4. Introduction to Ion-Solid
Interactions
(the two-body collision)
So, how much energy can be transferred from a projectile
particle, such as an ion, to a lattice atom in a target solid?
To solve this problem, it is useful to consider
details of the so-called Two-Body Interaction
or Two-Body Collision.
This problem was made famous by astronomers
when they worked out details concerning the
orbital motion of the planets about the Sun.
Two-Body Collision
Calculation of the energy
transferred during the scattering
interaction
Two-Body Collision
Energy transfer calculation
Ordinarily, in a first-year physics
class, you would be asked to solve
this problem using Conservation of
Energy
and
Conservation
of
Momentum considerations.
This problem is algebraically intractable using these
rules of conservation. However, the problem is solved
elegantly using a transformation of coordinates to the
Center-of-Mass (CM) Frame-of-Reference.
Proof that the net momentum in the Center-of-Mass
(CM) coordinate system is zero - at all times.
v
v
v
m1
m2
RCM ! m1 +m2 r1 + m1 +m2 r2
v
v
v
m1
If r2 = 0, then RCM = m1 +m2 r1
v v
v*
v v
v*
r1 = RCM + r1
r2 = RCM + r2
v
v
v v*
v
v
v v*
m1
m2
m1
m2
r1 = m1 +m2 r1 + m1 +m2 r2 + r1
r2 = m1 +m2 r1 + m1 +m2 r2 + r2
v*
v
v
v*
v
m1
m2
! m1 v
m2
r1 = 1! m1 +m2 r1 ! m1 +m2 r2
r2 = m1 +m2 r1 + 1! m1 +m2 r2
v*
v m1 m2 v
v * ! m1 m2 v
v
m12
m2 2
m1 r1 = m1 ! m1 +m2 r1 ! m1 +m2 r2 m2 r2 = m1 +m2 r1 + m2 ! m1 +m2 r2
(
(
v
m r = (m !
•
1
*
1
1
)
m12
m1 +m2
v•*
v•*
m1 r1 + m2 r2 =
(
)
v
)r!
•
1
(
m1 m2
m1 +m2
v•
r2
m1 (m1 +m2 )!m12 !m1m2
m1 +m2
v•*
m2 r2 =
) (
v•
r1 +
! m1 m2
m1 +m2
)
(
)
v
v
r + (m !
)r
v
)r =0
•
1
2
m2 (m1 +m2 )!m1m2 !m12
m1 +m2
m2 2
m1 +m2
•
2
•
2
Calculate the energy transfer from projectile m1 to target atom m2.
Step 1. Transform from laboratory to CM coordinates.
v
RCM =
m1
m1 +m2
v v•
u = RCM =
v
r1
m1
m1 +m2
v•
r1 =
m1
m1 +m2
v
v10
v
( v20 = 0 )
v
Subtract the velocity
of the CM ( u ) from vthe initial velocity of
v
projectile m1 ( v10 ) and target atom m2 ( v20 ).
v * v
v
v10 = v10 ! u
v * v
v
v
v
v20 = v20 ! u
1
= v10 ! m1m+m
v
10
2
v
= 0!u
v
m1
= 1! m1 +m2 v10
m1 +m2 !m1 v
= m1 +m2 v10
(
(
v *
v10 =
)
)
m2
m1 +m2
v
v10
v *
v
v20 = ! u
Calculate the energy transfer from projectile m1 to target atom m2.
Step 2. Perform the scattering interaction in CM coordinates.
Properties of the Center-of-Mass (CM) :
1. The net linear momentum in CM coordinates is zero.
2. In the absence of external forces (an inertial frame-ofreference), the velocity of the center of mass is constant
v
(equal to u ).
3. In a two-body system, the center-of-mass must lie on the
line adjoining the two particles at all times.
4. In a two-body system, the velocity vectors for the two
particles in CM coordinates are parallel at all times.
5. In a two-body system, the magnitude of the velocity of
each particle in CM coordinates is unchanged by the
collision.
Calculate the energy transfer from projectile m1 to target atom m2.
Step 3. Transform from CM to laboratory coordinates.
v
Add the velocity of the CM ( u ) to the final CM velocity of
v
v
projectile m1 ( v1 f *) and target atom m2 ( v2 f *).
v
v * v
v1 f = v1 f + u
v
v * v
v2 f = v2 f + u
v
Finally, solve for v2 f using trigonometry and convert to
kinetic energy using:
T2 f = 12 m2 v2 f 2
Trigonometry (Law of Cosines expression).
v2 f 2 = v2 f * 2 + u 2 ! 2 v2 f * u cos" CM
= u 2 + u 2 ! 2 u 2 cos" CM
= 2 u 2 (1! cos" CM )
Then:
1
2
m2 v2 f 2 = 12 m2 2 u 2 (1! cos" CM )
#
T2 f = m2
=
=
But: E10 =
1
2
m1 v10
2
(
)
2
m1
m1 +m2
m1 m2
(m1 +m2 )2
m1 m2
1 (m +m )2
1
2
2
!
v10 2 (1! cos" CM )
m1 v10 (1! cos" CM )
2
1
2
m1 v10 2 (1! cos" CM )
T2 f = ( m +m )2 E10 (1! cos" CM )
1
2
2 m1 m2
Maximum Kinetic Energy Transfer
Two-Body Collision
1 m2
T2 f = ( m2 m+m
2 E10 (1! cos " CM )
)
1
2
Head-on Collision:
! CM = "
cos ! = "1
# (1" cos$ CM ) = (1" "1) = 2
!
4 m1 m2
T2max
=
2 E10
f
( m1 +m2 )
max
2f
T
= ! E10
! = ( m +m )2
1
2
! " Kinetic Efficiency Factor
4 m1 m2
Kinetic Efficiency Factor, Λ
Kinetic Efficiency Factor
e- / e-, He, O, Cu, U
Kinetic Efficiency Factor
He / e-, He, O, Cu, U
Kinetic Efficiency Factor
O / e-, He, O, Cu, U
Rutherford Backscattering Spectrometry
(RBS)
Example: 2 MeV He+ ions into Sc4Zr3O12
m1 = He m1 = 4.00 amu
m1/m2 =0.25
Λ =0.64
m2 = O m2 = 15.9994 amu
T2f = 1.28 MeV
E1f = 0.72 MeV
m1 = He m1 = 4.00 amu
m1/m2 =0.089
Λ =0.30
m2 = Sc m2 = 44.956 amu
T2f = 0.60 MeV
E1f = 1.40 MeV
m1 = He m1 = 4.00 amu
m1/m2 =0.044
Λ =0.16
m2 = Zr m2 = 91.22 amu
T2f = 0.32 MeV
E1f = 1.68 MeV
m1 = He m1 = 4.00 amu
m1/m2 =0.022
Λ =0.086
m2 = Hf m2 = 178.49 amu
T2f = 0.17 MeV
E1f = 1.83 MeV
Kinetic Efficiency Factor
He / Sc4Zr3O12
E1Of = 0.72 MeV
E1Scf = 1.40 MeV
E1Zrf = 1.68 MeV
Hf
1f
E
= 1.83 MeV