Sept. 12, 2007
Chem. 1410
Problem Set 1, Solutions
(1) (4 points) Note: since ν = c / λ , then, given a small change in wavelength of δλ about a
central value of λ , the corresponding frequency change is δν = (dυ / d λ )δλ = -(c/ λ 2 )δλ . [If
frequency increases, wavelength decreases: hence the – sign.] Thus, appealing to the equation in
point (ii), i.e., Dλ (λ )∆λ = Dν (υ )∆ν , we can write
8π λ 2
⋅
∆υ = Dv (υ )∆ν
λ4 c
[1]
Note, either side of this equation gives the number of modes in the selected interval, and thus
∆υ and ∆λ are the magnitudes of the (small) frequency and wavelength intervals, respectively.
Finally, Eq. 1 above implies:
Dυ (υ ) =
8π
8πν 2
=
λ2c
c3
, QED
(2) (a) (2 points) In general, the average of a property A over a discrete probability distribution
is given by < A >= ∑ p j Aj , where p j is the normalized probability to be in state j, and Aj is
j
the value of the property A in associated with state j. In the case of interest here the states are
labeled by j=0,1,2,…∞ and the normalized probability to be in state j is the Boltzmann factor
∞
p j = e − jhν / kBT / ∑ e − jhν / kBT . Furthermore, the energy of state j (corresponding to j photons in an
j =0
electromagnetic mode of frequency υ ) is E j = jhυ . Thus:
∞
∞
j =0
j =0
< E (υ ) >= hν ∑ je − jhν / kBT / ∑ e − jhν / kBT .
1
for x < 1 (otherwise the series
1− x
diverges). In the case of our series expression for D(υ ) , we identify x = exp(−hυ / k BT ) < 1 , and
thus:
(b) (i) (1 point) The infinite geometric series 1 + x + x 2 + ... =
D(υ ) = [1 − e − hν / kBT ]−1
, QED
(ii) (2 points) Differentiating the series for D(υ ) term by term w.r.t. υ :
1
− jh − jhν / kBT
e
j =0 k BT
∞
∂D / ∂υ = ∑
and thus:
N (ν ) =
− k BT
∂D(υ ) / ∂υ , QED.
h
iii)(1 point) Given the expression in (ii) and the explicit form for D(υ ) in (i), we can evaluate:
N (υ ) =
e − hυ / kBT
(1 − e − hυ / kBT ) 2
and hence:
< E (ν ) >= hν N (υ ) / D (υ ) =
hv
[e hv / kBT − 1]
, QED.
2
P1.2)
P1.7)
P1.17) The observed lines in the emission spectrum of atomic hydrogen are given by
 1
1 
ν~ (cm −1 ) = RH (cm −1 ) 2 − 2 cm −1 , n > n1 . In the notation favored by spectroscopists,
 n1 n 
1 E
ν~ = = and RH = 109,677cm-1. The Lyman, Balmer, and Paschen series refers to n1 =
λ hc
1, 2, and 3, respectively, for emission from atomic hydrogen. What is the highest value of
ν~ and E in each of these series?
1
The highest value for ν~ corresponds to → 0 . Therefore,
n
1
 
ν~ = RH  2 cm −1 = 109,667cm −1 or Emax = 2.18×10-18 J for the Lyman series.
1 
 1 
ν~ = RH  2 cm −1 = 27419cm −1 or Emax = 5.45×10-19 J for the Balmer series, and
2 
 1
ν~ = RH  2 cm −1 = 12186cm −1 or Emax = 2.42×10-19 J for the Paschen series.
3 
P1.19) If an electron passes through an electrical potential difference of 1 V, it has an
energy of 1 electron-volt. What potential difference must it pass through in order to have
a wavelength of 0.100 nm?
2
 h 
h2
1
1
2
 =
E = m e v = m e ×
m
2
2
λ
2 me λ 2
 e 
(6.626 × 10 −34 J s) 2
1 eV
= 2.41 × 10 −17 J ×
= 150.4 eV
−31
−10
2
2 × 9.109 × 10 kg × (10 m)
1.602 × 10 −19 J
The electron must pass through an electrical potential of 150.4 V.
=