Chapter 1 - Solutions
1. a. Find the quantum number of the initial energy state
nf = 6 (fifth excited state),  = 5910 nm
c = 
 1
1 
 = R  2  2  = 3.29  1015 Hz
 n f n i 
 1
1 
 2  2
 n f n i 
 1
1 
c
= 3.29  1015 Hz  2  2 
λ
 n f n i 
1
3.00 108 ms 109 nm
1 
= 3.29  1015 Hz  2  2 

5910 nm
1m
n i 
 6
1
1

 0.0154
2
6
n i2
1
 0.0123 ;
ni = 9, therefore the emitted photon corresponds to the third
n i2
lowest-energy spectral line in the Humphreys series.
b.
B
A
Lymar
(nf = 1)
C
Balmer
(nf = 2)

Paschen
(nf = 3)
D
Humphreys
(nf = 6)
n=9
n=8
D
n=7
n=6
n=5
C
n=4
spectral line A
spectral line B
Eatom
spectral line C
n=3
A
spectral line D
B
n=2
n=1
2
2 

h 2  n 2x n y n Z 


8m  L2x L2y L2 
Z

3. a. Enxnynz =

(6.626  10 34 Js) 2 (1012 pm) 2
(8)(9.11  10 31 kg)

2
n 2y
n 2Z 
 nx 

 (550) 2 (700) 2 (550) 2 


 n2
n 2y
n 2Z 
x

= 6.024  1014 J 


 302500 490000 302500 
E211 = E112 = 1.12  1018 J  2nd excited state
E121 = 8.90  1019 J  1st excited state
E131 = 1.50  1018 J
b.
112, 211
E
121
111
c. There is one node in the first excited state.
d. There is one node in the second excited state.
5.
Atom
He
N
Na
P
Ground-state
configuration
1a2
1a21b42a1
1a21b42a22b3
1a21b42a22b42c3
Number of
unpaired electrons
0
1
1
3
7. a. Set (,) = 0 and solve for values of  and  that make this true.
sin() = 0 when  = 0 and   a line, not a nodal plane
cos(2) = 0 when 2 = /2, 3/2, 5/2, 7/2
and when  = /4, 3/4, 5/4, 7/4
Hence, the nodal planes are defined by  = /4 and 5/4 and  = 3/4 and 7/4,
but not the  = 0 and  line.
b. Set R(r) = 0 and solve for values of r that make this true.
R(r) = 0 = 2 (2  2) exp[/4] is true, when 2  2 = 0
The roots are  = 0 (which is not a radial node) and  = 2.
Hence, r = 2a0 is the only radial node
c.
l = # of angular nodes = 2
n – l – 1 = # of radial nodes; 1 = n – 2 – 1; n = 4
9. a. S 999.6 kJ/mol
Ca 589.8_______
Cl 1251______
Sc 631_________
Ar 1520______
Ti 658 kJ/mol___
b. S 2251 kJ/mol
Ca 1145______
Cl 2297_____
Sc 1235______
Ar 2666_____
Ti 1310 kJ/mol
K 418.8______
K 3051______
11. a. Zeff = Force of attraction between nucleus and electron
F = Q1 Q2
4or2
Where Q1 represents the charge of the nucleus
Where Q2 represents the charge of the electron
o is the permittivity of free space = 8.85419  1012 C2N1m2
r is the distance between charges (radius of the energy level)
for Be
F = Q1 Q2 = [4(1.6022  1019 C)] [1.6022  1019 C]
4or2

4(8.85419  1012 C2N1m2) 112 pm 

F = 8.24 
1018

1m

11012 pm 
N
For Mg
F = Q1 Q2 = [12(1.6022  1019 C)] [1.6022  1019 C]
4or2

4(8.85419  1012 C2N1m2) 160 pm 


1m

12
110 pm 
F = 1.73  1017 N
For Ca
F = Q1 Q2 = [20(1.6022  1019 C)] [1.6022  1019 C]
4or2

4(8.85419  1012 C2N1m2) 197 pm 


1m

11012 pm 
F = 2.34  1017 N
b. As the force of attraction between the nucleus and the outermost electron (Zeff)
increases because of the increase in the number of protons in the nucleus, the number of
electrons existing within the atom also increases. Electron-electron repulsion forces are
also increased within the atom. These electron-electron repulsion forces work against the
Zeff resulting in electrons in outer energy levels actually experiencing less attraction to the
nucleus, so the amount of energy required to remove the electron from the atom (the
ionization energy) is reduced as the number of electrons in the atom increases.