bohr and rydberg - atomic spectra problems

8/11/2019 Bohr and Rydberg - Atomic Spectra Problems

1/13

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

volume of the atom that is occupied by the nucleus.

2. In a Rutherford scattering experiment a target nucleus has a di

ameter of l X

10-14

m. The incoming Cl particle has a mass of

6.64 X 10-27 kg. What is the kinetic energy of an Cl particle that has

a 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 a

radius of about 1.0 X

IO-IY

m. The single electron in a hydrogen

atom 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 r e building a scale

model of the hydrogen atom, and the nucleus is represented by a ball

of 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 a

radius 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 is

at rest, to the surface of a copper nucleus?

*

6. There are Z protons in the nucleus of an atom, where Z is the

atomic number of the element. An Cl particle carries a charge of

+2e. In a scattering experiment, an Cl particle, heading directly to

ward a nucleus in a metal foil, will come to a halt when all the parti

cle s kinetic energy is converted to electr ic potential energy. In such

a situation, how close will an Cl particle with a kinetic energy of

5.0

X

10-13 J come to a gold nucleus (Z = 79)?

Section 30.2 Line Spectra

Section 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 problem

depends. 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 nth

orbit of a triply ionized beryllium atom

(BeH,

Z

=

4) to the energy

of the nth orbit of a hydrogen atom (H).

9. A singly ionized helium atom (He+) has only one electron in or

bit 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 ion

ized helium atom (He+, Z

=

2)? (b) What is the ionization energy

for He+?

11. ssm Find the energy (in joules) of the photon that is emitted

when the electron in a hydrogen atom undergoes a transition from

the n

=

7 energy level to produce a line in the Paschen series .

.-@ A hydrogen atom is in the ground state. It absorbs energy and

makes 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 ionized

l ithium Li2+ (Z

=

3). This expression predicts equal electron ener

gies for these two species for certain values of the quantum number

n (the quantum number is different for each species). For quantum

numbers 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 9

rilf.)In the hydrogen atom the radius of orbit B is sixteen times great

~ the radius of orbit A. The total energy of the electron in orbit A

- 3AO

eV.What is the total energy of the electron in orbi t B?

*

15.

ssm A wavelength of 410.2 nm is emitted by the hydroge

atoms in a high-vol tage discharge tube. What are the initial and fin

values of the quantum number n for the energy level transition th

produces this wavelength?

*@ The energy of the n = 2 Bohr orbit is - 30.6 eV for an unident

fied ionized atom in which only one electron moves about the n

cleus. What is the radius of the n

=

5 orbit for this species?

* 17. ssm For atomic hydrogen, the Paschen series of lines occu

when nf = 3, whereas the Brackett series occurs when nf = 4

Equation 30.14. Using this equation, show that the ranges of wav

lengths in these two series overlap.

*@ Interactive LearningWare 30.1 at www.wiley.comlcolIege/cutnell

reviews the concepts that play roles in this problem. A hydroge

atom emits a photon that has momentum with a magnitude

5A52

X 10-27 kg m/so This photon is emitted because the electro

in the atom falls from a higher energy level into the n

=

I lev

What is the quantum number of the level from which the electro

falls? 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 t

wavelengths in the Balmer series of atomic hydrogen. (Section 27

discusses diffraction gratings.) The grating and an observatio

screen (see Figure 27.32) are separated by a distance of 81.0 c

You may assume that () is small, so sin ()= when radian measu

is used for (). How many lines per centimeter should the gratin

have so that the longest and the next-to-the-Iongest wavelengths

the series will be separated by 3.00 cm on the screen?

** 20. A certain species of ionized atoms produces an emission li

spectrum according to the Bohr model, but the number of protons

in the nucleus is unknown. A group of lines in the spectrum forms

series in which the shortest wavelength is 22.79 nm and the longe

wavelength is 41.02 nm. Find the next-to-the-Iongest wavelength

the series of lines.

Section 30.5 The Quantum Mechanical Picture

of the Hydrogen Atom

@

The orbital quantum number for the electron in a hydrogen ato

IS

=

5. What is the smallest possible value (algebraically) for t

total energy of this electron? Give your answer in electron volts.

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

cording to quantum mechanics, (a) the total energy (in eV) of t

atom, (b) the magnitude of the maximum angular momentum

electron can have in this state, and (c) the maximum value that the

~mponent Lz of the angular momentum can have.

, 23. ssm www The principal quantum number for an electron in

a om is n

=

6, and the magnetic quantum number is me

=

2. Wh

possible values for the orbital quantum number could this electro

have?

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

A is me

=

2, while in state B it is me

=

I. What is the ratio LA/LB

the magnitudes of the orbital angular momenta of an electron

these 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 h

drogen atom, the z component of the angular momentum has a ma

mum

value of

L =

4.22 X

10-34 J .

s. Find the three smallest pos

ble values (algebraically) for the total energy (in electron volts) th

this atom could have.


2/13

972

I CHAPTER 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 mechanics

both 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 one

value 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 or

bital angular momentum and the orbital angular momentum?

Section 30.6 The Pauli Exclusion Principle and the Periodic

Table 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 the

atom 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-state

electronic configuration for arsenic As (2 = 33). Refer to Figure

30.17 for the order in which the subshells f ill.

30. Figure 30.17 was constructed using the Pauli exclusion principle

and indicates that the n = 1 shell holds 2 electrons, the n = 2 shell

holds 8 electrons, and the

n

= 3 shell holds 18 electrons. These num

bers can be obtained by adding the numbers given in'the figure for

the subshells contained within a given shell. How many electrons can

be 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 contains

the 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 the

Bohr model, estimate the wavelength of the Kcx X-ray.

34. Interactive LearningWare 30.2 at www.wiley.comlcollege/cutnell

reviews the concepts that are pertinent to this problem. By using the

Bohr model, decide which element is likely to emit a Kcx X-ray with

a wavelength of 4.5 X 10-9 m.

35. Interactive Solution 30.35 at www.wiley.comlcollege/cutnell

provides one model for solving problems such as this one. An X-ray

tube is being operated at a potential difference of 52.0 kV. What is

the Bremsstrahlung wavelength that corresponds to 35.0 of the ki

netic energy with which an electron collides with the metal target in

the tube?

I

ADDITIONAL PROBLEMS

44. Referring to Figure 30.17 for the order in which the sub shells

fill and following the style used in Table 30.3, determine the ground

state electronic configuration for cadmium Cd (2 = 48).

45. ssm www In the line spectrum of atomic hydrogen there is

also a group of lines known as the Pfund series. These lines are pro

duced when electrons, excited to high energy levels, make transi

tions to the n = 5 level. Determine (a) the longest wavelength and

36. What is the minimum potential difference that must be applied

to an X-ray tube to knock a K-shell electron completely out of an

atom 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 high

voltage 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 in

the X-ray spectrum.

*

38. Multiple-Concept Example 9 reviews the concepts that are

important 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, the

electron decelerates to one-quarter of its original speed, emitting an

X-ray in the process. What is the wavelength of the X-ray photon?

Section 30.8 The Laser

39.

r.

sm wwwAlaserisusedineyesurgerytoweldade

tached retina back into place. The wavelength of the

laser beam is 514 nm, and the power is 1.5 W. During

surgery, 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 wavelength

of 1.06

X

10-5 m. What is the minimum number of photons that the

carbon dioxide laser must produce to deliver at least as much or

more 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 hav

ing a duration of 25.0 ms. The average power of each pulse is

5.00 mW, and the wavelength of the light is 633 nm. Find (a) the

energy of each pulse and (b) the number of photons in each pulse.

42. T' 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, thereby

relieving the pressure buildup. In one application the laser delivers

4.1

X

10-3

J

of energy to the iris in creating the hole. How many

photons does the laser deliver?

*

43. Fusion is the process by which the sun produces energy. One

experimental technique for creating controlled fusion utilizes

a solid-state laser that emits a wavelength of 1060 nm and can pro

duce 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 pro

duces a power of about 1.0 X 10-3 W. How long (in days)

would the helium/neon laser have to operate to produce the

same number of photons that the solid-state laser produces in

1.1 X 1O-11 s?

(b) the shortest wavelength in this series. (c) Refer to Figure 24.10

and identify the region of the electromagnetic spectrum in which

these lines are found.

46. The atomic number of lead is 2 = 82. According to the Bohr

model, what is the energy (in joules) of a Kcx X-ray photon?

47. ssm When an electron makes a transition between energy levels

of an atom, there are no restrictions on the initial and final values of


3/13

Chapter 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. R SONING According to the Bohr model, the energy (in joules) of the nth orbit of an

atom 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 atom

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

=

[lli

n2

9

R SONING 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 ofthe second excited state is

(30.10)

10. R SONING

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

Z2

En =

-(13.6 eV)-2

n

(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 = 00) .


4/13

11,

1516 TH N TUR OF TH TOM

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

E, El

- 13.6

eV) 2)

[- 13.6

eV) 2) ]

2 -

12

= 140.8 eV

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

eV

11. SS

R SONING

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 >nf

A h3e

nl

ni2

where

2tr2mk2e4 / h3e

= 1.097 X

107

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

=-=-----------= 1.98xl0-19

J

A 1.005xl0-6

m ------


5/13


2 R SONING

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 ne

SOLUTION

The wavelengths of the photons are

1 ( 7 _1)()2(1 1)

6

-1

=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)

3 R SONING

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, the

quantum number n is different for the equal energy states for each species.

SOLUTION

For, equal energies, we can write

Simplifying, this becomes

or

Z2 Z2

-(13.6 eV)

~e

= -(13.6 eV)

ii

nHe

nLi

Thus,

or

4 9

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


6/13

---

--~- --- -- --

----

11 =

3, 4, S, ...

EA -3.40 eV =1-0.213 eVI

EB=16.0- 16.0

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:

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

solution.

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.

1518 TH N TUR OF TH TOM

14. R SONING

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:

IS. I

I R SONING

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 the

hydrogen 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.

nHe

Li

Energy

I

2

13.6 eV

4

3.40eV1.S1eV


7/13

Chapter Problems

1519

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

wavelength.

SOLUTION Solving for ni we find that

Therefore, the initial and final states are identified by

n

i

6 and nf

2

16 RE SONING

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 using

Equation 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.

SSMI RE SONING N 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


8/13

A=1.459x10 6 m

v

Longest wavelength in

Brackett series

~---~v~---~

L onges t 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 in

Brackett 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, the

wavelengths of the two series overlap.

1520

TH N TUR OF TH TOM

18. R SONING To obtain the quantum number of the higher level from which the electron

falls, 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 a

value for the magnitude

p

of the photon s momentum, and the momentum and the

wavelength are related according to Equation 29.6:

j~.

~

,[i

o

i

where his Planck s constant.

p

Using Equation 30.14 to substitute for ~, we obtain


9/13


h 1 1 J

p=-=hR 1

A nf nf

1 1

P

_ 1 5.452x10-27 kgm/s -0.2499 or

n

= lI

nj2 = ni - hR -12- 6.626X10-34 J.s 1.097X107 m-l

1

_1 __ 1 _L

nf

nf

hR

or

1 __ 1 _

P

nf nf hR

Thus, we find

SOLUTION

Rearranging Equation 1 gives

19. RE SONING N SOLUTION We need to use Equation 30.14 to find the spacing

between the longest and next-to-the longest wavelengths in the Ba1mer series. In order to do

this, 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 = Y

L

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


10/13

Chapter 3 Problems 1523

21 R SONING 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. This

knowledge 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

SOLUTION The smallest value of

E

(i.e., the most negative) occurs when

n

6. Thus,

using Z 1 for hydrogen, we find

Z2 12

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

I

22. R SONING

a. 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. The total energy (in eV) of a hydrogen atom in the n 3

state 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. The

discussion 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

SOLUTION

a. 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 eVI

E3 = 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:


11/13

1524 THE N TURE OF THE TOM

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.sl

z

2tr 2tr .------

23. I

SSMII

wwwl R SONING 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 for

the integers 2, 3, 4, and 5. Therefore, the possible values for the orbital quantum number I

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

24. R SONING 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 of

angular 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'3~11.7321

A

= h

V2

LB -J1 1

+ 1) 2tr

25. R SONING 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 calculate

the 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

E/1 = -(13.6 eV)2

n

(30.13)


12/13


?

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)

Lz

=

mf

2 C

SOLUTION Solving Equation 30.16 for mf gives

rn

21fL

21f(4.22X10-34 J,s)_4

h 6.63xl0-34 J. S -

As explained in the RE SONING this maximum value for mf indicates that

= 4.

Therefore, a minimum value for n is

n

=+1=4+1=5 or

n>5

mm

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

=

E5

= -(13.6 eV)~ = 1-0.544 eVj

5

[n

=

6]

E6

= -(13.6 eV)~ = 1-0.378 eV[

6

[n ]

E7

= -(13.6 eV)~ = 1-0.278 eV[

26

RE SONING ND SOLUTION

a. 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. For

n=I =O


13/13

e = cos -1

J4i5

=

I

26.6

I

r

cos e = e

re

~.e .e+l

~+1

3 .e = 2]

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

the largest value for cos

e

corresponds to the largest value for rn But the largest p

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

SOLUTION The smallest value for e corresponds to the largest value for I. Wh

electron is in the n = 5 state, the largest allowed value of I is I = 4 ; therefore, we se

[n=3 e=1]

= 3 .e = ]

while quantum mechanics gives

b. For n = 3; Bohr theory gives

27. ISSMII wwwl R SONING 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

1526

THE N TURE OF THE TOM

bohr and rydberg - atomic spectra problems

Documents