atom, (b) the volume- of the nucleus, and' (c) the percentage of the

atom, (b) the volume- of the nucleus, and' (c) the percentage of thevolume of the atom that is occupied by the nucleus.

2. In a Rutherford scattering experiment a target nucleus has a diameter of lA X 10-14 m. The incoming Cl particle has a mass of6.64 X 10-27 kg. What is the kinetic energy of an Cl particle that hasa de Broglie wavelength equal to the diameter of the target nucleus?

Ignore relativistic effects.

3. The nucleus of a hydrogen atom is a single proton, which has aradius of about 1.0 X IO-IY m. The single electron in a hydrogenatom normally orbits the nucleus at a distance of 5.3 X 10-11 m.

What is the ratio of the density of the hydrogen nucleus to the den

sity of the complete hydrogen atom?

4. Review Conceptual Example I and use the information therein as

an aid in working this problem. Suppose you're building a scalemodel of the hydrogen atom, and the nucleus is represented by a ballof radius 3.2 cm (somewhat smaller than a baseball). How many

miles away (I mi = 1.61 X 105 cm) should the electron be placed?

* 5. ssm The nucleus of a copper atom contains 29 protons and has aradius of 4.8 X 10-15 m. How much work (in electron volts) is done

by the electric force as a proton is brought from infinity, where it isat rest, to the "surface" of a copper nucleus?

* 6. There are Z protons in the nucleus of an atom, where Z is theatomic number of the element. An Cl particle carries a charge of+2e. In a scattering experiment, an Cl particle, heading directly toward a nucleus in a metal foil, will come to a halt when all the particle's kinetic energy is converted to electric potential energy. In sucha situation, how close will an Cl particle with a kinetic energy of5.0 X 10-13 J come to a gold nucleus (Z = 79)?

Section 30.2 Line SpectraSection 30.3 The Bohr Model of the Hydrogen Atom

7. ssm www Concept Simulation 30.1 at www.wiley.comlcollege/cutnell reviews the concepts on which the solution to this problemdepends. The electron in a hydrogen atom is in the first excited state,when the electron acquires an additional 2.86 eV of energy. What is

the quantum number n of the state'into which the electron moves?

8. Using the Bohr model, determine the ratio of the energy of the nthorbit of a triply ionized beryllium atom (BeH, Z = 4) to the energyof the nth orbit of a hydrogen atom (H).

9. A singly ionized helium atom (He+) has only one electron in orbit about the nucleus. What is the radius of the ion when it is in the

~ond excited state?(!!J: (a) What is the minimum energy (in electron volts) that is re

quired to remove the electron from the ground state of a singly ionized helium atom (He+, Z = 2)? (b) What is the ionization energyfor He+?

11. ssm Find the energy (in joules) of the photon that is emittedwhen the electron in a hydrogen atom undergoes a transition fromthe n = 7 energy level to produce a line in the Paschen series .

.-@ A hydrogen atom is in the ground state. It absorbs energy andmakes a transition to the n = 3 excited state. The atom returns to the

ground state by emitting two photons. What are their wavelengths?

13. Consider the Bohr energy expression (Equation 30.13) as it ap

plies to singly ionized helium He+ (Z = 2) and doubly ionizedlithium Li2+ (Z = 3). This expression predicts equal electron ener

gies for these two species for certain values of the quantum numbern (the quantum number is different for each species). For quantumnumbers less than or equal to 9, what are the lowest three energies(in electron volts) for which the helium energy level is equal to the

lithium energy level?

PROBLEMS I 971

rilf.)In the hydrogen atom the radius of orbit B is sixteen times greater~ the radius of orbit A. The total energy of the electron in orbit A is- 3AO eV. What is the total energy of the electron in orbit B?

* 15. ssm A wavelength of 410.2 nm is emitted by the hydrogenatoms in a high-voltage discharge tube. What are the initial and final

values of the quantum number n for the energy level transition thatproduces this wavelength?

*@ The energy of the n = 2 Bohr orbit is - 30.6 eV for an unidentified ionized atom in which only one electron moves about the nu

cleus. What is the radius of the n = 5 orbit for this species?

* 17. ssm For atomic hydrogen, the Paschen series of lines occurswhen nf = 3, whereas the Brackett series occurs when nf = 4 in

Equation 30.14. Using this equation, show that the ranges of wavelengths in these two series overlap.

*@ Interactive LearningWare 30.1 at www.wiley.comlcolIege/cutnellreviews the concepts that play roles in this problem. A hydrogenatom emits a photon that has momentum with a magnitude of5A52 X 10-27 kg' m/so This photon is emitted because the electron

in the atom falls from a higher energy level into the n = I leveLWhat is the quantum number of the level from which the electronfalls? Use a value of 6.626 X 10-34 J. s for Planck's constant.

** 19. A diffraction grating is used in the first order to separate thewavelengths in the Balmer series of atomic hydrogen. (Section 27.7discusses diffraction gratings.) The grating and an observationscreen (see Figure 27.32) are separated by a distance of 81.0 cm.You may assume that () is small, so sin () = ()when radian measureis used for (). How many lines per centimeter should the gratinghave so that the longest and the next-to-the-Iongest wavelengths inthe series will be separated by 3.00 cm on the screen?

** 20. A certain species of ionized atoms produces an emission linespectrum according to the Bohr model, but the number of protons Zin the nucleus is unknown. A group of lines in the spectrum forms aseries in which the shortest wavelength is 22.79 nm and the longestwavelength is 41.02 nm. Find the next-to-the-Iongest wavelength inthe series of lines.

Section 30.5 The Quantum Mechanical Pictureof the Hydrogen Atom

"@ The orbital quantum number for the electron in a hydrogen atomIS £ = 5. What is the smallest possible value (algebraically) for thetotal energy of this electron? Give your answer in electron volts.

22. A hydrogen atom is in its second excited state. Determine, ac

cording to quantum mechanics, (a) the total energy (in eV) of theatom, (b) the magnitude of the maximum angular momentum theelectron can have in this state, and (c) the maximum value that the z

~mponent Lz of the angular momentum can have., 23. ssm www The principal quantum number for an electron in an

a om is n = 6, and the magnetic quantum number is me = 2. Whatpossible values for the orbital quantum number £ could this electronhave?

24. The maximum value for the magnetic quantum number in state

A is me = 2, while in state B it is me = I. What is the ratio LA/LB ofthe magnitudes of the orbital angular momenta of an electron inthese two states?

* 25. Interactive Solution 30.25 at www.wiley.comlcollege/cutnell

offers one approach to problems of this type. For an electron in a hydrogen atom, the z component of the angular momentum has a maxi

mum value of L, = 4.22 X 10-34 J . s. Find the three smallest possible values (algebraically) for the total energy (in electron volts) thatthis atom could have.

972 I CHAPTER 30 I THE NATURE OF THE ATOM

ff6J Review Conceptual Example 6 as backg;ound for this problem.~ the hydrogen atom, the Bohr model and quantum mechanicsboth give the same value for the energy of the nth state. However,

they do not give the same value for the orbital angular momentum L.(a) For n = I, determine the values of L [in units of h/(27T)] pre

dicted by the Bohr model and quantum mechanics. (b) Repeat part(a) for n = 3, noting that quantum mechanics permits more than onevalue of £ when the electron is in the n = 3 state.

** 27. ssm www An electron is in the n = 5 state. What is the small

est possible value for the angle between the z component of the orbital angular momentum and the orbital angular momentum?

Section 30.6 The Pauli Exclusion Principle and the PeriodicTable of the Elements

28. Two of the three electrons in a lithium atom have quantum num

bers of n = 1, £ = 0, me = 0, m, = +! and n = 1, £ = 0, me = 0,m, = -!. What quantum numbers can the third electron have if theatom is in (a) its ground state and (b) its first excited state?

29. In the style shown in Table 30.3, write down the ground-stateelectronic configuration for arsenic As (2 = 33). Refer to Figure30.17 for the order in which the sub shells fill.

30. Figure 30.17 was constructed using the Pauli exclusion principleand indicates that the n = 1 shell holds 2 electrons, the n = 2 shellholds 8 electrons, and the n = 3 shell holds 18 electrons. These num

bers can be obtained by adding the numbers given in'the figure forthe subshells contained within a given shell. How many electrons canbe put into the n = 5 shell, which is only partly shown in the figure?

31. Write down the fourteen sets of the four quantum numbers that

correspond to the electrons in a completely filled 4f subshell.

* 32. What is the atom with the smallest atomic number that containsthe same number of electrons in its s subshells as it does in its d sub

shell? Refer to Figure 30.17 for the order in which the subshells fill.

Section 30.7 X-Rays

33. ssm Molybdenum has an atomic number of 2 = 42. Using theBohr model, estimate the wavelength of the Kcx X-ray.

34. Interactive LearningWare 30.2 at www.wiley.comlcollege/cutnellreviews the concepts that are pertinent to this problem. By using theBohr model, decide which element is likely to emit a Kcx X-ray witha wavelength of 4.5 X 10-9 m.

35. Interactive Solution 30.35 at www.wiley.comlcollege/cutnellprovides one model for solving problems such as this one. An X-raytube is being operated at a potential difference of 52.0 kV. What is

the Bremsstrahlung wavelength that corresponds to 35.0% of the kinetic energy with which an electron collides with the metal target inthe tube?

I ADDITIONAL PROBLEMS44. Referring to Figure 30.17 for the order in which the sub shellsfill and following the style used in Table 30.3, determine the groundstate electronic configuration for cadmium Cd (2 = 48).

45. ssm www In the line spectrum of atomic hydrogen there isalso a group of lines known as the Pfund series. These lines are produced when electrons, excited to high energy levels, make transitions to the n = 5 level. Determine (a) the longest wavelength and

36. What is the minimum potential difference that must be appliedto an X-ray tube to knock a K-shell electron completely out of anatom in a copper (2 = 29) target? Use the Bohr model as needed.

* 37. ssm An X-ray tube contains a silver (2 = 47) target. The highvoltage in this tube is increased from zero. Using the Bohr model,find the value of the voltage at which the Kcx X-ray just appears inthe X-ray spectrum.

* 38. Multiple-Concept Example 9 reviews the concepts that areimportant in this problem. An electron, traveling at a speed of

6.00 X 107 m/s, strikes the target of an X-ray tube. Upon impact, theelectron decelerates to one-quarter of its original speed, emitting anX-ray in the process. What is the wavelength of the X-ray photon?

Section 30.8 The Laser

39. r.'ssm wwwAlaserisusedineyesurgerytoweldadeR tached retina back into place. The wavelength of the

laser beam is 514 nm, and the power is 1.5 W. Duringsurgery, the laser beam is turned on for 0.050 s. During this time,how many photons are emitted by the laser?

40. ~ The dye laser used in the treatment of the port-wine stain

8 in Figure 30.30 (see Section 30.9) has a wavelength of~ 585 nm. A carbon dioxide laser produces a wavelengthof 1.06 X 10-5 m. What is the minimum number of photons that thecarbon dioxide laser must produce to deliver at least as much ormore energy to a target as does a single photon from the dye laser?

41. A pulsed laser emits light in a series of short pulses, each having a duration of 25.0 ms. The average power of each pulse is5.00 mW, and the wavelength of the light is 633 nm. Find (a) theenergy of each pulse and (b) the number of photons in each pulse.

42. T'A laser peripheral iridotomy is a procedure for treating. an eye condition known as narrow-angle glaucoma, in

• which pressure buildup in the eye can lead to loss of vi-sion. A neodymium YAG laser (wavelength = 1064 nm) is used in

the procedure to punch a tiny hole in the peripheral iris, therebyrelieving the pressure buildup. In one application the laser delivers4.1 X 10-3 J of energy to the iris in creating the hole. How manyphotons does the laser deliver?

* 43. Fusion is the process by which the sun produces energy. Oneexperimental technique for creating controlled fusion utilizesa solid-state laser that emits a wavelength of 1060 nm and can produce a power of 1.0 X 1014 W for a pulse duration of 1.1 X 10-11 s.In contrast, the helium/neon laser used at the checkout counter

in a bar-code scanner emits a wavelength of 633 nm and produces a power of about 1.0 X 10-3 W. How long (in days)would the helium/neon laser have to operate to produce thesame number of photons that the solid-state laser produces in1.1 X 1O-11 s?

(b) the shortest wavelength in this series. (c) Refer to Figure 24.10and identify the region of the electromagnetic spectrum in whichthese lines are found.

46. The atomic number of lead is 2 = 82. According to the Bohrmodel, what is the energy (in joules) of a Kcx X-ray photon?

47. ssm When an electron makes a transition between energy levelsof an atom, there are no restrictions on the initial and final values of

Chapter 30 Problems 1515

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.

8. REASONING According to the Bohr model, the energy (in joules) of the nth orbit of anatom containing a single electron is

(30.12)En = -(2.18x10-18 J) Z:n

SOLUTION Taking the ratio of the energy E B 3+ of the nth orbit of a beryllium atomn, e

(ZBe3+ = 4) to the energy En, H of the nth orbit of a hydrogen (ZH = I) atom gives

En, B,;' -(2.18xlO-18 J) Z~,;.= 2 Z2

En.H "2 =~= (4)2-(2.18x 10-18 J) ZH Z~ (1)2 = [llin2

9. 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. WithZ= 2 and n = 3, we can use Equation 30.10 to find the radius of the ion.

SOLUTION The radius ofthe second excited state is

(30.10)

10. REASONING

a. The total energy En for a single electron in the nth state is given by

Z2En = -(13.6 eV)-2n

(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 whichn = 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 theelectron from the n = 1 orbit to the highest possible excited state (n = 00) .

11,

1516 THE NATURE OF THE ATOM

SOLUTIONa. 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 = E, _ El = -(13.6 eV)(2)' [-(13.6 eV) (2)' ]22 - 12 = 140.8 eVI

b. The ionization energy is the difference between the ground-state energy (n = 1) and the

energy in the highest possible excited state (n = 00) . Thus,

Ionization energy ~ E__ El = -(13.6 e~) (2)' [-(13.6 eV)(2)']_(00) 12 -154.4 eVI

11. I SSMI REASONING According to Equation 30.14, the wavelength A emitted by the

hydrogen atom when it makes a transition from the level with nj to the level with nf is given

by

1 2tr2mk2e4 ( 1 1 J .-=---- (Z2) --- wIth nj,nf =1,2,3, ... and nj >nfA h3e nl ni2

where 2tr2mk2e4 /(h3e) = 1.097 X107 m-I and Z= 1 for hydrogen. Once the wavelength

for the particular transition in question is determined, Equation 29.2 (E = hf = he / A) 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, nj = 7

and nf= 3, we find that

~ = (1.097 X 107 m-I)(I') (3~- 71,) or A=1.005xlO-6 m

The photon energy is

he (6.63XlO-34 J.s)(3.00XI08 rnls) I IE=-=-----------= 1.98xl0-19 JA 1.005xl0-6 m '------


12. REASONING Since the atom emits two photons as it returns to the ground state, one isemitted 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 duringthese transitions are given by Equation 30.14 with the appropriate values for the initial andfinal numbers, ni and ne

SOLUTION The wavelengths of the photons are

1 ( 7 _1)()2(1 1) 6 -1n =3 ton =2 ,.1,= 1.097x10 m 1 "22-"32 =1.524xlO m

,.1,= !6.56XlO-7ml

(30.14)

n = 2 to n = 1~ =(1.097X107 m-1)(1)2(*_ 212)=8.228X106 m-I

A=I1.22X10-7 m!

(30.14)

13. REASONING The Bohr expression as it applies to anyone-electron species of atomic

number Z, is given by Equation 30.13: En = -(13.6 eV)(Z2 / n2). 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, thequantum number n is different for the equal energy states for each species.

SOLUTION For, equal energies, we can write

Simplifying, this becomes

orZ2 Z2

-(13.6 eV) ~e = -(13.6 eV) iinHe nLi

Thus,

or4 92 2

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

--- --~- --- -- -- ----

11 = 3, 4, S, ...

EA _ -3.40 eV =1-0.213 eVIEB=16.0- 16.0

SOLUTION We know that the radius of orbit B is sixteen times greater than the radius oforbit A. Since the total energy is inversely proportional to the radius, it follows that the totalenergy of the electron in orbit B is one-sixteenth of the total energy in orbit A:

Thus, the energy is inversely proportional to the radius, and it is on this fact that we base oursolution.

En = -13.6 = (-13.6)(S.29X10-11)rnl(S.29X10-11) rn

Solving the radius equation for n2 and substituting the result into the energy equation gives

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.


14. 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 byEquation 30.10:

IS. I SSM I REASONING A wavelength of 410.2 nm is emitted by the hydrogen atoms in a

high-voltage discharge tube. This transition lies in the visible region (380-7S0 nm) of thehydrogen spectrum. Thus, we can conclude that the transition is in the Balmer series and,

therefore, that nf = 2. The value of ni can be found using Equation 30.14, according to which

the Ba1mer series transitions are given by

-I' j.

nHenLiEnergy,

I. 23-13.6 eVI 4

6-3.40eV6

9-1.S1eV


This expression may be solved for ni for the energy transition that produces the givenwavelength.

SOLUTION Solving for ni' we find that

Therefore, the initial and final states are identified by I n i = 6 and nf = 2 I·

16. REASONING The energy levels and radii of a hydrogenic species of atomic number Z are

given by Equations30.13 and 30.10, respectively: En=-(13.6eV)(Z2/n2) and

rn = (5.29 x 10-11 m)( n2 /Z) . We can use Equation 30.13 to find the value of Z for the

unidentified ionized atom and then calculate the radius of the n = 5 orbit usingEquation 30.10.

SOLUTION Solving Equation 30.13 for atomic number Z of the unknown species, we have

Z=Enn2 = {(-30.6 eV)(2)2 = 3-13.6 eV ~ -13.6 eV

Therefore, the radius of the n = 5 orbit is

17. I SSMI REASONING AND SOLUTION For the Paschen series, nf = 3. The range of

wavelengths occurs for values of nj = 4 to ni= 00. Using Equation 30.14, we find that the

shortest wavelength occurs for nj = 00 and is given by

/L=8.204x10-7 m~---~v~---~Shortest wavelength in

Paschen series

A=1.459x10-6 m

vLongest wavelength in

Brackett series

~---~v·~---~Longest wavelength in

Paschen series

A = 1.875 X 10-6 m

A = 4.051 X 10-6 m

or

or~ =(l.097Xl07 m-I)( 41,- 5; J

The longest wavelength in the Paschen series occurs for ni= 4 and is given by

For the Brackett series, nf= 4. The range of wavelengths occurs for values of ni= 5 to

ni= 00, Using Equation 30.14, we find that the shortest wavelength occurs for ni= 00 and is

given by

v

Shortest wavelength inBrackett series

The longest wavelength in the Brackett series occurs for ni= 5 and is given by

Since the longest wavelength in the Paschen series falls within the Brackett series, thewavelengths of the two series overlap.


18. REASONING· To obtain the quantum number of the higher level from which the electronfalls, we will use Equation 30.14 for the reciprocal of the wavelength A of the photon:

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 avalue for the magnitude p of the photon's momentum, and the momentum and thewavelength are related according to Equation 29.6:

j~.'~',[ioi where his Planck's constant.

hp=-

A

Using Equation 30.14 to substitute for ~, we obtainA


h (1 1 Jp=-=hR --- (1)

A nf nf

1 1 P _ 1 5.452x10-27 kg·m/s -0.2499 or n =!lInj2 = ni - hR -12- (6.626X10-34 J.s)(1.097X107 m-l) 1

_1 __ 1 _Lnf - nf hR

or_1 __ 1 _ P

nf nf - hR

Thus, we find

SOLUTION Rearranging Equation (1) gives

19. REASONING AND SOLUTION We need to use Equation 30.14 to find the spacingbetween the longest and next-to-the longest wavelengths in the Ba1mer series. In order to dothis, we need to first find these two wavelengths.

Longest:

1 ( 1 1 J ( 7 -1)( 1 1 )

-=R --- = 1.097x10 m ---/4 nf ni2 22 32

or /4 = 656.3 nm

Next-to-Iongest:

Equation 27.7 states that sin B= mAid. Using the small angle approximation, we have

_I =R( ~_~ J~(1.097XI07 m-I)(_l __1 )~ nf nj 22 42

or ~ = 486.2 nm

sin B"'" tan B"'" B = YL

so that y/L = mAid. The position of the fringe due to the longest wavelength is Yl = mAlL/d.

For the next-to-Iongest, Y2 = m~L/d. The difference in the positions on the screen is,

therefore, Yl - Y2 = (mL/d)(Al -~) which gives

d= mL(/4 -~)Yj- Y2


21. REASONING The orbital quantum number .e has values of 0, 1, 2, ..... , (n -1), according

to the discussion in Section 30.5. Since.e = 5, we can conclude, therefore, that n ;? 6. Thisknowledge about the principal quantum number n can be used with Equation 30.13,

E = -(13.6 eV)Z2/n2, to determine the smallest value for the total energy E .n n

SOLUTION The smallest value of E (i.e., the most negative) occurs when n = 6. Thus,n

using Z = 1 for hydrogen, we find

Z2 12

En = -(13.6 eV)~ = -(13.6 eV)"62 = 1-0.378 eV I

22. REASONINGa. The ground state is the n = 1 state, the first excited state is the n = 2 state, and the secondexcited state is the n = 3 state. The total energy (in eV) of a hydrogen atom in the n = 3state is given by Equation 30.13.

b. According to quantum mechanics, the magnitude L of the angular momentum is given by

Equation 30.15 as L=~.e(.e+l)(h/21r), where .e is the orbital quantum number. Thediscussion in Section 30.5 indicates that the maximum value that .e can have is one less

than the principal quantum number, so that .e max = n - 1.

c. Equation 30.16 gives the z-component Lz of the angular momentum as Lz = m£ (h / 21r) ,

where me is the magnetic quantum number. According to the discussion in Section 30.5,

the maximum value that m£ can attain is when it is equal to the orbital quantum number,

which is .e max'

SOLUTIONa. The total energy of the hydrogen atom is given by Equation 30.13. Using n = 3, we have

(13.6 eV)(1)2 = 1-1.51 eVIE3 = 32

b. The maximum orbital quantum number is .e max = n - 1 = 3 - 1 = 2. The maximum

angular momentum Lmax has a magnitude given by Equation 30.15:


c. The maximum value for the z-component Lz of the angular momentum is

(withmf =J!max=2)

L =m ~=(2)6.63XlO-34 ],s=12.11X10-34 J.slz £ 2tr 2tr .------

23. ISSMII wwwl REASONING The values that I can have depend on the value of n, and

only the following integers are allowed: I = 0, 1, 2, ... (n -1). The values that ml can

have depend on the value of I , with only the following positive and negative integers being

permitted: mf = -J!, ... -2, -1,0, +1, +2, .. .+e.

SOLUTION Thus, when n = 6, the possible values of I are 0, 1,2,3, 4,5. Now when

mf = 2 , the possible values of I are 2, 3, 4, 5, ... These two series of integers overlap forthe integers 2, 3, 4, and 5. Therefore, the possible values for the orbital quantum number I

that this electron could have are 11 = 2, 3, 4, 51.

24. REASONING The maximum value for the magnetic quantum number is ml = I ; thus, in

state A, I = 2, while in state B, I = 1. According to the quantum mechanical theory ofangular momentum, the magnitude of the orbital angular momentum for a state of given I is

L = -JJ!(J! + 1) (h / 2tr) (Equation 30.15). This expression can be used to form the ratio

LA / ~ of the magnitudes of the orbital angular momenta for the two states.

SOLUTION Using Equation 30.15, we find that

h

.j2(2+ 1) 2& ~ @" ~v'3~11.7321LA = h V2LB -J1(1 + 1) 2tr

25. 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:

Thus, we need to determine values for the principal quantum number n if we are to calculatethe three smallest possible values for E. Since the maximum value of the orbital quantum

number J! is n - 1, we can obtain a minimum value for n as nmin = J! + 1. But how to obtain

1E/1 = -(13.6 eV)2n

(30.13)


£? 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

where mf is the magnetic quantum number and his Planck's constant. For a given value of £

the allowed values for mf are as follows: -£, .. " -2, -1, 0, +1, +2, ... , +£. Thus, the

maximum value ofmf is £, and we can use Equation 30.16 to calculate the maximum value of

mf from the maximum value given for Lz·

(30.16)h

Lz = mf 21C

SOLUTION Solving Equation 30.16 for mf gives

rn, ~ 21fL, ~ 21f(4.22X10-34 J,s)_4h 6.63xl0-34 J. S -

As explained in the REASONING, this maximum value for mf indicates that £ = 4.

Therefore, a minimum value for n is

n . =£+1=4+1=5 or n>5mm -

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

[n = 5] E5 = -(13.6 eV)~ = 1-0.544 eVj5

[n = 6] E6 = -(13.6 eV)~ = 1-0.378 eV[6

[n=7] E7 = -(13.6 eV)~ = 1-0.278 eV[7

26. REASONING AND SOLUTIONa. For the angular momentum, Bohr's value is given by Equation 30.8, with n = 1,

According to quantum theory, the angular momentum is given by Equation 30.15. Forn=I,£=O

"1i,ji;1

!"

e = cos-1 (J4i5) = I 26.6° Ior

cos e =.e re~.e(.e+l) ~£+1

[n = 3, .e = 2]

The smallest value for e corresponds to the largest value of cos e. For a given value of I ,

the largest value for cos e corresponds to the largest value for rn,. But the largest possible

value for rnl is rnl = I . Therefore, we find that

SOLUTION The smallest value for e corresponds to the largest value for I. When theelectron is in the n = 5 state, the largest allowed value of I is I = 4 ; therefore, we see that

[n=3,.e=1]

[n = 3, .e = 0]

while quantum mechanics gives

b. For n = 3; Bohr theory gives

27. I SSMII wwwl REASONING Let e denote the angle between

the angular momentum L and its z-component Lz· We can see

from the figure at the right that Lz = L cos e . Using

Equation 30.16 for Lz and Equation 30.15 for L, we have


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atom, (b) the volume- of the nucleus, and' (c) the percentage of the