Tutorial Assignment 5 - Chapter 4 PDF

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CHEM 1001 A

7:30 PM - Wednesdays October 9th

Fall 2019

Tutorial Assignment 5: Chapter 4 1. Convert the following to the indicated unit/measurement: a. What is the frequency (in Hz) of a microwave with λ = 4.33x10-3m? c = λν c

ν= = λ

3.00x108 m s−1 4.33x10 −3 m

= 𝟔. 𝟗𝟑𝐱𝟏𝟎𝟏𝟎 𝐇𝐳

b. What is the wavelength (in m) of an AM radio wave with frequency ν = 1150 kHz? c = λν c

λ= = ν

3.00x108 m s−1

1150 𝑘𝐻𝑧∗

1𝐻𝑧 1000 𝑘𝐻𝑧

= 𝟐. 𝟔𝟏𝐱𝟏𝟎𝟖 𝐦

c. What is the energy (in kJ/mol) of photons of X rays with λ = 5.49nm? Ephoton =

hc λ

=

(6.626x10 −34 J s)∗(3.00x108 m s−1 ) 5.49nm∗

1𝑚 109 𝑛𝑚

= 3.6208x10−17 J

Emolar (photons) = Ephoton ∗ NAv Emolar (photons) = (3.6208x10−17J) ∗ (6.022x1023mol−1 )

Emolar (photons) = 2.1804x107 J mol−1 ∗ 𝐄𝐦𝐨𝐥𝐚𝐫 (𝐩𝐡𝐨𝐭𝐨𝐧𝐬) = 𝟐. 𝟏𝟖𝐱𝟏𝟎𝟒 𝐤𝐉 𝐦𝐨𝐥−𝟏

1kJ

1000J

d. What is the wavelength (de Broglie wavelength, in m) of a pitched baseball with a mass of 129g and a speed of 44.7m/s? λ=

h mv

=

6.626x10−34J s

1kg )∗(44.7 m s −1) (129g∗ 1000g

=

6.626x10 −34(kg m2s −2) s

(0.129kg)∗(44.7 m s−1)

= 𝟏. 𝟏𝟓𝐱𝟏𝟎−𝟑𝟒 𝐦

CHEM 1001 A

7:30 PM - Wednesdays October 9th

Fall 2019

2. Assuming the maximum excitation level for hydrogen is 6, what is the shortestwavelength line (in nanometers) in the Paschen series of the hydrogen spectrum? Paschen series has m = 3 (returning to third energy level) (see Figure 4-13 in Textbook) Shortest-wavelength of Paschen series would be largest energy transition (from n = 6 to m = 3). Ephoton = |ΔE atom | = |Efinal − Einitial | = |E3 − E6 | −2.18x10 −18 J

−2.18x10 −18 J

)| Ephoton = |( )−( 62 32 −19 Ephoton = |(−2.422x10 J) − (−6.056x10−20 J)| Ephoton = | − 1.8166x10−19 J| Ephoton = 1.8166x10−19J Ephoton = λ=

hc

hc λ

Ephoton

(6.626x10 −34 J s)∗(3.00x108 m s−1 )

=

1.8166x10−19 J

= 1.094x10−6 m = 𝟏. 𝟎𝟗𝐱𝟏𝟎𝟑 𝐧𝐦

3. In a photoelectric effect experiment, a photon with energy of 1.05x10-18J is absorbed, causing ejection of an electron with kinetic energy of 2.05x10-19J. a. Determine the wavelength (in nm) associated with the absorbed particle. E= λ=

hc

λ hc E

=

(6.626x10 −34 J s)∗(3.00x108 m s−1 ) 1.05x10−18 J

= 1.893x10−7 m = 𝟏𝟖𝟗𝐧𝐦

b. Determine the wavelength (in nm) associated with the emitted particle. 1

EKinetic = mv2 2 2∗EKinetic m

v= √

λ=

h mv

=

=√

2∗2.05x10−19J

9.109x10−31 kg

= √4.50104x1011 m2 s−2 = 6.70898x105 m s−1

6.626x10−34J s

(9.109x10 −31kg)∗(6.70898x105 m s−1)

= 1.084x10−6 m = 𝟏. 𝟎𝟖𝐱𝟏𝟎𝟑 𝐧𝐦

CHEM 1001 A

7:30 PM - Wednesdays October 9th

Fall 2019

4. Considering the usual quantum number rules: a. Give the orbital notation and possible orientations for electrons in an orbital with the following quantum numbers: n = 4, l = 2, ml = 0 4d; dx2 −y2 , dz2 , dxy , dxz , dyz b. List all the possible combinations of quantum numbers for the following orbitals: A 4p orbital n = 4, l = 1, ml = -1, ms = + ½ n = 4, l = 1, ml = -1, ms = - ½ n = 4, l = 1, ml = 0, ms = + ½ n = 4, l = 1, ml = 0, ms = - ½ n = 4, l = 1, ml = 1, ms = + ½ n = 4, l = 1, ml = 1, ms = - ½

5. Sketch the shapes of all orientations of s, p, and d orbitals (match label to image in online submission).

CHEM 1001 A

7:30 PM - Wednesdays October 9th

Fall 2019

6. The Sun’s atmosphere contains vast quantities of He+ cations. These ions absorb some of the Sun’s thermal energy, promoting electrons from the He+ ground state to various excited states. A He+ ion in the fifth energy level may return to the ground state by emitting three successive photons: an infrared photon (λ = 1014 nm), a green photon (λ = 469 nm), and an X ray (λ = 26 nm). Calculate the excitation energy (in J) for the He+ ion immediately before it emits the green photon.

Since green is second photon being emitted, right before this emission is the He+ ion’s excitation of X-ray + green photon: Egreen photon = EX ray =

hc λ

=

hc

=

(6.626x10−34 J s)∗(3.00x108 m s−1)

1m 109 nm −34 8 (6.626x10 J s)∗(3.00x10 m s−1 )

λ

469nm∗

26nm∗

Eexcitation before green emission Eexcitation before green emission Eexcitation before green emission Eexcitation before green emission

1m 109 nm

= 7.6454x10−18 J

= Egreen + E X ray = (4.2384x10−19J) + (7.6454x10−18 J) = 8.0692x10−18J = 𝟖. 𝟏𝐱𝟏𝟎−𝟏𝟖𝐉

Notes and Tools Unit Conversions: Equations: 1J = 1kg m2 s-2

c = λν Ephoton = hν

Constants:

h mv

NAv = 6.022x1023 mol-1

λ=

h = 6.626x10-34 J s

ΔEatom = ±Ephoton

8

c = 3.00x10 m s me = 9.109x10

-1

-31

kg

= 4.2384x10−19J

1

EKinetic = mv 2 2...