Teoria Quântica e Estrutura Eletrônica

Prévia do material em texto

CHAPTER 7: QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 
 
204 
 
7.31 In this problem ni = 5 and nf = 3. 
 
 18 19
H 2 2 2 2
i f
1 1 1 1(2.18 10 J) 1.55 10 J
5 3
− −⎛ ⎞ ⎛ ⎞
⎜ ⎟Δ = − = × − = − ×⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
E R
n n
 
 
 The sign of ΔE means that this is energy associated with an emission process. 
 
 
34 8
19
(6.63 10 J s)(3.00 10 m/s)
1.55 10 J
−
−
× ⋅ ×
= = = =
Δ ×
6 31.28 10 m 1.28 10 nmhc
E
−λ × × 
 
 Is the sign of the energy change consistent with the sign conventions for exo- and endothermic processes? 
 
7.32 Strategy: We are given the initial and final states in the emission process. We can calculate the energy of 
the emitted photon using Equation (7.6) of the text. Then, from this energy, we can solve for the frequency of 
the photon, and from the frequency we can solve for the wavelength. The value of Rydberg's constant is 
2.18 × 10−18 J. 
 
 Solution: From Equation (7.6) we write: 
 
 H 2 2
i f
1 1⎛ ⎞
⎜ ⎟Δ = −
⎜ ⎟
⎝ ⎠
E R
n n
 
 18
2 2
1 1(2.18 10 J)
4 2
− ⎛ ⎞
Δ = × −⎜ ⎟
⎝ ⎠
E 
 ΔE = −4.09 × 10−19 J 
 
 The negative sign for ΔE indicates that this is energy associated with an emission process. To calculate the 
frequency, we will omit the minus sign for ΔE because the frequency of the photon must be positive. We 
know that 
 ΔE = hν 
 
 Rearranging the equation and substituting in the known values, 
 
 
19
34
(4.09 10 J)= or
(6.63 10 J s)
−
−
Δ ×
= =
× ⋅
14 1 146.17 10 s 6.17 10 HzE
h
−ν × × 
 
 We also know that λ =
ν
c . Substituting the frequency calculated above into this equation gives: 
 
8
7
14
m3.00 10
s 4.86 10 m
16.17 10
s
−
×
= = × =
⎛ ⎞×⎜ ⎟
⎝ ⎠
486 nmλ 
 
 Check: This wavelength is in the visible region of the electromagnetic region (see Figure 7.4 of the text). 
This is consistent with the fact that because ni = 4 and nf = 2, this transition gives rise to a spectral line in the 
Balmer series (see Figure 7.6 of the text). 
 
7.33 This problem must be worked to four significant figure accuracy. We use 6.6256 × 10−34 J⋅s for Planck’s 
constant and 2.998 × 108 m/s for the speed of light. First calculate the energy of each of the photons. 
 
 
34 8
19
9
(6.6256 10 J s)(2.998 10 m/s) 3.372 10 J
589.0 10 m
−
−
−
× ⋅ ×
= = = ×
λ ×
hcE