Chem 151
Honors Chemistry
Fall 2007
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4: 7, 18, 22, 30, 35
7. The speed of sound in dry air at 20C is 343.5 m/s and the frequency of the sound from
the middle C note on a piano is 261.6 cps. Calculate the wavelength and time it takes to
travel 30 m across a concert hall.
!" = speed
" = speed / !
"=
343.5ms #1
= 1.313m
261.6s #1
dis tan ce = speed * time
30m = 343.5m / s * t
30m
t=
= 87.3ms
343.5ms !1
18. In a Franck-Hertz experiment on H atoms, the first two excitation threshold occur at
10.1 eV and 11.9 eV. Three optical emission lines are associated with these levels.
Sketch an energy-level diagram for H atoms based on this information. Identify the three
transitions associated with these emission lines. Calculate the wavelength of each
emitted line.
The wavelength of these lines follows from the relationship that ∆E=hc/λ
6.63x10 "34 x3x10 8
= 1230 Å, 1045Å, AND 691Å. The first two lines are in the
So = ! =
1.6x10 "19 ! E(eV )
Lyman series, and the last in the Balmer.
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22. The Be3+ ion has a single electron. Calculate the frequencies and wavelengths of
light in the ion’s emission spectrum for the first three lines in each of the series that are
analogous to the Lyman and Balmer series of neutral hydrogen. In what region of the
spectrum does this radiation lie?
Since there is only one electron, we can use the Bohr equations, but must now
realize that Z=4. Thus eq 15.11 becomes
# 1
# 1
1&
1&
! = 3.29x1015 *16 % 2 " 2 ( = 5.26x1016 % 2 " 2 (
$ n f ni '
$ n f ni '
For the Lyman lines, nf=1, and ni=2,3,4 (for the first three lines). For the Balmer series,
nf=2, (see fig 15.19) and ni=3,4,5.
Thus for the Lyman lines, the frequencies are (3.9450 4.6756 4.9313)x1016 giving
wavelengths of (7.65, 6.41 & 6.08) nm. For the Balmer, the frequencies are ( 0.7306
0.9862 1.1046)x1016. The wavelengths are 41.06, 30.4 & 27.2 nm. These are
wavelengths that lie in the X-ray region of the spectrum.
30. Suppose we picture an electron in a chemical
bond as being a wave with fixed ends. Take the
length of the bond to be 1Å (0.1nm)
a) Calculate the wavelength of the electron
wave in its ground state and in its first excited state.
The groundstate has wavelength 2Å and the first
excited state 1Å
b) How many nodes does the excited state
have? One. (we won’t count the ends because all of
the modes have fixed ends.) (see fig)
35. a) The position of an electron is known to be within 10Å. What is the minimum
uncertainty in the velocity?
∆p∆x>h/4π ∆x≈10Å (10x10-10m), so ∆p>6.63x10-34Js/(4πx10-9m)=
5.27x10-24Js/m=∆(mv) m=9.1x10-31kg, so ∆v=5.27x10-24/9.1x10-31 = 5.8x104m/s
b) Repeat this calculation for a helium atom.
∆p>6.63x10-34Js/(4πx10-9m)=
5.27x10-24Js/m=∆(mv) m=4x1.66x10-27kg, so ∆v=5.27x10-24/4x1.66x10-27= 7.9m/s