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Chemistry for Engineers
Homework EXTRA
Atomic Structure
Answer Key
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Hand in by Friday
13:00
Chemistry for Engineers
Homework Extra: Atomic Structure
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Q2: Redraw this table neatly and fill in every blank box
regarding the atoms or ions of the following isotopes:
Q1: a) Write the electron configurations of the following
species e.g. Si is [Ne]2s22p2
Species
i) Sr2+
iii) Auv) Pb2+
ii) the hydride ion
iv) Ga3+
vi) Fr
239
Pu
∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )]
4
[Rn]5f
4
(9 marks)
Q3: Using the Rydberg equation:
No. of
neutrons
2+
1
(where R = 3.29 x 1015 Hz,
h = 6.63 x 10-34 Js, c = 3 x 108 m/s,
NA = 6.022 x 1023).
platinum-197
52 2+
V
[He]
1
2
1s
197
23
42
thallium-202
9
33
[Kr]
81
(10.5 marks)
a) Calculate the photon energy and wavelength corresponding to:
i) the three lowest energy lines of the Balmer series and
ii) the convergence limit of the Balmer series.
b) sketch the appearance of the Balmer series of lines in relation to the visible region of the electromagnetic spectrum
Far UV
100
UV
200
visible
400
IR
800 nm
c) Calculate the wavelength and photon energy of both the lowest and highest energy lines of the Paschen series (transitions down to n = 3
quantum state)
d) Add and label the Paschen series lines to your sketch of the Balmer Series. In which region of the electromagnetic spectrum do the
Paschen lines lie? Do the Paschen and Balmer series overlap?
e) By considering the convergence limit of the Lyman series, calculate the ionisation energy (in kJ/mol) of the hydrogen atom.
(17 marks)
Why is the energy required to remove an electron from He much greater (2370 kJ/mol)?
Q4: a) A 1.0 mg sample of silver was found to contain atoms of two isotopes. Lighter 107Ag (106.905095 Da) accounted
for 0.51377 mg of metal. Given that the atomic weight of Ag is 107.8682, calculate the exact mass (to 6 s.f.) of the
heavier isotope. How many neutrons are contained in each nucleus of the heavier isotope?
(8.5 marks)
Total: 45 marks
1
Q1: a) Write the electron configurations of the following species e.g. Si is
[Ne]2s22p2
i) Sr2+
[Kr]
ii) H-
[He] or 1s2
iii) Au-
[Xe]6s25d10
iv) Ga3+
[Ar]3d10
v) Pb2+
[Xe]6s25d10
vi) Fr
[Rn]7s1
4f14 electrons omitted in iii) and v)
(6 x 2 marks)
Q2: Fill in every blank box regarding the atoms or ions of the
following isotopes:
(10.5 marks - 0.5 marks per box)
Species
239
Pu
No. of
neutrons
2+
[Rn]5f
4
1
platinum-197
52 2+
V
9
4
[He]
1
2
1s
197
23
42
thallium-202
33
[Kr]
81
Species
No. of
neutrons
No. of
protons
No. of
nucleons
Electron
configuration
239
Pu
2+
4
145
94
239
[Rn]5f
2+
5
4
9
[He]
+
1
2
3
1s
platinum-197
52 2+
V
119
29
78
23
197
52
[Xe]6s 5d
3
[Ar]3d
75
42
33
75
[Kr]
121
81
202
[Xe]6s 5d 6p
9
Be
3
He
3-
As
thallium-202
1
2
8
2
10
1
2
Q3: Using the Rydberg equation:
∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )]
for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz,
h = 6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023).
a) Calculate the photon energy and wavelength corresponding to i) the
three lowest energy lines of the Balmer series and ii) the convergence
limit of the Balmer series.
Lowest energy line: nL = 2, nU = 3 so [(1/nL2 ) – (1/nU2 )] = 0.1389
∆E = 3.03 x 10-19 J
λ = 656.4 nm
second lowest energy line: nL = 2, nU = 4 so [(1/nL2 ) – (1/nU2 )] = 0.1875
∆E = 4.09 x 10-19 J
λ = 486.3 nm
third lowest energy line: nL = 2, nU = 5 so [(1/nL2 ) – (1/nU2 )] = 0.21
∆E = 4.58 x 10-19 J
λ = 434.3 nm
the convergence limit: nL = 2, nU = ∞ so [(1/nL2 ) – (1/nU2 )] = 0.25
∆E = 5.45 x 10-19 J
λ = 364.7 nm
(6 + 2 + 2 + 3 + 4 marks)
Q3: Using the Rydberg equation:
∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )]
for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h =
6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023).
b) sketch the appearance of the Balmer series of lines in relation to the
visible region of the electromagnetic spectrum
Far UV
100
UV
200
Balmer convergence limit
lies in near UV
visible
400
IR
800
(6 + 2 + 2 + 3 + 4 marks)
3
Q3: Using the Rydberg equation:
∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )]
for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h =
6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023).
c) Calculate the photon energy and wavelength of both the lowest and
highest energy lines of the Paschen series (transitions down to n = 3
quantum state)
Highest energy Paschen line: nL = 3, nU = ∞ so [(1/nL2 ) – (1/nU2 )] = 0.111
∆E = 2.424 x 10-19 J
λ = 821 nm (convergence limit in near IR)
Lowest energy Paschen line: nL = 3, nU = 4 so [(1/nL2 ) – (1/nU2 )] = 0.0486
∆E = 1.06 x 10-19 J
λ = 1876 nm (1.9 µm = infrared)
(6 + 2 + 2 + 3 + 4 marks)
Q3: Using the Rydberg equation:
∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )]
for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h =
6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023).
d) Add and label the Paschen series lines to your sketch of the Balmer
Series. In which region of the electromagnetic spectrum do the
Paschen lines lie? Do the Paschen and Balmer series overlap?
Far UV
100
UV
200
visible
400
IR
800
Paschen convergence limit
lies in near IR
(No overlap of two series of lines)
Lowest energy Paschen
line in middle IR
4
Q3: Using the Rydberg equation:
∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )]
for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h =
6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023).
e) By considering the convergence limit of the Lyman series, calculate
the ionisation energy (in kJ/mol) of the hydrogen atom. Why is the
energy required to remove an electron from He much greater (2370
kJ/mol)?
the Lyman convergence limit:
∆E = 2.181 x 10-18 J/photon
∆E = 2.181 x 10-18 x 6.022 x 1023 = 1.31 x 106 J/mol
= 1310 kJ/mol
1st I.E. of He is higher due to greater nuclear charge and smaller
atomic radius
(6 + 2 + 2 + 3 + 4 marks)
Q4: a) A 1.0 mg sample of silver was found to contain atoms of two
isotopes. Lighter 107Ag (106.905095 Da) accounted for 0.51377 mg of
metal. Given that the atomic weight of Ag is 107.8682, calculate the
exact mass (to 6 s.f.) of the heavier isotope (xAg). How many neutrons
are contained in each nucleus of the heavier isotope?
calculate mol% of each isotope from the mass composition:
107Ag
Mass composition
Moles
Moles
xAg
0.51377 mg
0.48623 mg
-3
0.51377 x 10 /106.905095 0.48623/x
4.805851 x 10-6
0.48623 x 10-3 /x
Total moles Ag = 0.001/ 107.8682 = 9.270573 x 10-6 mol
Moles (xAg) = (9.270573 - 4.805851) x 10-6 = 4.46472 x 10-6 mol
x = mass (xAg)/ moles (xAg) = 0.48623 x 10-3/4.46472 x 10-6
(7 + 1.5 marks)
= 108.905 Da (contains 62 neutrons)
5