫ 42.31. Solve: From Equation 42.7, the 2p radial wave function of the hydrogen atom can be written ( ) B22 2 B2 r a p p r R r A e a −⎛ ⎞= ⎜ ⎟⎝ ⎠ The normalization condition for the three-dimensional hydrogen atom is ( ) ( ) ( ) B B 22 2 22 2 2 2 20 0 0 B 2 2 2 24 3 2 B 2 252 2 30 B B BB 4 4 4 4! 1 24 1 241 r a r nl p p pr a p p r P r dr r R r dr A e r dr a A A r e dr a A A a a aa π π ππ π π −∞ ∞ ∞ ∞ − ⎛ ⎞= = ⎜ ⎟⎝ ⎠ ⎡ ⎤= = = = ⇒ =⎢ ⎥⎢ ⎥⎣ ⎦ ∫ ∫ ∫ ∫ 42.32. Solve: Using Equations 42.10 and 42.7, the radial probability density for the 1s state is ( ) ( ) B2 22 2 21 14 4 r ar s sP r r R r A r eπ π −= = This radial probability density peaks at the point where dPr/dr = 0. The derivative is B B B 2 2 2 22 2 1 1 B B 2 4 2 8 1r a r a r ar s s dP r r A re e A re dr a a π π− − −⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ The derivative is zero at r = aB, so Pr(r) is a maximum at this value of r. 42.33. Solve: (a) From Equation 42.7, the 2p radial wave function is ( ) B2 22 B2 p r a p A R r re a −= The graph of R2p(r) is seen to have a single maximum. (b) R2p(r) is a maximum at the point where dR2p/dr = 0. The derivative is B B B2 2 22 2 2 B B B B 1 2 2 2 2 p p pr a r a r adR A Ar re e e dr a a a a − − −⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ The derivative is zero at r = 2aB, so R2p(r) is a maximum at r = 2aB . (c) The radial wave function R2p and the probability density P2p(r) is smaller at r = 4aB than it is at r = 2aB. We are less likely to find the electron at a point with r = 4aB than at a point with r = 2aB. However, there are many more points with r = 4aB than with r = 2aB. The increased number of points more than compensates for the decreased probability per point. As a result, it is more probable to find the electron at distance r = 4aB than it is at distance r = 2aB. 42.34. Solve: (a) For a hydrogen atom in the p-state, l = 1. Because Lz = m= and 12 ,zS = ± = the three possible values of Lz are =, 0, and −=, and the two possible values of Sz are 12 = and 12 .− = We can now compute Jz using Jz = Lz + Sz. The results are presented in the following table. Lz Sz Jz mj = 12+ = 32 = 32− = 12− = 12 = 12− 0 1 2+ = 12 = 12− 0 1 2− = 12− = 12− −= 12+ = 12− = 12− −= 12− = 32− = 32− (b) The values of Jz found above can be divided into two groups: ( )3 31 12 2 2 2, , ,− −= = = = and ( )1 12 2, −= = . Because Jz = mj= and mj = −j to j, the j values of the above two groups are 32 and 12 . 42.35. Solve: The electron configuration in the ground state of K (Z = 19) is 1s22s22p63s23p64s1 This means that all the states except the 4s state are completely filled. For Ti (Z = 22), all the states up to the 4s state are completely filled. The electron configuration is 1s22s22p63s23p64s23d2 The d subshell has only two electrons. In the case of Fe (Z = 26), the 3d subshell has six electrons and all the lower level states are completely filled. The electron configuration is 1s22s22p63s23p64s23d6 The ground-state electron configurations of Ge (Z = 32) and Br (Z = 35) are 1s22s22p63s23p64s23d104p2 1s22s22p63s23p64s23d104p5 42.36. Solve: The electron configuration in the ground state of Ca (Z = 20) is 1s22s22p63s23p64s2. The configuration for V (Z = 23) is 1s22s22p63s23p64s23d3. Ni (Z = 28), As (Z = 33), and Kr (Z = 36) are 1s22s22p63s23p64s23d8 1s22s22p63s23p64s23d104p3 1s22s22p63s23p64s23d104p6 42.37. Visualize: Please refer to Figure 42.25. Solve: (a) The allowed transitions are those with ∆l = ±1. Starting from an s-state, the only allowed transitions are to p-states. (b) Once the states are identified, the wavelength is λ = hc/∆E = 1240 eV nm/∆E. Transition ∆E λ 6s → 5p 6s → 4p 6s → 3p 0.17 eV 0.76 eV 2.41 eV 7290 nm 1630 nm 515 nm 42.38. Visualize: Please refer to Figure 42.25. Solve: For sodium, the energy of the 3p state (which is the first excited state) is E3p = 2.104 eV, compared to the ground state at E = 0 eV. We need to add the transition energy E5d →3p to E3p to obtain the energy of the 5d state. The transition energy is 5 3 5 3 5 3 1240 eV nm 2.485 eV 2.104 eV 2.485 eV 4.59 eV 499 nmd p d p d p hc E E E Eλ→ →= = = ⇒ = + = + = Assess: E5d is larger than E6s, E5p, and E4d and less than the ionization energy limit of 5.14 eV. This is reasonable. 42.39. Visualize: Please refer to Figure 42.25. Solve: A photon wavelength of 818 nm corresponds to an energy of 1240 eV nm 1.516 eV 818 nm hc E λ= = = From Figure 42.25, the transition that obeys the selection rule ∆l = 1 and has a magnitude around 1.5 eV is 3p → 3d. Note that E3d – E3p = 3.620 eV – 2.104 eV = 1.516 eV, which is exactly equal to the photon energy. The atom was excited to the 3d state from the ground state. Thus the minimum kinetic energy of the electron was ( )( )192 6 31 2 3.62 eV 1.60 10 J/eV1 3.62 eV 1.13 10 m/s 2 9.11 10 kg mv v − − ×= ⇒ = = ×× 42.40. Model: Allowed transitions are those with ∆l = ±1. Visualize: For the fictitious atom, our task is to draw an energy level diagram that will have four emission wavelengths 310.0 nm, 354.3 nm, 826.7 nm, and 1240.0 nm. These wavelengths correspond to energies of 4.00 eV, 3.50 eV, 1.50 eV, and 1.00 eV. A ground p-state (l = 1), an excited s-state (l = 0) at 3.50 eV, an excited d-state (l = 2) at 4.0 eV, and an excited p-state (l = 1) at 5.0 eV will lead to the four energies or transitions. The excited p-state can decay only to the s and d states, and s and d states can decay only to the ground state. So, the four transitions make a diamond pattern on the diagram. 42.41. Visualize: Please refer to Figure P42.41. Solve: We need to use the condition ∆l = 1 to determine the allowed transitions of the emission spectrum, and the equation 1240 eV nmhc E λ λ= = to find the wavelengths. Visible wavelengths are in the range 400–700 nm, with infrared being greater than 700 nm and ultraviolet being less than 400 nm. The absorption spectrum has only transitions from the ground state, so the wavelengths of the emission spectrum that are in the absorption spectrum belong to transitions that end on the 2s ground state. Transition (a) Wavelength (b) Type (c) Absorption 2p → 2s 3s → 2p 3p → 2s 3p → 3s 3d → 2p 3d → 3p 4s → 2p 4s → 3p 670 nm 816 nm 324 nm 2696 nm 611 nm 24,800 nm 498 nm 2430 nm VIS IR UV IR VIS IR VIS IR Yes No Yes No No No No No 42.42. Visualize: Please refer to Figure P42.42. Solve: (a) We need to use the condition ∆l = 1 to determine the allowed transitions, and the equation 1240 eV nmhc E λ λΔ = = to find the wavelengths. Transition ∆E (eV) λ (nm) 6p → 6s 6d → 6p 7s → 6p 7p → 6s 7p → 7s 8s → 6p 8s → 7p 8p → 6s 8p → 7s 8p → 8s 8p → 6d 6.70 2.14 1.22 8.84 0.92 2.52 0.38 9.53 1.61 0.31 0.69 185 579 1016 140 1350 492 3260 130 770 4000 1800 (b) From part (a), the 492-nm-wavelength blue emission line in the Hg spectrum corresponds to the transition 8s → 6p. The energy of the 8s state is 9.22 eV. The atom is excited to the 8s state from the ground state. Thus the minimum kinetic energy of the electron must be ( )( )192 6 31 2 9.22 eV 1.60 10 J/eV1 9.22 eV 1.80 10 m/s 2 9.11 10 kg mv v − − ×= ⇒ = = ×× 42.43. Model: We have a one-dimensional rigid box with infinite potential walls and a length 0.50 nm. Solve: (a) From Equation 41.22, the lowest energy level is ( ) ( )( ) 2342 1 22 1931 10 6.63 10 J s 1 eV 1.51 V 8 1.60 10 J8 9.11 10 kg 5.0 10 m h E mL − −− − ×= = =×× × The next two levels are E2 = 4E1 = 6.04 eV and E3 = 9E1 = 13.6 eV. The Pauli principle allows only two electrons in each of these energy levels, one with spin up and one with spin down. So five electrons fill the n = 1 and n = 2 levels, with the fifth electron going to n = 3. (b) The ground-state energy of these five electrons is E = 2E1 + 2E2 + E 3 = 28.7 eV 42.44. Model: The electrons are in a one-dimensional rigid box. Solve: (a) The energy levels in a rigid box are En = n2E1 where E1 = h2/8mL2 = 1.51 eV for L = 0.50 nm. The figure below shows the energy-level diagram. The electron in the n = 6 state can decay to lower states, but the transitions must obey the selection rule ∆n = odd. In principle, this allows transitions to n = 5, 3, and 1. However, the n = 1 state is full since the Pauli principle allows only two electrons, one with spin up and the other with spin down. Consequently, the only possible emission transitions are 6 → 5 and 6 → 3. (b) The wavelengths of the photon are calculated from 2 photon electron 6 1 2 1 (36 ) (36 )n hc hc E E E E n E n E λλ= = Δ = − = − ⇒ = − The two allowed transitions are 6 → 5 with λ65 = 75 nm and 6 → 3 with λ63 = 30 nm. 42.45. Solve: (a) From Table 42.3, the lifetime of the 2p state of hydrogen is τ = 1.6 ns. The decay rate is 8 1 9 1 1 6.25 10 s 1.6 10 s r τ − −= = = ×× (b) From Equation 42.25, the number of excited atoms left at time t is Nexc = N0e−t/τ. If 10% of a sample decays, then 90% of the atoms in the sample are still excited. That is, Nexc = 0.90N0. The time for this to occur is calculated as follows: Nexc = 0.90N0 = N0e−t/τ⇒ e−t/τ = 0.90⇒ t = −τ ln0.90 = 0.17 ns 42.46. Solve: For very small time intervals ∆t, the probability that an atom decays is P = r∆t. The time interval for a 1% probability of decay is ( )( )0.01 1.6 ns 0.016 nsPt P r τΔ = = = = where we used τ = 1/r and took the value of τ from Table 42.3. 42.47. Solve: The number of excited atoms left at time t is given by Equation 42.25: Next = N0e−t/τ. If 1% of the atoms in the excited state decay in t = 0.20 ns, then 99% of the atoms remain in the excited state. So, 0.99N0 = N0e−0.20 ns/τ ⇒ 0.99 = e−0.20 ns/τ ⇒ ln(0.99) = −0.20 ns/τ ⇒τ = 19.90 ns Having determined τ, we can now find the time during which 25% of the sample of excited atoms would decay, leaving 75% still excited. Applying Equation 42.25 once again, 0.75N0 = N0e−t/19.90 ns ⇒ t = 5.7 ns 42.48. Solve: (a) The half-life t1/2 is the time needed for half of the initially excited atoms to decay. This means that half of the excited atoms are still present at t = t1/2. Using the exponential decay law, ( ) 1/ 2 1/ 21 1exc 1/ 2 0 02 2t tN t N N e eτ τ− −= = ⇒ = ⇒ t1/2 = τ ln 2 = 0.693τ (b) The half-life is 69.3% of the lifetime τ. From Table 42.3, we find τ = 17 ns for the 3p state of sodium. The half life is t1/2 = (0.693)(17 ns) = 12 ns. 42.49. Solve: If there are Nexc atoms in the excited state, the rate of decay is rNexc, where r = 1/τ is the decay rate. Each decay generates one photon, so the rate of photon emission (photons per second) is the same as the rate of decay (decays per second). Hence, the number of photons emitted per second is 9 16exc exc 9 1.0 10 5.0 10 20 10 s N rN τ − ×= = = ×× 42.50. Solve: 100 MW is the peak power (P = ∆E/∆t). The energy in the laser pulse is ∆E = P∆t = (1.0 × 108 W)(1.0 × 10–8 s) = 1.0 J This light energy is due to N photons, each of energy Eph = hf = hc/λ. That is, ( )( ) ( )( ) 18ph 34 8ph 690 nm 1.0 J 3.5 10 6.63 10 J s 3.0 10 m/s E E E E NE N E hc hc λ λ − Δ Δ ΔΔ = ⇒ = = = = = ×× × Each photon is emitted when a chromium atom undergoes stimulated emission. So a total of 3.5 × 1018 atoms undergo stimulated emission to generate this laser pulse. 42.51. Solve: (a) To derive the formula, consider the change in the energy of the photon as the wavelength changes. The energy of an emitted photon is 2 2 hc dE hc E hf E d hc λλλ λ λ= = ⇒ = − ⇒ Δ = Δ (b) The energy difference between the two states is 21 20 2 2 1 1 13.6 eV 21 20 E E E ⎛ ⎞Δ = − = − +⎜ ⎟⎝ ⎠ = 3.16 × 10 −3 eV To obtain λΔ from the formula in part (a), we need λ. The wavelength is calculated as follows: ( )20 1 2 21 1 1240 eV nm 1240 eV nm13.6 eV 13.56 eV 91.4 nm1 20 13.56 eV hc E λλ λ→ ⎛ ⎞= − = = = ⇒ = =⎜ ⎟⎝ ⎠ The wavelength of the 21 → 1 transition is almost identical. The two wavelengths differ by ( ) ( ) 22 391.4 nm 3.16 10 eV 0.021 nm 1240 eV nm E hc λλ −Δ = Δ = × = 42.52. Solve: The probability of finding a 1s electron at r > aB is ( ) ( ) ( ) B B B B 2 22 2 B 1 3 B 4 Prob 4 r ar sa a ar a P r dr r R r dr r e dra π∞ ∞ ∞ −> = = =∫ ∫ ∫ where we used the expression for ( )1sR r found in Equation 42.7. Change the variable to u = 2r/aB. This means 2r dr = ( )3 28Ba u du and u = 2 when r = aB. The probability is Prob(r > aB) 2 2 1 2 uu e du ∞ −= ∫ ( ) ( )2 2 221 12 2 4 4 2 5 0.677 67.7%2 2uu u e e e ∞− − −⎡ ⎤ ⎡ ⎤= − + + = + + = = =⎣ ⎦⎣ ⎦ 42.53. Solve: The probability of finding a 1s electron at 1 B2r a< is ( ) ( ) ( )B B B B2 22 21 B 12 30 0 0 B 1 1 1 2 2 24Prob 4 a a a r a r sr a P r dr r R r dr r e dra π −< = = =∫ ∫ ∫ where we used the expression for ( )1sR r found in Equation 42.7. Change the variable to u = 2r/aB. This means 2r dr = ( )3 28 .Ba u du Also, the upper limit of the integral changes to u = 1 when r 1 B2 a= . The probability is ( ) ( ) 31 12 2B1 B2 3 0 0 B 1 2 1 1 0 4 1 Prob 8 2 1 1 5 2 2 5 2 1 0.0803 8.03% 2 2 2 u u u a r a u e du u e du a u u e e e − − − − − ⎛ ⎞< = =⎜ ⎟⎝ ⎠ ⎡ ⎤ ⎡ ⎤= − + + = − + = − = =⎣ ⎦⎣ ⎦ ∫ ∫ 42.54. Solve: The radial probability density for the 2s state is ( ) ( ) B B 2 3 4 22 2 2 2 2 2 2 B B B 4 4 1 2 4 r a r a r s s r r r P r r R r A r e C r e a a a π π − −⎛ ⎞ ⎛ ⎞= = − = − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ where 224 sC Aπ= is a constant. This is a maximum where dPr/dr = 0. The derivative is B B B 2 3 3 4 2 2 2 B B B B B 2 3 2 3 B B B 3 1 2 4 4 2 2 4 r a r ar r a dP r r r r C r e r e dr a a a a a r r r C re a a a − − − ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + − − +⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎡ ⎤= − + −⎢ ⎥⎣ ⎦ The derivative is zero, giving the maxima and minima of Pr(r), when 2 3 B B B 1 2 4 2 0 4 r r r a a a ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ This is a cubic equation, which will have three roots. This is expected because we know from the graph in Figure 42.8 that Pr(r) has not only two maxima but also, in between, a minimum. Although the cubic equation can be solved numerically, here we just need to verify that r = 5.236 aB is a solution. Substituting into the equation, 2 1 42 4(5.236) 2(5.236)− + − = 3(5.236) 0.= Thus B5.236r a= is the most probable distance of the electron in the 2s state. 42.55. Solve: The radial probability density for the 2s state is ( ) ( ) B B 2 3 4 22 2 2 2 2 2 2 B B B 4 4 1 2 4 r a r a r s s r r r P r r R r A r e C r e a a a π π − −⎛ ⎞ ⎛ ⎞= = − = − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ where 224 sC Aπ= is a constant. This is a maximum where dPr/dr = 0. The derivative is B B B 2 3 3 4 2 2 2 B B B B B 2 3 2 3 B B B 3 1 2 4 4 2 2 4 r a r ar r a dP r r r r C r e r e dr a a a a a r r r C re a a a − − − ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + − − +⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎡ ⎤= − + −⎢ ⎥⎣ ⎦ The derivative is zero, giving the maxima and minima of Pr(r), when 2 3 B B B 1 2 4 2 0 4 r r r a a a ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ This is a cubic equation, which will have three roots. This is expected because we know from the graph of Figure 42.8 that Pr(r) has not only two maxima but also, in between, a minimum. The cubic equation can be solved numerically to give r = 0.764aB, 2.00aB, and 5.236aB. By referring to the graph, we see that the outer two roots correspond to the maxima and the middle root is the minimum. The separation between the peaks is ( )B B B5.236 0.764 4.472r a a aΔ = − = 42.56. Solve: For an electron in the 1s and 2p states of hydrogen, the radial wave functions are given by Equation 42.7 and the radial probability density for a state nl is given by Equation 42.10. For the 1s state, ( ) ( ) ( )B B 4 2 22 3 B avg B43 3 3 30 0 0 B B B BB 4 4 3! 24 4 1.5 162 r a r a r r a r rP r dr e r dr r e dr a a a a aa ππ ∞ ∞ ∞− −= = = = = =∫ ∫ ∫ where we used formula for the definite integral given in Problem 42.33. For the 2p state, ( )B B 2 2 2 2 5 avg B63 2 5 50 0 B B B B B 1 1 5! 4 5 24 4 24 24 1 r a r ar rr e r dr r e dr a a a a a a ππ ∞ ∞− −⎛ ⎞= = = =⎜ ⎟⎝ ⎠∫ ∫ 42-1 42.57. Solve: (a) The atoms leave the oven at 500 m/s, so their momentum is patom = mv = (1.4 × 10–25 kg)(500 m/s) = 7.0 × 10–23 kg m/s The momentum of a photon traveling in the –x-direction is 34 photon 28 photon 9 6.63 10 J s 8.5 10 kg m/s 780 10 m E hf h p c c λ − − − ×= − = − = − = − = − ×× The negative sign is due to the direction of the photon. (b) As the photon approaches the atom, the total momentum of the atom+photon system is (patom)i + pphoton. The photon is absorbed and disappears, so the final momentum is (patom)f. Momentum is conserved in the absorption process, so atom f atom i photon atom photon( ) ( )p p p p p= + ⇒ Δ = ∆patom is negative because the atom loses momentum. Since ∆patom is the same for every absorption, independent of the atom’s velocity, the number of absorptions that will bring the atom to rest is 23 atom i atom i absorb 28 atom photon ( ) ( ) 7.0 10 kg m/s 82,400 photons | | | | 8.5 10 kg m/s p p N p p − − ×= = = =Δ × (c) An atom in the ground state instantly absorbs a photon then, on average, spends 15 ns in the excited state before it drops back to the ground state where it can absorb another photon. That is, the average time between photon absorptions is 15 ns. The time to absorb 82,400 photons is ∆t = (82,400 photons)(15 × 10–9 s/photon) = 1.24 × 10–3 s = 1.24 ms (d) The atom’s momentum changes by ∆patom = pphoton = –8.5 × 10–28 kg m/s during each ∆t = 15 ns. The force exerted on the atom and the consequent acceleration are 28 20atom 9 20 5 2 25 8.5 10 kg m/s 5.67 10 N 15 10 s 5.67 10 N 4.05 10 m/s 1.4 10 kg p F t F a m − − − − − Δ − ×= = = − ×Δ × − ×⇒ = = = − ×× (e) From kinematics, 22 2 2 i f i 5 2 (500 m/s) 0 2 0.31 m = 31 cm 2 2( 4.05 10 m/s ) v v v a x x a = = + Δ ⇒ Δ = − = − =− ×