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c _
9
d _
d _
7.3. d = 6.00 × 10 km v = ∴ t = Step 1: c = λ f ∴ f =

λ
v
12
= 6.00 × 10 m t 8 m _
12
__ 3.00 × 10
6.00 × 10 m

8
c = 3.00 × 10 m/s t = 8 m _ = __
s

3.00 × 10 –7
t = ? s 5.60 × 10 m
14
12
_ s _ = 5.36 × 10 Hz
6.00 × 10

m

= ×
m
3.00 × 10 8
Step 2: E = hf
4
= 2.00 × 10 s –34 14
= (6.63 × 10 J·s)(5.36 × 10 Hz)
4
__
2.00 × 10 s

–19
= s _ = 3.55 × 10 J
3,600

h Step 3: Since one photon carries an energy of 3.55 × 10 J and
–19
4
_ h _ the overall intensity is 1,000.0 W, each square meter
2.00 × 10

s

= ×
s
3.600 × 10 3 must receive an average of
= 5.56 h _ J

1,000.0
__
s

–19 _
7.4. From equation 7.1, note that both angles are measured from the 3.55 × 10 J
normal and that the angle of incidence (θ i ) equals the angle of photon
3
refl ection (θ r ), or __ _ _
photon
1.000 × 10 J

–19 s ×
θ i = θ r ∴ θ i = 10° 3.55 × 10 J
21 _
8

7.5. v = 2.20 × 10 m/s c _ 2.82 × 10 photon
n = s
v
8
c = 3.00 × 10 m/s 8 m _ c _
14
3.00 × 10 7.8. (a) f = 4.90 × 10 Hz c = λ f ∴ λ =

s
n = ? __ f
= 8 m _ 8
2.20 × 10 c = 3.00 × 10 m/s
8 m _
s 3.00 × 10

s

= 1.36 λ = ? λ = __

14 1 _

4.90 × 10
s
According to Table 7.1, the substance with an index of
8
3.00 × 10 m _ _
refraction of 1.36 is ethyl alcohol. _ s

= 14 s

×

4.90 × 10 1
7.6. (a) From equation 7.3: –7
–7
λ = 6.00 × 10 m = 6.12 × 10 m
8
c = 3.00 × 10 m/s (b) According to Table 7.2, this is the frequency and
wavelength of orange light.
f = ?
20
c _ 7.9. f = 5.00 × 10 Hz

c = λ f ∴ f =
λ h = 6.63 × 10 J·s
–34
8 m _
3.00 × 10

__ E = ?
s

f =
–7
6.00 × 10 m E = hf
8
_ 1 –34 20 1 _
3.00 × 10 m _ _

×

= –7 s m = (6.63 × 10 J·s)(5.00 × 10 )

s
6.00 × 10 1 _
20
–34

14 1 _ = (6.63 × 10 ) (5.00 × 10 ) J·s ×
= 5.00 × 10 s

s
–13
= 3.32 × 10 J
14
= 5.00 × 10 Hz
c _
7.10. λ = 1.00 mm Step 1: c = λ f ∴ f =
(b) From equation 7.4: λ
= 0.001 m
8 m _
14
f = 5.00 × 10 Hz 3.00 × 10

s
f = ? __

–34
h = 6.63 × 10 J·s 8 f = 1.00 × 10 m
–3
c = 3.00 × 10 m/s
E = ? _ 8 1
3.00 × 10 m _ _
–34
h = 6.63 × 10 J·s = ×

E = hf 1.00 × 10 –3 s m
14 1 _
–34

= (6.63 × 10 J·s)(5.00 × 10 ) E = ? = 3.00 × 10 Hz
11
s
1 _
–34
14
= (6.63 × 10 ) (5.00 × 10 ) J·s × Step 2: E = hf

s
11 1 _
–19
= 3.32 × 10 J = (6.63 × 10 –34 J·s) ( 3.00 × 10 )
s
1 _

7.7. First, you can find the energy of one photon of the peak = (6.63 × 10 ) (3.00 × 10 ) J·s ×
–34
11

–7
intensity wavelength (5.60 × 10 m) by using equation 7.3 to s
–22

find the frequency, then equation 7.4 to find the energy: = 1.99 × 10 J
662 APPENDIX E Solutions for Group A Parallel Exercises E-20