College Physics Student Solutions Manual Chapter 30 CHAPTER 30: ATOMIC PHYSICS 30.1 DISCOVERY OF THE ATOM 1. Using the given charge-­‐to-­‐mass ratios for electrons and protons, and knowing the magnitudes of their charges are equal, what is the ratio of the proton’s mass to the electron’s? (Note that since the charge-­‐to-­‐mass ratios are given to only three-­‐digit accuracy, your answer may differ from the accepted ratio in the fourth digit.) Solution We can calculate the ratio of the masses by taking the ratio of the charge to mass q
q
= 1.76 × 1011 C/kg and ratios given: = 9.57 ×10 7 C/kg , so that me
mp
mp
q / me 1.76 × 1011 C/kg
=
=
= 1839 = 1.84 × 10 3 . 7
me q / m p 9.57 × 10 C/kg
The actual mass ratio is: mp
me
=
1.6726 × 10 −27 kg
= 1836 = 1.84 × 10 3 , so to three −31
9.1094 × 10 kg
digits, the mass ratio is correct. 30.3 BOHR’S THEORY OF THE HYDROGEN ATOM 12. Solution A hydrogen atom in an excited state can be ionized with less energy than when it is in its ground state. What is n for a hydrogen atom if 0.850 eV of energy can ionize it? Using E n =
− 13.6 eV
, we can determine the value for n , given the ionization energy: n2
1/ 2
n=
− 13.6 eV ⎛ − 13.6 eV ⎞
= ⎜
⎟
En
⎝ − 0.85 eV ⎠
= 4.0 = 4 (Remember that n must be an integer.) 214 College Physics 18. Student Solutions Manual Chapter 30 (a) Which line in the Balmer series is the first one in the UV part of the spectrum? (b) How many Balmer series lines are in the visible part of the spectrum? (c) How many are in the UV? Solution (a) We know that the UV range is from λ = 10 nm to approximately λ = 380 nm. Using the equation 1
⎛ 1
1 ⎞
= R⎜⎜ 2 − 2 ⎟⎟ , where nf = 2 for the Balmer series, we can λ
⎝ nf ni ⎠
solve for ni . Finding a common denominator gives ni2 nf2 = λR(ni2 − nf2 ), or ni = nf
1 ni2 − nf2
= 2 2 , so that λR
ni nf λR
. The first line will be for the lowest λR − nf2
energy photon, and therefore the largest wavelength, so setting λ = 380 nm gives ni = 2
(3.80 ×10 −7 m)(1.097 ×10 7 m -1 )
= 9.94 ⇒ ni = 10 will be the first. (3.80 ×10 −7 m)(1.097 ×10 7 m -1 ) − 4
(b) Setting λ = 760 nm allows us to calculate the smallest value for ni in the visible range: ni = 2
(7.60 ×10 −7 m)(1.097 ×10 7 m -1 )
= 2.77 ⇒ ni = 3 so ni = 3 to 9 are (7.60 ×10 −7 m)(1.097 ×10 7 m -1 ) − 4
visible, or 7 lines are in the visible range. (c) The smallest λ in the Balmer series would be for ni = ∞ , which corresponds to a value of: ⎛ 1 1 ⎞ R
n2
4
= R⎜⎜ 2 − 2 ⎟⎟ = 2 ⇒ λ = f =
= 3.65 ×10 -7 m = 365 nm , which 7
−1
λ
R 1.097 ×10 m
⎝ nf ni ⎠ nf
is in the ultraviolet. Therefore, there are an infinite number of Balmer lines in the ultraviolet. All lines from ni = 10 to ∞ fall in the ultraviolet part of the spectrum. 1
23. 2
h
n2
= 0.529 ×10 −10 m using the aB and a B = 2
2
4π me kqe
Z
approach stated in the text. That is, equate the Coulomb and centripetal forces and then insert an expression for velocity from the condition for angular momentum quantization. 1. Verify Equations rn =
215 College Physics Solution Student Solutions Manual Using Fcoulomb = Fcentripetal
Chapter 30 kZqe2 kZqe2 1
kZqe2 me v 2
⇒ 2 =
, so that rn =
=
. rn
rn
me v 2
me v 2
kZqe2 4π 2 me2 rn2
h
Since me vrn = n
⋅
, we can substitute for the velocity giving: rn =
me
2π
n2h2
so that rn =
n2
h2
n2
h2
w
here =
a
,
a
=
. B
B
Z 4π 2 me kq e2
Z
4π 2 me kq e2
30.4 X RAYS: ATOMIC ORIGINS AND APPLICATIONS 26. Solution A color television tube also generates some x rays when its electron beam strikes the screen. What is the shortest wavelength of these x rays, if a 30.0-­‐kV potential is used to accelerate the electrons? (Note that TVs have shielding to prevent these x rays from exposing viewers.) Using the equations E = qV and E =
hc
λ
gives E = qV =
hc
λ
, which allows us to calculate the wavelength: λ=
(
)(
)(
)
hc
6.626 ×10 −34 J ⋅ s 2.998 ×108 m/s
=
= 4.13 ×10 −11 m −19
4
qV
1.602 ×10 C 3.00 ×10 V
(
)
30.5 APPLICATIONS OF ATOMIC EXCITATIONS AND DE-­‐EXCITATIONS 33. (a) What energy photons can pump chromium atoms in a ruby laser from the ground state to its second and third excited states? (b) What are the wavelengths of these photons? Verify that they are in the visible part of the spectrum. Solution (a) From Figure 30.64, we see that it would take 2.3 eV photons to pump chromium atoms into the second excited state. Similarly, it would take 3.0 eV photons to pump chromium atoms into the third excited state. 216 College Physics Student Solutions Manual Chapter 30 hc 1.24 × 10 −6 eV ⋅ m
=
= 5.39 × 10 −7 m = 5.4 × 10 2 nm , which is yellow-­‐
E
2.3 eV
green. (b) λ2 =
λ2 =
hc 1.24 × 10 −6 eV ⋅ m
=
= 4.13 × 10 −7 m = 4.1× 10 2 nm , which is blue-­‐violet. E
3.0 eV
30.8 QUANTUM NUMBERS AND RULES 40. Solution (a) What is the magnitude of the angular momentum for an l = 1 electron? (b) Calculate the magnitude of the electron’s spin angular momentum. (c) What is the ratio of these angular momenta? h
, we can calculate the angular momentum of 2π
(a) Using the equation L = ( + 1)
h
= 1(2)
an  = 1 electron: L = ( + 1)
2π
⎛ 6.626 × 10 −34 J ⋅ s ⎞
⎜⎜
⎟⎟ = 1.49 × 10 −34 J ⋅ s 2π
⎝
⎠
h
, we can determine the electron’s spin 2π
angular momentum, since s = 1 2 :
(b) Using the equation S = s( s + 1)
S = s( s + 1)
h
1 ⎛ 3 ⎞ 6.626 ×10 −34 J ⋅ s
=
= 9.13 ×10 −35 J ⋅ s ⎜ ⎟
2π
2 ⎝ 2 ⎠
2π
( + 1) h
L
2π =
(c) =
S
s( s + 1) h
2π
2
3
= 1.63 4
30.9 THE PAULI EXCLUSION PRINCIPLE 55. Integrated Concepts Calculate the velocity of a star moving relative to the earth if you observe a wavelength of 91.0 nm for ionized hydrogen capturing an electron directly into the lowest orbital (that is, a ni = ∞ to nf = 1, or a Lyman series transition). 217 College Physics Student Solutions Manual Chapter 30 Solution We will use the equation ΔE = Ef − Ei to determine the speed of the star, since we are given the observed wavelength. We first need the source wavelength: ΔE = Ef − Ei =
so that λs =
⎛ Z 2 ⎞ ⎛ Z 2 ⎞
⎡ 12
⎤
= ⎜⎜ − 2 E0 ⎟⎟ − ⎜⎜ − 2 E0 ⎟⎟ = 0 − ⎢− 2 (13.6 eV)⎥ = 13.6 eV , λs ⎝ nf
⎣ 1
⎦
⎠ ⎝ ni
⎠
hc
hc 1.24 ×10 3 eV ⋅ nm
1+ v / c
=
= 91.2 nm. Therefore, using λobs = λs
,
ΔE
13.6 eV
1− v / c
1 + v / c λ2obs
v λ2obs ⎛ v ⎞
we have =
, so that 1 + = 2 ⎜1 − ⎟ and thus, 1 − v / c λs2
c λs ⎝ c ⎠
v λ2obs / λs2 − 1 (91.0 nm/91.2 nm) 2 − 1
= 2
=
= −2.195 ×10 −3. 2
2
c λobs / λs + 1 (91.0 nm/91.2 nm) + 1
So, v = (−2.195 ×10 −3 )(2.998 ×108 m/s) = −6.58 ×10 5 m/s . Since v is negative, the star is moving toward the earth at a speed of 6.58 × 10 5 m/s. 59. Solution Integrated Concepts Find the value of l , the orbital angular momentum quantum number, for the moon around the earth. The extremely large value obtained implies that it is impossible to tell the difference between adjacent quantized orbits for macroscopic objects. d
, we can get an expression for the velocity in t
2πR
terms of the period of rotation of the moon: v =
. Then, from L = Iω for a point T
v
object we get the angular momentum: L = Iω = mR 2ω = mR 2 = mRv . Substituting R
h
for the velocity and setting equal to L = ( + 1)
gives: 2π From the definition of velocity, v =
L = mvR =
=
2π mR 2 h
2π mR 2
h
= ( + 1)
≈
. Since l is large : , so T
2π
T
2π
4π 2 mR 2 4π 2 (7.35 ×10 22 kg )(3.84 ×108 m ) 2
=
= 2.73 ×10 68 . 6
−34
Th
(2.36 ×10 s )(6.63 ×10 J.s)
218 College Physics 66. Student Solutions Manual Chapter 30 Integrated Concepts A pulsar is a rapidly spinning remnant of a supernova. It rotates on its axis, sweeping hydrogen along with it so that hydrogen on one side moves toward us as fast as 50.0 km/s, while that on the other side moves away as fast as 50.0 km/s. This means that the EM radiation we receive will be Doppler shifted over a range of ± 50.0 km/s . What range of wavelengths will we observe for the 91.20-­‐nm line in the Lyman series of hydrogen? (Such line broadening is observed and actually provides part of the evidence for rapid rotation.) Solution We will use the Doppler shift equation to determine the observed wavelengths for the Doppler shifted hydrogen line. First, for the hydrogen moving away from us, we use u = +50.0 km/s, so that: λobs
(
(
)
)
1 + 5.00 × 10 4 m/s/2.998 × 108 m/s
= (91.20 nm )
= 91.22 nm
1 − 5.00 × 10 4 m/s/2.998 × 108 m/s
Then, for the hydrogen moving towards us, we use u = −50.0 km/s, so that: λobs = (91.20 nm )
(
(
)
)
1 − 5.00 × 10 4 m/s/2.998 × 108 m/s
= 91.18 nm
1 + 5.00 × 10 4 m/s/2.998 × 108 m/s
The range of wavelengths is from 91.18 nm to 91.22 nm. 219