Name _________________________________________ Date _________________________ Period ____

HomeworkChapter 6: Electronic Structure of Atoms

Exercises: Sections 6.3 and 6.4: Line Spectra and the Bohr Model The Wave Behavior of Matter

1. (a) In terms of the Bohr theory of the hydrogen atom, what process is occurring when excited hydrogen atoms emit radiant energy of certain wavelengths and only those wavelengths?

electrons are moving from a higher allowed state to a lower allowed state; since only certain energy states are allowed only certain energy changes can occur and their λ be emitted; these emitted λ correspond to the lines in the emission spectrum of hydrogen

(b) What kind of process corresponds to the absorption of light of certain wavelengths by hydrogen atoms?

electrons are moving from a lower allowed state to a higher allowed state; only certain energy states are allowed, so only certain wavelengths are absorbed; these absorbed λ correspond to the dark lines seen on a continuous emission spectrum

(c) Does a hydrogen atom “expand” or “contract” as it mores from its ground state to an excited state? Expand

2. Indicate whether energy is emitted or absorbed when the following electronic transitions occur in hydrogen: (a) from n = 2 to n = 6; (b) from an orbit of radius 4.77 Å to one of radius 0.530 Å; (c) from the n = 6 to the n = 9 state.

absorbed (a) emitted (b) absorbed (c)

3. For the following electronic transitions in the hydrogen atom, calculate the energy, frequency, and wavelength of the associated radiation and determine whether the radiation is emitted or absorbed during the transition: Follow ALL math work rules!

(a) from n = 1 to n = 3; * = RH ch2.18 x 10−18 J

h 6.626 x10−34 J−s = 3.29 x 1015 s-1 ν = 3.29 x 1015Hz (

1

n i2 −

1

n f 2 )

i) v = 3.29 x 1015Hz* ( 1

12 − 1

32 ) → 3.29 x 1015Hz ( 11 − 1

9 )

→ 3.29 x 1015Hz (0.889) = 2.92 x 1015 Hz = v

ii) E = hv → (6.626 x 10-34 J-s)( 2.92 x 1015 s-1) = 1.94 x 10-18J = E

iii) λ = ch → 3.00 x 108 m s−1

2.92 x1015 s−1 = 1.02 x 10-7m = 102 nm = λ3. Continued:

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(b) from n = 9 to n = 3;

i) v = 3.29 x 1015Hz ( 1

92 − 1

32 ) → 3.29 x 1015Hz ( 181 − 1

9 )

→ 3.29 x 1015Hz (−0.0988) = − 3.25 x 1014 Hz = v*

ii) E = hv → (6.626 x 10-34 J-s)( 3.25 x 1014 s-1) = 2.15 x 10-19 J = E

iii) λ = ch → 3.00 x 108 m s−1

3.25 x1014 s−1 = 9.23 x 10-7m = 923 nm = λ

* Remember: the negative sign only means that energy was emitted

(c) from n = 7 to n = 4.

i) v = 3.29 x 1015Hz ( 1

72 − 1

42 ) → 3.29 x 1015Hz ( 149 − 1

16 )

→ 3.29 x 1015Hz (−0.0421) = − 1.38 x 1014 Hz = v*

ii) E = hv → (6.626 x 10-34 J-s)( 3.25 x 1014 s-1) = 9.18 x 10-20 J = E

iii) λ = ch → 3.00 x 108 m s−1

1.38 x1014 s−1 = 2.17 x 10-6m = 2170 nm = λ

(d) Do any of these transitions emit or absorb visible light? NO If yes, which ones? _____________________

X-ray = 10-10 m UV = 10-9 thru 10-7 m Visible = = 4 x 10-7 thru 7.5 x 10-7 m

IR = = 10-3 thru high to mid 10-7 m Microwave = 10-2 m TV, Radio = 1 mBalmer → n f = 2 (visible) Lyman → nf = 1 (UV) Pachen → nf = 3 (IR)

−RHch* = − 2.18 x 10-18 J = energy of each allowed orbit, n = principle quantum number (orbit)

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*RH = Rydberg constant c = speed of light h = Planck's constant

(1.10 x 107 waves m-1) (3.00 x 108 ms-1) (6.626 x 10-34 J-s)

4. The Lyman series of emission lines of the hydrogen atom are those for which nf = 1.

(a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed.

i) En = −RHch( 1

n2 ) → (− 2.18 x 10-18 J) ( 1

12 ) → (− 2.18 x 10-18 J) ( 11 )

= − 2.18 x 10-18 J = En

ii) E = hv → v = En

h → −2 .18 x10−18 J6 .626 x 10−34 J−s

= − 3.29 x 1015 Hz

iii) c = λν → λ = 3 . 00 x108 m s−1

−3 .29 x1015 s−1 = −9.11 x 10-8 m = 91 nm

91 nm falls between UV and X-ray region

(b) Calculate the wavelengths of the first three lines in the Lyman series-those for which ni = 2, 3, and 4. Follow ALL math work rules!

En = hv and c = λν so....... En = hcλ → λ =

hcEn

also, ΔEn = −RHch( 1

ni2− 1

nf2 ) so....... λn =

hc

−RH ch (ni2−n f

2) →

λn = 1.99 x10−25 J−m

−2.18 x10−18 J (ni2−nf

2)

4. (b) Continued:

Since this is the Lyman Series, in each case nf = 1

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i) ni = 2 → λn = 1 .99 x 10−25 J−m

−2 .18 x10−18 J ( 122 −

112 )

→ −1.21 x 10-7 m* = 121 nm

ii) ni = 3 → λn = 1 .99 x 10−25 J−m

−2 .18 x10−18 J ( 132 −

112 )

→ −1.03 x 10-7 m = 103 nm

iii) ni = 4 → λn = 1 .99 x10−25 J−m

−2 .18 x10−18 J ( 142 −

112 )

→ −9.75 x 10-8 m = 20.4 nm

* Remember: the negative sign only means that energy was emitted

5. Among the elementary subatomic particles of physics is the muon, which decays within a few nanoseconds after formation. The muon has a rest mass 206.8 times that of an electron. Calculate the de Broglie wavelength associated with a muon traveling at a velocity of 8.85 x 105 cm/ s. Follow ALL math work rules!

deBroglie wavelength

λ = h

mv

i) 6.626 x10−34 J−s(9.10 x 10−28 g)(8.85 x 105 cm s−1) → 3.99 x 10-10 m

(Chapter 2)

ii) 3.99 x 10-10 m 1 Å = 3.99 Å

10-10 m6. The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of 9.38 x 106 m/ s.

(a) Calculate the characteristic wavelength of this electron? Follow ALL math work rules!

deBroglie wavelength

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λ = h

mv

i) 6.626 x10−34∗kg

m2

s2 ∙ s

(9.10 x 10−31 k g)(9.38 x 106 m s−1) → 7.77 x 10-11 m

* Remember: 1 Joule = 1 kgm2

s2

ii) 7.77x 10-11 m 1 Å = 0.777 Å

10-10 m

(b) Is the wavelength comparable to the size of atoms?

atomic radii and inter-atomic distance ≈ 1-5 Å, (section 2.3)so the wavelength of this electron IS comparable to the size of atoms

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