1. APHO II 2001 Theoretical Question 3 2013/09/28 00:19 A9/P9 p. 1 / 4
[Solution] Theoretical Question 3
Thermal Vibration of Surface Atoms
(1) (a) The wavelength of the incident electron is

A53.11053.1
0.641060.11011.92
1063.6
2
10
1931
34
=×=
×××××
×
=
==
−
−−
−
m
meV
h
p
h
λ
(b) Consider the interference between the atomic rows on the surface as shown in
Fig. 3c.
The path difference between electron beam 1 and 2 is
λφφ nb =−=∆ )sin(sin 
Given 
 0.15=φ , λ=

A35.1 and

A77.2
2
92.3
2
===
a
b , two solutions
are possible.
(i) When n = 0, φ φ= o = 15.0
(Answer 1)
(ii) When n =1
53.11)15sin(sin77.2 ×=−=∆ 
 φ
812.0
77.2
72.053.1
sin =
+
=φ
[Solution] (continued) Theoretical Question 3
Vibration of Surface Atoms

3.54=φ (Answer 2)
1
Fig.3c
φ
φ
e-
1
2
[011]b
e-
φ
φ
n

n


2. APHO II 2001 Theoretical Question 3 2013/09/28 00:19 A9/P9 p. 2 / 4
For n = 2, no solution exists as 531215772 .)sin(sin. ×=−= 
 φ∆ and φsin > 1.
(2) 〉⋅〈−= 2
0 )Ku(expII

∆
For the specularly reflected beam, we have from Fig. 3d
θcos2KKKK =−′=∆

xˆ
where xˆ is the unit vector in the direction of the surface normal. Take the x-
component of u

, we then obtain
><−>⋅<−
== )t(ucosKcosK)t(u xx
eIeII
222222
4
0
4
0
θθ (2)
The vibration in the direction of the surface normal of the surface atoms is
simple harmonic, take
tAtux ωcos)( =
 22
11 2
0 0
2
2222 AA
dttcosAdtu)t(ux =⋅===〉〈 ∫ ∫
τ
τ
ω
ττ
τ τ
∴ 〉〈= )(2
22
tuA x
The total energy E is thus given by
><⋅== )(2
2
1
2
1 22
tuCCAE x = >< )(
2
tuC x
= ><′ )(
22
tum xω
2
Fig. 3d
surface
θK

K

′
K∆

3. APHO II 2001 Theoretical Question 3 2013/09/28 00:19 A9/P9 p. 3 / 4
[Solution] (continued) Theoretical Question 3
Vibration of Surface Atoms
Therefore, one obtains
>< )(
2
tux
= E / ( 2
ωm′ )
Tku'mE Bx =><=
22
ω
where 'm is the mass of the atom. From either of the above two equations,
one then has the following equality
222
2
4 fm
Tk
m
Tk
u BB
x
πω ′
=
′
>=< (3)
From eq. (3) and eq. (2), one obtains
22
22
4
cos4
0
fm
Tk
K B
eII π
θ
′
−
=
where
λ
ππ 22
==
h
p
K . Accordingly,
TM
T
fm
k
eIeII
B
′−
−
== 0
'
cos4
0
22
2
λ
θ
(4)
and
TM
I
I
n ′−=
0

From the plot of
0I
I
n versus T，one obtains the slope
22
2
cos4
λ
θ
fm
k
M B
′
=′ (5)
The slope of the curve can be estimated from Fig. 3b and leads to the result
3
103.2 −
×=′M .
Using the following data in Eq.(5),
K/J.kB
23
10381 −
×=
m10
1053.1 −
×=λ
m′ = 195.1 3
10−
× /(6.02 23
10× ) = 3.24 25
10−
× kg/atom
[Solution] (continued) Theoretical Question 3
Vibration of Surface Atoms
3

4. APHO II 2001 Theoretical Question 3 2013/09/28 00:19 A9/P9 p. 4 / 4
one finds
2102
23
3
223
3
10531
10026
101195
15103814
1032
).(f
.
.
cos.
.
−
−
−
−
×××
×
×
⋅××
=×

The solution for frequency is then
242
100.3 ×=f (new) Hzf 12
107.1 ×=⇒ Answer (a)
From 22
2
4 f'm
Tk
u B
x
π
=>< ， KT 300= , one finally obtains
222
242
23
3
23
2
101.1
100.34
1002.6
101.195
3001038.1
mux
−
−
−
×=
×××
×
×
⋅×
>=<
π
(new)
and

A10.0100.1 112
=×=>< −
mux
(new)
Ans
4