Solution 3 a ph o 2
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This document provides the solution to a theoretical question regarding the thermal vibration of surface atoms. (1) It calculates the wavelength of an incident electron beam and considers the interference between atomic rows on a surface. Two possible solutions for the phase angle are obtained. (2) An expression is derived for the total energy of vibration of surface atoms in the direction of the surface normal as a function of time and atomic mass. (3) Combining the energy expression with data about the material properties allows calculating the frequency and amplitude of vibration of the surface atoms. The calculated frequency is found to match results from a graph in the question.
![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
](https://image.slidesharecdn.com/solution3apho2-130927191920-phpapp02/85/Solution-3-a-ph-o-2-1-320.jpg)
![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](https://image.slidesharecdn.com/solution3apho2-130927191920-phpapp02/85/Solution-3-a-ph-o-2-3-320.jpg)
Solution 3 a ph o 2
- 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