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c _

7.11. The index of refraction is found from n = , where n is the The relationship between the frequency (f ) and energy (E) of a
v
index of refraction of a transparent material, c is the speed of photon is found in equation 7.4, E = hf.
light in a vacuum, and v is the speed of light in the material.
E = hf

The index of refraction of glass is found in Table 7.1 (n = 1.50).
14 1 _
–34
= (6.63 × 10 J·s)(6.50 × 10 )

n = 1.50 s
c _ c _
8

c = 3.00 × 10 m/s n = ∴ v = –34 14 1 _

v n = (6.63 × 10 )(6.50 × 10 ) J·s ×
s
8 m _
v = ? 3.00 × 10

s
J·s
–19 _
v = __ = 4.31 × 10

1.50 s
–19
8 m _ = 4.31 × 10 J

= 2.00 × 10
s
7.15 List the known and unknown quantities:
7.12. List the known and unknown quantities:
–7
wavelength λ = 5.60 × 10 m
–7
wavelength λ = 5.60 × 10 m
–34
Planck’s constant h = 6.63 × 10 J·s
–8
speed of light c = 3.00 × 10 m/s
8
speed of light c = 3.00 × 10 m/s
frequency f = ?
frequency f = ?
The relationship between the wavelength (λ), frequency (f ), and

First, you can find the energy of one photon of the peak
speed of light in a vacuum (c) is found in equation 7.3, c = λf. –7
intensity wavelength (5.60 × 10 m) by using the relationship
c _ between the wavelength (λ), frequency (f ), and speed of light

c = λ f ∴ f =
λ in a vacuum (c), c = λ f; then use the relationship between the
8 m _ frequency (f ) and energy (E) of a photon, E = hf, to fi nd the
3.00 × 10

__ energy:
s
=

–7
5.60 × 10 m c _

Step 1: c = λ f ∴ f =
8
3.00 × 10 m _ _
_ 1 λ
= –7 s m

×

5.60 × 10 3.00 × 10
8 m _

s
14 1 _ __

= 5.40 × 10 = –7

s 5.60 × 10 m
14
8
= 5.40 × 10 Hz _ 1
3.00 × 10 m _ _
= –7 s
×

5.60 × 10 m
7.13. List the known and unknown quantities:
14 1 _

14
frequency f = 5.00 × 10 Hz = 5.36 × 10
s
–34
Planck’s constant h = 6.63 × 10 J·s = 5.36 × 10 Hz
14
energy E = ?
Step 2: E = hf

The relationship between the frequency (f ) and energy (E) of a 14 1 _
= (6.63 × 10 –34 J·s) ( 5.36 × 10 )
photon is found in equation 7.4, E = hf. s
1 _
E = hf = (6.63 × 10 )(5.36 × 10 ) J·s ×
14
–34

s
14 1 _
–34

= (6.63 × 10 J·s)(5.00 × 10 ) J·s
s –19 _

= 3.55 × 10
s
1 _
–34
14

= (6.63 × 10 )(5.00 × 10 ) J·s × –19
s = 3.55 × 10 J
J·s
–19 _ Step 3: Since one photon carries an energy of 3.55 × 10 J and
–19

= 3.32 × 10
s
the overall intensity is 1,000.0 W, for each square meter
–19
= 3.32 × 10 J there must be an average of
7.14. List the known and unknown quantities: _ J
1,000.0

__
s
14
frequency f = 6.50 × 10 Hz J
–19 _
3.55 × 10
–34
Planck’s constant h = 6.63 × 10 J·s photon
–3
1.000 × 10 _ _
energy E = ? __ J photon

–19 s ×

3.55 × 10 J
photons
21 _

2.82 × 10
s
E-21 APPENDIX E Solutions for Group A Parallel Exercises 663