chapter 28 cp

Physics 112 Homework 12 (solutions) (2004 Fall)

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Solutions to Homework Questions 12

Chapt28, Problem-3: The “size” of the atom in Rutherford’s model is about 1.0x10–10 m.

(a) Determine the attractive electrical force between an electron and a proton separated by this distance.(b) Determine (in eV) the electrical potential energy of the atom.

Solut ion:

(a) From Coulomb’s law,

F =ke q

1q

2

r2=

8.99! 109 N "m

2C

2( ) 1.60! 10#19

C( )2

1.0! 10#10

m( )2

= 2 .3! 10

"8 N

(b) The electrical potential energy V =

keq1q

2

r=

8.99 !109 N "m

2C

2( ) #1.60! 10#19

C( ) 1.60! 10#19

C( )1.0! 10

#10 m

= !2 .3" 10!18

J1 eV

1.60"10-19

J

#

$ %

&

' ( =

!14 eV

Chapt28, Problem-6: In a Rutherford scattering experiment, an !-particle (charge = +2e) heads

directly toward a gold nucleus (charge = +79e). The !-particle has a kinetic energy of 5.0 MeV when very far

(r " !) from the nucleus. Assuming the gold nucleus to be fixed in space, determine the distance of closest

approach.

(Hint: Use conservation of energy with r

qqkPE e 21= .)

Solut ion:

Assuming a head-on collision, the a-particle comes to rest momentarily at the point of closest approach. Let

this point be the ‘final’ point. From conservation of energy,

KEf +PE f = KEi +PEi , or

0+ke 2 e( ) 79e( )

rf

= KEi +ke 2 e( ) 79e( )

ri

With ri !" , then the right-most term is zero, hence we can rearrange the resulting expression to gives the

distance of closest approach:

rf =158 ke e2

KEi

=158 8.99! 10

9 N "m

2C

2( ) 1.60! 10#19

C( )2

5.0 MeV 1.60! 10-13

J MeV( ) = 4.5! 10

"14 m =

45 fm


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Chapt28, Problem-7: A hydrogen atom is in its first excited state (n = 2). Using the Bohr theory of

the atom, calculate (a) the radius of the orbit, (b) the linear momentum of the electron, (c) the angularmomentum of the electron, (d) the kinetic energy, (e) the potential energy, and (f) the total energy.

Solut ion:

(a) rn = n

2a0 yields

r2 = 4 0.0529 nm( )=

0.212 nm

(b) With the electrical force supplying the centripetal acceleration,

mev

n

2

rn

=k

ee

2

rn

2, giving

vn =k

ee

2

mer

n

and pn =mevn =

mekee2

rn

Thus,

p2 =

me ke e2

r2

=9.11! 10"31 kg( ) 8.99! 109 N #m 2 C2( ) 1.6 !10"19 C( )

2

0.212!10"9 m

= 9.95! 10"25 kg #m s

(c)

Ln = nh

2!

"

# $

%

& ' ( L2 = 2

6.63) 10*34

J+s

2!

"

# $

%

& ' =

2 .11! 10"34

J #s

(d)

KE2 =1

2me v2

2 =p2

2

2 me

=9.95 !10"25 kg#m s( )

2

2 9.11! 10"31 kg( )= 5.43 !10"19 J =

3.40 eV

(e)

PE2 =ke !e( ) e

r2

= !8.99" 10

9 N #m

2C

2( ) 1.60" 10!19

C( )2

0.212"10!9

m( ) = !1.09" 10

!18 J =

! 6.80 eV

(f) E2 = KE2 +PE2 = 3.40 eV ! 6.80 eV =

! 3.40 eV

Chapt28, Problem-12: Four possible transitions for a hydrogen atom are listed below:

I. ni = 2; nf = 5 II. ni = 5; nf = 3III. ni = 7; nf = 4 IV. ni = 4; nf = 7

(a) Which transition will emit the shortest-wavelength photon? (b) For which transition will the atom gainthe most energy? (c) For which transition(s) does the atom lose energy?

Solut ion:

The change in the energy of the electron is

!E = E f "Ei = 13.6 eV1

ni2"

1

nf2

#

$ %

&

' (

Transition I:

!E = 13.6 eV1

4"

1

25

# $ %

& ' ( = 2.86 eV (absorption)

Transition II:

!E = 13.6 eV1

25"

1

9

# $ %

& ' ( = "0.967 eV (emission)

Transition III:

!E = 13.6 eV1

49"

1

16

# $ %

& ' ( = "0.572 eV (emission)

Transition IV:

!E = 13.6 eV1

16"

1

49

# $ %

& ' ( = 0.572 eV (absorption)

(a) Since

! =hc

E"

=hc

# $E,

transition II emits the shortest wavelength photon.

(b) The atom gains the most energy in transition I .

(c) The atom loses energy in transitions II and III .


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Chapt28, Problem-23: Consider a hydrogen atom. (a) Calculate the frequency f of the n = 2 to

n = 1 transition and compare with the frequency forb of the electron orbital motion in the n = 2 state. (b)Make the same calculation for the n = 10 000 to n = 9 999 transition. Comment on the results.

Solut ion:

(a) The wavelength emitted in the ni = 2 ! nf = 1 transition is

! =1

RH

ni2nf

2

ni2 " n f

2

#

$ %

&

' ( =

1

1.09737 )107 m

-1( )4( ) 1( )4"1

#

$ %

&

' ( = 1.22) 10

"7 m

and the frequency is f =

c

!=

3.00 "108 m s

1.22" 10#7

m=

2 .47 ! 10

15 Hz

From Ln =mevnrn = nh, the speed of the electron is

vn = nh mern

Hence, with rn = n

2a0 , the orbital frequency is

forb =1

T=

vn

rn

=nh

2!me rn2

=h

4! 2 me a0

2

"

# $

%

& '

1

n3=

6.59( 1015

Hz

n3

For the n = 2 orbit,

forb =6.59 !10

15 Hz

2( )3

= 8.23! 10

14 Hz

(b) For the ni = 10 000! n f = 9 999 transition,

! =1

1.09737 "107 m

-1( )10 000( )

2

9 999( )2

10 000( )2

# 9 999( )2

$

%

& &

'

(

) )

= 4.56" 104 m

and f =

c

!=

3.00 "108 m s

4.56" 104 m

= 6.59 !10

3 Hz

For the n = 10 000 orbit,

forb =6.59 !10

15 Hz

10 000( )3

= 6.59 !10

3 Hz

For small n, significant differences between classical and quantum results appear. However, as n becomes

large, classical theory and quantum theory approach one another in their results. (correspondence principle)

Chapt28, Problem-30: Construct an energy level diagram like that in Figure 28.7 for doubly

ionized lithium (Li+), for which Z = 3

Solut ion:

We use

En =!Z

213.6 eV( )

n2

with Z= 3 to give:

En =!122 eV

n2

n = × ___________________ E = 0n = 5 ___________________ –4.90 eVn = 4 ___________________ –7.65 eV

n = 3 ___________________ –13.6 eV

n = 1 ___________________ –122 eV

n = 2 ___________________ –30.6 eV


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Chapt28, Problem-33: List the possible sets of quantum numbers for electrons in the 3p subshell.

Solut ion:

In the 3p subshell, n = 3 and l = 1. The 6 possible quantum states are

n = 3 l = 1 ml = +1 ms = ±

1

2

n = 3 l = 1 ml = 0 ms = ±

1

2

n = 3 l = 1 ml = !1

ms = ±

1

2

Chapt28, Problem-36: (a) Write out the electronic configuration of the ground state for oxygen (Z

= 8). (b) Write out values for the set of quantum numbers n, , lm , and ms for each of the electrons in

oxygen.

Solut ion:

(a) The electronic configuration for oxygen Z = 8( ) is

1s

22s

22p

4.

(b) The quantum numbers for the 8 electrons can be:

1s states n = 1 l = 0 ml = 0 ms = ±

1

2

2s states n = 2 l = 0 ml = 0 ms = ±

1

2

2p states n = 2 l = 1

ml = 0

ml = 1 ms = ±

1

2

ms = ±

1

2

Chapt28, Problem-38: How many different sets of quantum numbers are possible for an electron

for which (a) n = 1, (b) n = 2, (c) n = 3, (d) n = 4, and (e) n = 5? Check your results to show that they agreewith the general rule that the number of different sets of quantum numbers is equal to 2n2.

Solut ion:

(a) For n = 1, l = 0 and there are 2 (2l +1) states = 2(1)=

2 sets of quantum numbers

(b) For n = 2, l = 0 for 2 (2l +1) states = 2(0+ 1)= 2 sets

and l = 1 for 2 (2l +1) states = 2(2+ 1)= 6 sets total number of sets =

8

(c) For n = 3, l = 0 for 2 (2l +1) states = 2(0+ 1)= 2 sets

and l = 1 for 2 (2l +1) states = 2(2+ 1)= 6 sets

and l = 2 for 2 (2l +1) states = 2(4 +1) = 10 sets total number of sets =

18

(d) For n = 4, l = 0 for 2 (2l +1) states = 2(0+ 1)= 2 sets


and l = 2 for 2 (2l +1) states = 2(4 +1) = 10 sets

and l = 3 for 2 (2l +1) states = 2(6+ 1)= 14 sets total number of sets =

32

(e) For n = 5, l = 0 for 2 (2l +1) states = 2(0+ 1)= 2 sets


and l = 2 for 2 (2l +1) states = 2(4 +1) = 10 sets


and l = 4 for 2 (2l +1) states = 2(8 + 1)= 18 sets total number of sets =

50

For n = 1: 2n

2= 2 . For

n = 2 : 2n

2= 8 .

For n = 3: 2n

2= 18 . For

n = 4: 2n

2= 32 .

For n = 5 : 2n

2= 50 . Thus, the number of sets of quantum states agrees with the 2n

2 rule.


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Chapt28, Conceptual-1: In the hydrogen atom, the quantum number n can increase without limit.

Because of this, does the frequency of possible spectral lines from hydrogen also increase without limit?

Solut ion:

The energy of the atom is inversely proportional to n2. Thus, as n grows to infinity, the energy ofthe atom approaches a value that is above the ground state by a finite amount, namely the ionizationenergy 13.6eV. As the electron falls from one bound state to another, its enrgy loss is always lessthan the ionization energy. The energy and frequency of the meitted photon are finite.

Chapt28, Conceptual-2: Does the light emitted by a neon sign constitute a continuous spectrum or

only a few colors? Defend your answer.

Solut ion:

Neon signs do not emit a continuous spectrum. They emit many discrete wavelengths as could bedetermined by observing the light from the sign through a spectrometer (in fact ‘mysterium’ wasNeonn for some of you in Expt#8) However, they do not emit all wavelengths. The specificwavelengths and intensities account for the color of the sign.

Chapt28, Conceptual-3: In an X-ray tube, if the energy with which the lectrons strike the metal

target is increased, the wavelength of the characteristic X-rays do not change. Why?

Solut ion:

The characteristic X-rays originate from transitions within the atom of the target, such as an L-shell electron making a transition to a vacancy in the K-shell. This vacancy in the K-shell is causedwhen an accelerated electron in the X-ray tube supplies energy the the K-shell electron to eject itfrom the atom. If the energy of the bombarding electron were to be increased, the K-shell electronwould be ejected with more kinetic energy. But the energy difference between the K- and L-shellshas not changed, so the emitted X-ray has exactly the same wavelength.

Chapt28, Conceptual-8: If matter has a wave nature, why is this not observable in our dailyexperience?

Solut ion:

The de Broglie wavelength of macroscopic objects such as a baseball moving with a typical speedsuch as 30 m/s is very small and impossible to measure. That is,

! = h mv , is a very small number for

macroscopic objects. We are not able to observe diffraction effects because the wavelength ismuch smaller than any aperture through which the object could pass.


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Chapt28, Conceptual-14: The ionization energies for Li, Na, K, Rb, and Cs are 5.390, 5.138,

4.339, 4,.176, and 3.893 eV respectively. Explain why this patter of decrease should be expected in terms ofthe atomic structure.

Solut ion:

Each of the given atoms has a single electron in an l = 0 or s( ) state outside a fully closed-shell core,

shielded from all but one unit of the nuclear charge. Since they reside in very similar environments,one would expect these outer electrons to have nearly the same electrical potential energies andhence nearly the same ionization energies. This is in agreement with the given data values. Also,since the distance of the outer electron from the nuclear charge should tend to increase with Z (toallow for greater numbers of electrons in the core), one would expect the ionization energy todecrease somewhat as atomic number increases. This is also in agreement with the given data.

chapter 28 cp

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