CHAPTER 37

Atoms 1* As n increases, does the spacing of adjacent energy levels increase or decrease? The spacing decreases. 2 The energy of the ground state of doubly ionized lithium (Z = 3) is ______ ,where E0 = 13.6 eV. (a) 9E0, (b) 3E0, (c) E0/3, (d)E0/9. (a) 3 Bohrs quantum condition on electron orbits requires

(a) that the angular momentum of the electron about the hydrogen nucleus equal nh_. (b) that no more than one electron occupy a given stationary state. (c) that the electrons spiral into the nucleus while radiating electromagnetic waves. (d) that the energies of an electron in a hydrogen atom be equal to nE0, where E 0 is a constant energy and n is an integer. (e) none of the above. (a) 4 If an electron moves to a larger orbit, does its total energy increase or decrease? Does its kinetic energy increase or decrease? Its total energy increases; its kinetic energy decreases. 5* The kinetic energy of the electron in the ground state of hydrogen is 13.6 eV = E0. The kinetic energy of the electron in the state n = 2 is _____. (a) 4E0, (b) 2E0, (c) E0/2, (d) E0/4. (d) 6 The radius of the n = 1 orbit in the hydrogen atom is a0 = 0.053 nm. What is the radius of the n = 5 orbit? (a) 5a0, (b) 25a0, (c) a0, (d) (1/5)a0, (e) (1/25)a0. (b) 7 Use the known values of the constants in Equation 37-11 to show that a0 is approximately 0.0529 nm.

Evaluate Equ. 37-12, a0 = h_ 2/mke2

a0 = m )10(1.6 108.99 109.11

)10(1.05219931

234

--

-

= 5.261011 m = 0.0526 nm

8 The longest wavelength of the Lyman series was calculated in Example 37-2. Find the wavelengths for the transitions (a) n1 = 3 to n2 = 1 and (b) n1 = 4 to n2 = 1.

Chapter 37 Atoms

? = (1240/?E) nm, where ?E = Ei Ef and the energies are in eV (see Equs. 17-5 and 17-21); here Ef = -13.6 eV. (a) E3 = 13.6/9 eV = 1.51 eV (b) E4 = 13.6/16 eV = 0.85 eV

?E = 12.09 eV; ? = 102.6 nm ?E = 12.75 eV; ? = 97.25 nm

9* Find the photon energy for the three longest wavelengths in the Balmer series and calculate the wavelengths. For the Balmer series, Ef = E(n = 2) = 3.40 eV. Use Equs. 17-5 and 17-21, i.e., ? = (1240 eV.nm)/(?E eV). 1. ?E = E3 E2; E3 = 13.6/9 eV = 1.51 eV 2. ?E = E4 E2; E4 = 13.6/16 eV = 0.85 eV 3. ?E = E5 E2; E5 = 13.6/25 eV = 0.544 eV

?E = 1.89 eV; ?3,2 = 656.1 nm ?E = 2.55 eV; ?4,2 = 486.3 nm ?E = 2.856 eV; ?5,2 = 434.2 nm

10 (a) Find the photon energy and wavelength for the series limit (shortest wavelength) in the Paschen series (n2 = 3). (b) Calculate the wavelengths for the three longest wavelengths in this series and indicate their positions on a horizontal linear scale. (a) For ?min, n = and Ei = 0

h_f = (13.6/9) eV = 1.51 eV; ?min = 1240/1.51 nm = 820.6 nm

(b) For the three longest ?, ni = 4, 5, and 6 ?E4,3 = 0.66 eV; ?4,3 = 1876 nm ?E5,3 = 0.967 eV; ?5,3 = 1282 nm ?E6,3 = 1.133 eV; ?6,3 = 1094 nm

The positions of these lines on a horizontal linear scale are shown below with the wavelengths indicated.

11 Repeat Problem 10 for the Brackett series (n2 = 4). (a) For ?min, n = and Ei = 0 (b) For the three longest ?, ni = 5, 6, and 7

h_f = (13.6/16) eV = 0.85 eV; ?min = (1240/0.85) nm = 1459 nm ?E5,4 = 0.306 eV; ?4,3 = 4052 nm ?E6,4 = 0.472 eV; ?5,3 = 2627 nm ?E7,4 = 0.572 eV; ?6,3 = 2166 nm

The positions of these lines on a horizontal linear scale are shown below with the wavelengths indicated.

Chapter 37 Atoms

12 A hydrogen atom is in its tenth excited state according to the Bohr model (n = 11). (a) What is the radius of

the Bohr orbit? (b) What is the angular momentum of the electron? (c) What is the electrons kinetic energy? (d) What is the electrons potential energy? (e) What is the electrons total energy?

(a) Use Equs. 37-11 and 37-12

(b) L = nh_ (c) K = E0/n2 (d) U = 2K (e) E = E0/n2

r11 = 121a0 = 6.40 nm L = 111.051034 J.s = 11.61034 J.s K = 13.6/121 eV = 0.112 eV U = 0.225 eV E = 0.112 eV

13* The binding energy of an electron is the minimum energy required to remove the electron from its ground state to a large distance from the nucleus. (a) What is the binding energy for the hydrogen atom? (b) What is the binding energy for He+? (c) What is the binding energy for Li2+? [Singly ionized helium (He+) and doubly ionized lithium (Li2+) are hydrogen-like in that the system consists of a positively charged nucleus and a single bound electron.] (a) BE = E E(n = 1) = E0 (b) Note that En Z2; Z for He+ = 2 (c) For Li2+, Z = 3

BE = 13.6 eV BE = 413.6 eV = 54.4 eV BE = 913.6 eV = 122.4 eV

14 The electron of a hydrogen atom is in the n = 2 state. The electron makes a transition to the ground state.

(a) What is the energy of the photon according to the Bohr model? (b) The linear momentum of the emitted photon is related to its energy by p = E/c. If we assume conservation of linear momentum, what is the recoil velocity of the atom? (c) Find the recoil kinetic energy of the atom in electron volts. By what percentage must the energy of the photon calculated in part (a) be corrected to account for this recoil energy?

(a) hf = E2 E1 = 3E0/4 (b) hf/c = prec; vrec = prec/m = hf/mc (c) Erec = prec2/2m = (hf)2/2mc2

hf = 10.2 eV = 1.6321018 J vrec = 1.6321018/(1.671027 3108 ) m/s = 3.26 m/s Erec = (10.2)2/(2938.3 106 ) eV = 5.54108 eV Percent correction = 5.54106/10.2 = 5.44107%

15 Show that the speed of an electron in the nth Bohr orbit of hydrogen is given by vn = e2/2e0hn.

Ln = nh_ = mvnrn = n2h

_2mvn/mke2; solving for vnwe obtain vn = ke2/nh_. But k =1/4pe0, so vn = e2/2e0hn.

16 In this problem you will estimate the radius and the energy of the lowest stationary state of the hydrogen atom using the uncertainty principle. The total energy of the electron of momentum p and mass m a distance r from the proton in the hydrogen atom is given by E = p2/2m ke 2/r, where k is the Coulomb constant. Assume that the

minimum value of p2 is p2 (?p)2 = h2/r2, where ?p is the uncertainty in p and we have taken ?r ~ r for the order of magnitude of the uncertainty in position; the energy is E = h2/2mr2 ke2/r. Find the radius rm for which this energy is a minimum, and calculate the minimum value of E in electron volts.

With E = h2/2mr2 ke2/r, dE/dr = ke2/r2 - h2/mr3. Set dE/dr = 0 and solve for r = rm; rm = h2/ke2m. This is just the Bohr radius a0. Consequently, the energy is the ground state energy of the hydrogen atom, namely ,13.6 eV. 17* In a reference frame with the origin at the center of mass of an electron and the nucleus of an atom, the

Chapter 37 Atoms

electron and nucleus have equal and opposite momenta of magnitude p. (a) Show that the total kinetic energy of the electron and nucleus can be written K = p2/2mr, where mr = meM/(M + me) is called the reduced mass, me is the mass of the electron, and M is the mass of the nucleus. It can be shown that the motion of the nucleus can be accounted for by replacing the mass of the electron by the reduced mass. In general, the reduced mass for a two-body problem with masses m1 and m2 is given by

(Eq. 37-47)

(b) Use Equation 37-14 with m replaced by mr to calculate the Rydberg constant for hydrogen (M = mp) and for a very massive nucleus (M = ). (c) Find the percentage correction for the ground-state energy of the hydrogen atom due to the motion of the proton. (a) Both the electron and nucleus have the same momentum, so K = Ke + Kn = p2/2me + p2/2M = (p2/2)[(me + M/meM)]. Defining mr = meM/(me + M), K = p2/2mr.

(b) For H, RH = 1.096776107 m1 = Cmr, where C = k2e4/4pch_3, and mr = mpme/(me + mp) = me/(1 + me/mp) =

me/(1 + 5.447104). If instead of mp we have an infinite nuclear mass, then mr = me, and the Rydberg constant R is then greater than RH by the factor (1 + 5.447104), or R = 1.097373107 m1.

(c) The energy correction is just 5.447104 or 0.0545%. That is, the correct energy is slightly less than that calculated neglecting the motion of the nucleus. 18 For the principal quantum number n = 4, how many different values can the orbital quantum number l have? (a) 4, (b) 3, (c) 7, (d) 16, (e) 6. (b) 19 For the principal quantum number n = 4, how many different values can the magnetic quantum number m

have? (a) 4, (b) 3, (c) 7, (d) 16, (e) 6. (c) 20 For l = 1, find (a) the magnitude of the angular momentum L and (b) the possible values of m. (c) Draw to scale a vector diagram showing the possible orientations of L with the z axis.

(a) Use Equ. 37-24 L =

2 h_ = 1.491034 J.s

(b) m = l ...0...+ l m = 1, 0, +1

(c) The vector diagram is shown on the right. Note that since Lz =

mh_ and L = 2 h_, the vectors for m = 1 and m = 1 must make angles of 45 with the z axis.

m+m

mm = m21

21r

Chapter 37 Atoms

21* Work Problem 20 for l = 3.

(a) Use Equ. 37-24 L = 2 3 h_ (b) m = l ... 0 ...+ l m = 3, 2, 1, 0, +1, +2, +3

(c) The vector diagram is shown on the right. Note that since Lz = mh

_

and 32 = L

h

_, the angles between the vectors and the z axis are given by cos ?m = 3m/2 . Thus, ? 3 = 30, ? 2 = 54.7, and ? 1 = 73.2. As shown, the spacing between the allowed

values of Lz is constant and equal to h_.

22 A compact disk has a moment of inertia of about 2.3 105 kgm2. (a) Find its angular momentum L when it is rotating at 500 rev/min. (b) Find the approximate value of the quantum number l for this angular momentum. (a) L = I?

(b) l @ L/h_

? = (2p500/60) rad/s; L = 1.20103 J.s l @ 1.20103/1.0551034 @ 1031

23 If n = 3, (a) what are the possible values of l ? (b) For each value of l in (a), list the possible values of m. (c) Using the fact that there are two quantum states for each value of l and m because of electron spin, find the total number of electron states with n = 3. (a), (b) See Equ. 37-23 l = 0: m = 0; l =1: m = 1, 0, 1; l = 2: m = 2, 1,

0, 1, 2 (c) We will first derive the number of electron states for an arbitrary value of n and then substitute the specific

value of n. The number of m states for a given n is given by (1). + 2 = 1) + (2 = N1 n

0 =

1 n

0 =

1 n

0 = m

---

lll

ll

The sum of all integers from 0 to p is p(p + 1)/2, so the first term is 2[(n 1)n/2] = n2 n. The second term is just equal to n. So Nm = n2. Since N, the number of electron states is twice the number of m states, the number of electron states is N = 2n2.

For n = 3, N = 18.

Chapter 37 Atoms

24 Find the total number of electron states with (a) n = 2 and (b) n = 4. (See Problem 23.) (a), (b) Use the result of Problem 23(c) (a) N = 8; (b) N = 32 25* Find the minimum value of the angle ? between L and the z axis for (a) l = 1, (b) l = 4, and (c) l = 50. We consider the general case. The minimum angle between the z axis and L is the angle between the L vector

for m =l and the z axis. In this case, Lz = mh_ =l h_ and L = 1) + ( ll h_. Consequently, cos ? = 1) + /( ll .

(a) For l = 1, cos ? = 1/ 2 (b) For l = 4, cos ? = 2/ 5 (c) For l = 50, cos ? = 50/51

? = 45 ? = 26.6 ? = 8.05

26 What are the possible values of n and m if (a) l = 3, (b) l = 4, and (c) l = 0? (a), (b), (c) See Equ. 37-23 (a) n 4, m = 3, 2, 1, 0, 1, 2, 3

(b) n 5, m = 4, 3, 2, 1, 0, 1, 2, 3, 4 (c) n 1, m = 0

27 What are the possible values of n and l if (a) m = 0, (b) m = -1, and (c) m = 2? (a), (b), (c) See Equ. 37-23 (a) n 1, l 0; (b) n 2, l 1; (c) n 3, l 2 28 For the ground state of the hydrogen atom, find the values of (a) ?, (b) ? 2, and (c) the radial probability density P(r) at r = a0. Give your answers in terms of a0. (a) Use Equ. 37-33 with Z = 1 (b) Take the square of ?(a0) (c) Use Equ. 37-34 and result of (b)

?(a0) = 1/(ea0 a 0p )

?2(a0) = 1/(e2pa03) P(a0) = 4/a0e2

29* (a) If spin is not included, how many different wave functions are there corresponding to the first excited energy level n = 2 for hydrogen? (b) List these functions by giving the quantum numbers for each state. (a), (b) For n = 2, l = 0 or 1; for l = 1, there are three values of m, namely m = 1, 0, 1

There are four functions; they are listed by (n, l , m): (2,0,0), (2,1,1), (2,1,0), (2,1,1)

30 For the ground state of the hydrogen atom, find the probability of finding the electron in the range ?r = 0.03a0 at (a) r = a0 and (b) r = 2a0. (a) 1. Write P(a0), see Problem 28 2. For ?r

Chapter 37 Atoms

m = 0 in hydrogen. Give your answers in terms of a0. (a) 1. Use Equ. 37-36 and C2,0,0 as given; Z = 1 2. Set r = a0 and evaluate (b) Square the result of (a) (c) P(r) = 4pr2?2(r), Equ. 37-34

e a

r2

a

1

4 4

1 = ar/2

00

3/2

2,0,00-

-

py .

?2,0,0(a0) = 0.0605/a03/2 (?2,0,0(a0))2 = 0.00366/a03 P(a0) = 0.046/a0

32 Show that the radial probability density for the n = 2, l = 1, m = 0 state of a one-electron atom can be written as P(r) = e r A aZr/42 0-qcos , where A is a constant. The wave function for the state (2,1,0), ?2,1,0, is given by Equ. 37-37, and the probability density by Equ. 37-34. Thus, e ? r = A e ? r )a (Z/C r p P(r) = aZr/aZr/,, 00 coscos4 2422

20

2012

2 -- , where a/Z C p A = ,,20

220124 .

33* Calculate the probability of finding the electron in the range ?r = 0.02a0 at (a) r = a0 and (b) r = 2a0 for the state n = 2, l = 0, m = 0 in hydrogen. (See Problem 31 for the value of C2,0,0.) In this instance, P(r) dr extends over a sufficiently narrow interval ?r ==