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Material Type: Assignment; Professor: Castellani; Class: Principles of Chemistry I; Subject: Chemistry; University: Marshall ; Term: Fall 2008;
Typology: Assignments
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Chapter 6 Homework Solutions
s
= 3.00 x 10^17 s-^1
b) λ = (^)
s
10 - 1 = 0.0039 m (3.9 mm)
c) (a) = yes, (b) = no
d) distance = (25.5 fs)
1 s 3 00 10 fs
x 10 m (^15) s
8
= 7.65 x 10-^6 m
s
300 x 10 m 1 m
10 nm 589 nm
= 5.09 x 10^14 s-^1
b) E = (6.6262 x 10-^34 J•s)(5.09 x 10^14 s-^1 ) (^)
mol
c) E = (6.6262 x 10-^34 J•s) (^)
s
300 x 10 m 1 m
10 nm 589 nm
= 3.37 x 10-^19 J/photon
d) Probably not. Each element has a unique set of lines. It would be random chance that strontium had a line at 589 nm.
6.022x 10 molecules
1 mol kJ
941 kJ/mol
= 2.36 x 10^15 s-^1
λ = (^)
s
2 .998x 10 m
15 - 1 = 1.27 x 10
4
6.626x 10 J s
2.18 x 10 J = 3.08 x 10^15 s-^1
λ 41 = (^)
m
10 nm 3.08x 10 s
3.00 x 10 m/s^9 15 - 1
8 = 97.3 nm (ultraviolet light)
E 41 = (6.626 x 10-^34 J•s)(3.08 x 10^15 s-^1 ) = 2.04 x 10-^18 J (released) In the book, the constant is shown with a negative value. The difference between the book & these problems is that here you first calculate the frequency of the light, which must have a positive value. Thus, the sign convention used by the problems must be ignored initially. Plugging into nearly identical equations. ν 52 = 6.91 x 10^14 s-^1 λ 52 = 434 nm (visible light) E 52 = 4.58 x 10-^19 J (released)
ν 36 = 2.74 x 10^14 s-^1 λ 36 = 1090 nm (infrared light) E 36 = 1.82 x 10-^19 J (absorbed)
1000 m
1 km 1 hr
3600 s 50 km
1 hr 85 kg
b) λ = (^)
1 kg
1000 g (10.0g)(250m/s)
c) λ = (^)
1 kg
1000 g 1.66x 10 g
1 amu 2.5x 10 m
1 s 6.941amu
d) λ = (^)
1 kg
1000 g 1.66x 10 g
1 amu 550 m
1 s 48.00amu
= 1.5 x 10-^11 m
b) Ni: [Ar] 4 s^2 3 d^8 d) Cd: [Kr] 5 s^2 4 d^10 f) Pb: [Xe] 6 s^2 4 f^14 5 d^10 6 p^2