Answers to Chapter 9

Please speak up if you believe your answer is more correct than answers listed below!

30.
a 7.160 x10^14 Hz
b visible
c violet

36.

a. minimum wavelength 91.174 nm

maximum wavelength 121.56 nm

b. n=5

c n = 2.502. This is not an integer so there is no line with this wavelength.

40 3.4x10^-22 J/photon

42 One sufficiently energetic photon can dislodge one and only one electron.
Two photons cannot combine to contribute their joint energy.
Conversely, an excessively energetic photon cannot eject more than one electron.

48
n=8

50
700 m/s (1sf)

58
The probability density of finding an electron at the nucleus is high, but the volume of the nucleus is very small. The probability density is lower further from the nucleus, but there are many "unit volumes" at a specified distance from the nucleus.
Remember the dartboard analogy.

60
a l=2
b n=3 or greater
c ms = +1/2 or -1/2
d n=1 ml=0 ms =1/2 or -1/2

64
a one
b two
c ten
d 18
e five

66
b--make sure you have 42 e [Kr] 4d-5 5s-1

70
a 5 --n s 2 and np 3
b n=4 is completely full with 32 e
c O,S, Se,Te, Po
d 2
e10+14 = 24

76.
The longest wavelength will be the smallest frequency and energy.
This will be a transition from level n+1 to n.
We can calculate that (1/n^2 -1/(n+1)^2) = 0.0123.
Unless you are an algebraic whiz, just start picking values for n and n+1.
For this transition n=5 and n+1=6.
The series is then n=6,7,8,9 etc down to n=5.

88.
From Pressure, Temperature and Volume calculate 0.041 moles air.
(we'll do this in Chap. 6)
Using ppm volume find 1x10^-8 mole ozone.
4.8 x 10^-3 J or 4.7 x 10^-3 J.
Using photons to drive chemical reactions is very important!