# Homework Solutions Problem 7 - 37

Document Sample

```					                                     Homework Solutions

Problem 7 – 37

2                                                  -7
λ = 7.80 x 10 nm               1m               = 7.80 x 10          m
9
1 x 10 nm

8               -7            14
ν = c/λ = 2.9979 x 10 m/s / 7.80 x 10 m = 3.84 x 10 Hz

Problem 7 – 38

6              7
ν = 99.5 MHz         1 x 10 Hz = 9.95 x 10 Hz

1 MHz

8               7
λ = c/ν = 2.9979 x 10 m/s / 9.95 x 10 Hz = 3.01 m

Problem 7 – 39

8                        10
λ = 1.0 cm = 0.010 m          ν = c/λ = 2.9979 x 10 m/s / 0.010 m = 3.0 x 10 Hz

-34                      10                      -24
E = hν = 6.626 x 10           J sec x 3.0 x 10        Hz = 2.0 x 10            J/photon

-24                            23
2.0 x 10         J/photon x 6.022 x 10         photons/mol = 12 J/mol

Problem 7 – 43

-19                    -34                        15
E = hν ; n = E/h = 7.21 x 10            J / 6.626 x 10         J sec = 1.09 x 10          Hz

8               15             -7
λ = c/ν = 2.9979 x 10 m/s / 1.09 x 10 Hz = 2.76 x 10 m = 276 nm
Problem 7 – 45

8                        8
a) v = 2.9979 x 10 m/s x 0.90 = 2.6981 x 10 m/s

-34                       -27                   8
λ = h/mv = 6.626 x 10              J sec/(1.67 x 10         kg x 2.6981 x 10 m/s)

-15
λ = 1.47 x 10         m

-34                                           -34
b) λ = h/mv = 6.626 x 10             J sec/(0.15 kg x 10. m/s) = 4.4 x 10           m

Problem 7 – 112

-19                    -34                       14
E = hν ; n = E/h = 3.90 x 10            J / 6.626 x 10         J sec = 5.89 x 10        Hz

8               14             -7            -5
λ = c/ν = 2.9979 x 10 m/s / 5.89 x 10 Hz = 5.09 x 10 m = 5.09 x 10 cm

GREEN !

Problem 7 – 49

-18         2     2                -19
a) ΔE = -2.178 x 10           J (1/2 – 1/3 ) = -3.025 x 10     J

E = hν = hc/λ

-34                     8                  -19
λ = hc/E = [(6.626 x 10             J sec)(2.9979 x 10 m/s)]/ 3.025 x 10     J

-7
λ = 6.567 x 10        m = 656.7 nm

-18         2     2                -19
b) ΔE = -2.178 x 10           J (1/2 – 1/4 ) = -4.084 x 10     J

E = hν = hc/λ

-34                     8                  -19
λ = hc/E = [(6.626 x 10             J sec)(2.9979 x 10 m/s)]/ 4.084 x 10     J

-7
λ = 4.864 x 10        m = 486.4 nm
-18         2     2                -18
c) ΔE = -2.178 x 10             J (1/1 – 1/2 ) = -1.634 x 10     J

E = hν = hc/λ

-34                     8                  -18
λ = hc/E = [(6.626 x 10                 J sec)(2.9979 x 10 m/s)]/ 1.634 x 10     J

-7
λ = 1.216 x 10          m = 121.6 nm

Problem 7 – 51

n=5

n=4

E                                       n=3
a        b

n=2

c
n=1

Problem 7 – 53

n = 1 to n = 5

-18         2     2                -18
ΔE = -2.178 x 10           J (1/5 – 1/1 ) = -2.019 x 10     J

E = hν = hc/λ

-34                     8                  -18
λ = hc/E = [(6.626 x 10                 J sec)(2.9979 x 10 m/s)]/ 2.019 x 10     J

-8
λ = 9.500 x 10          m = 95.00 nm ; the light is not energetic enough
n = 2 to n = 6

-18         2     2                -19
ΔE = -2.178 x 10         J (1/6 – 1/2 ) = -4.840 x 10     J

E = hν = hc/λ

-34                     8                  -19
λ = hc/E = [(6.626 x 10           J sec)(2.9979 x 10 m/s)]/ 4.840 x 10     J

-7
λ = 4.104 x 10        m = 410.4 nm; the light is energetic enough

Problem 7 – 55

n = 1 to n = ∞

-18         2                -18
ΔE = -2.178 x 10         J (1/1 ) = -2.178 x 10     J

E = hν = hc/λ

-34                     8                  -18
λ = hc/E = [(6.626 x 10           J sec)(2.9979 x 10 m/s)]/ 2.178 x 10     J

-8
λ = 9.120 x 10        m = 91.20 nm

n = 2 to n = ∞

-18         2                -19
ΔE = -2.178 x 10         J (1/2 ) = -5.445 x 10     J

E = hν = hc/λ

-34                     8                  -19
λ = hc/E = [(6.626 x 10           J sec)(2.9979 x 10 m/s)]/ 5.445 x 10     J

-7
λ = 3.648 x 10        m = 364.8 nm
Problem 7 – 59

n = 1, 2, 3, 4, ……. ;  = 0, 1, 2, 3,….(n-1) ; m = -…0…+

Problem 7 – 60

1p, 3f, 2d

Problem 7 – 61

b) If n = 1, the only value of  is 0.

d) If  = 0, the only value of m is 0.

Problem 7 – 62

b) If  = 3, m can not equal 4.

c) n ≠ 0.

d)  ≠ -1.

Problem 7 – 65
2
5p :3 orbitals; 3dz :1 orbital; 4d :5 orbitals; n = 5 :25 orbitals; n = 4 :16 orbitals

Problem 7 – 66

1p: 0 electrons; 6dx2-y2: 2 electrons; 4f: 14 electrons; 7py: 2 electrons;
2s: 2 electrons; n = 3: 18 electrons

Problem 7 – 67

a) 32 electrons b) 8 electrons c) 25 electrons d) 10 electrons e) 6 electrons

Problem 7 – 68

a) 0 electrons b) 1 electron c) 18 electrons d) 0 electrons e) 2 electrons
Problem 7 – 69

2 2 6 2 2
Si = 1s 2s 2p 3s 3p

2 2 6 2 6 2 10 1
Ga = 1s 2s 2p 3s 3p 4s 3d 4p

2 2 6 2 6 2 10 3
As = 1s 2s 2p 3s 3p 4s 3d 4p

2 2 6 2 6 2 10 2
Ge = 1s 2s 2p 3s 3p 4s 3d 4p

2 2 6 2 1
Al = 1s 2s 2p 3s 3p

2 2 6 2 6 2 10 6 2 10
Cd = 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d

2 2 6 2 4
S = 1s 2s 2p 3s 3p

2 2 6 2 6 2 10 4
Se = 1s 2s 2p 3s 3p 4s 3d 4p

Problem 7 – 71

2 2 6 2 6 2 1
Sc = 1s 2s 2p 3s 3p 4s 3d

2 2 6 2 6 2 6
Fe = 1s 2s 2p 3s 3p 4s 3d

2 2 6 2 3
P = 1s 2s 2p 3s 3p

2 2 6 2 6 2 10 6 2 10 6 1
Cs = 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s

2 2 6 2 6 2 10 6 2 10 6 2 7
Eu = 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f

2 2 6 2 6 2 10 6 2 10 6 2 14 8
Pt = 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d

2 2 6 2 6 2 10 6 2 10 6
Xe = 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p

2 2 6 2 6 2 10 5
Br = 1s 2s 2p 3s 3p 4s 3d 4p
Problem 7 – 75

2 2 5
a) F = 1s 2s 2p

2 2 6 2 6 1
b) K = 1s 2s 2p 3s 3p 4s

2 2          2 2 6 2            2 2 6 2 6 2
c) Be = 1s 2s ; Mg = 1s 2s 2p 3s ; Ca = 1s 2s 2p 3s 3p 4s

2 2 6 2 6 2 10 6 2 10 1
d) In = 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p

2 2 2           2 2 6 2 2
e) C = 1s 2s 2p ; Si = 1s 2s 2p 3s 3p

f) element # 118 =

2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 10 6
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p

Problem 7 – 83

a) Be < Mg < Ca       b) Xe < I < Te   c) Ge < Ga < In

Problem 7 – 85

a) Ca < Mg < Be       b) Te < I < Xe   c) In < Ga < Ge

Problem 7 – 87

+
a) Li b) P c) O        d) Cl e) Cu

Problem 7 – 94

The electron affinities become more negative (more exothermic) across the
period because the effective nuclear charge is increasing due to more protons
in the nucleus.

The electron affinity for phosphorus is out of line because in order for
phosphorus to gain an additional electron it must be placed in an orbital that
already contains an electron causing the electrons to pair up. This causes a
decrease in the effective nuclear charge.
Problem 7 – 95

C and N            Ar and Br

a)        C                    Br

b)        N                    Ar

c)        C                    Br

Problem 7 – 135

a) Each time an electron is removed the effective nuclear charge increases
because the number of protons exceeds the number of electrons. Electrons
are therefore more difficult to remove (ionization energy increases).

b) The fourth electron to be removed from aluminum will have to come from
n = 2 which in a core electron that is closer to the nucleus.

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 22 posted: 5/12/2010 language: English pages: 8
How are you planning on using Docstoc?