1

Sept. 12, 2007 Chem. 1410

Problem Set 1, Solutions

(1) (4 points) Note: since /cν λ= , then, given a small change in wavelength of δλ about a central value of λ , the corresponding frequency change is ( / )d dδν υ λ δλ= = -(c/ 2 )λ δλ . [If frequency increases, wavelength decreases: hence the – sign.] Thus, appealing to the equation in point (ii), i.e., ( ) ( )D Dλ νλ λ υ ν∆ = ∆ , we can write

2

4

8( )vD

c

π λ υ υ νλ

⋅ ∆ = ∆ [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:

2

2 3

8 8( )D

c cυ

π πνυλ

= = , QED

(2) (a) (2 points) In general, the average of a property A over a discrete probability distribution is given by j j

j

A p A< >=∑ , where jp is the normalized probability to be in state j, and jA is

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

/ /

0

/B Bjh k T jh k Tj

j

p e eν ν∞

− −

=

= ∑ . Furthermore, the energy of state j (corresponding to j photons in an

electromagnetic mode of frequency υ ) is jE jhυ= . Thus:

/ /

0 0

( ) /B Bjh k T jh k T

j j

E h je eν νυ ν∞ ∞

− −

= =

< >= ∑ ∑ .

(b) (i) (1 point) The infinite geometric series 2 11 ...

1x x

x+ + + =

− for 1x < (otherwise the series

diverges). In the case of our series expression for ( )D υ , we identify exp( / ) 1Bx h k Tυ= − < , and

thus: / 1( ) [1 ]Bh k TD e νυ − −= − , QED (ii) (2 points) Differentiating the series for ( )D υ term by term w.r.t. υ :

2

/

0

/ Bjh k T

j B

jhD e

k Tνυ

∞−

=

−∂ ∂ =∑

and thus: ( ) ( ) /Bk TN D

hν υ υ−= ∂ ∂ , QED.

iii)(1 point) Given the expression in (ii) and the explicit form for ( )D υ in (i), we can evaluate:

/

/ 2( )

(1 )

B

B

h k T

h k T

eN

e

υ

υυ−

−=−

and hence:

/( ) ( ) / ( )

[ 1]Bhv k T

hvE h N D

eν ν υ υ< >= =

− , QED.

P1.17) The observed lines in the emission spectrum of atomic hydrogen are given by

11

221

11 ,cm11

)cm()cm(~ nnnn

RH >

−= −−−ν . In the notation favored by spectroscopists,

hc

E==λ

ν 1~ and RH = 109,677cm-1. The Lyman, Balmer, and Paschen series refers to n1 =

1, 2, and 3, respectively, for emission from atomic hydrogen. What is the highest value of ν~ and E in each of these series?

The highest value for ν~ corresponds to 01 →n

. Therefore,

112

cm667,109cm1

1~ −− =

= HRν or Emax = 2.18×10-18 J for the Lyman series.

112

cm27419cm2

1~ −− =

= HRν or Emax = 5.45×10-19 J for the Balmer series, and

112

cm12186cm3

1~ −− =

= HRν or Emax = 2.42×10-19 J for the Paschen series.

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

22

2

22

1v

2

1

λλ eeee m

h

m

hmmE =

×==

eV 4.150J 10602.1

eV 1J 1041.2

)m10(kg 10109.92

)s J 10626.6(19

1721031

234

=×

××=×××

×= −−

−−

−

The electron must pass through an electrical potential of 150.4 V.