Updated: January 23, 2008
Raman cross section from silicon nanobumps
Assume the incident laser beam is 1.0 mW with a square cross section of 1.0 μm.
So the incident power density is (1 mW)/(1.0 μm)2 = 1.0×109 W/m2.
Dividing by c, I get 3.3×100 J/m3.
Since half the radiant energy is in the electric field this corresponds to an
electric field of E = 6.1×105 V/m.
Assume the temperature is 300 K.
So the energy of a normal mode is kB T =
(1.38×10-23 )(300K)= 4.14×10-21 J.
The size of the oscillating object (a silicon "nanobump") is 10 nm × 10 nm × 10 nm, and has
density 2329 kg/m3 So its mass is 2.329×10-21 kg.
As it oscillates in a given single vibrational mode, half of its energy is kinetic
energy, so the characteristic speed is v = sqrt(2E/m) =
sqrt(2(0.5)(4.14×10-21 J)/(2.329×10-21 kg) ),
or 1.33 m/s. Given the (longitudinal) speed of sound in silicon is 5840 m/s,
the characteristic vibrational frequency of this object is around
f = (5840 m/s)/(10 nm) = 5.8×1011 Hz. So that means
that the amplitude of the vibration is on the order of
d = (1.33 m/s)/(5.8×1011 Hz)
= 2.3×10-12 m.
Consider the electric dipole induced by the incident electric
field. Silicon has a high dielectric constant, so the simplest approximation
is to assume an induced surface charge density
σ = εo E
= 5.4×10-6 C/m2. This leads to an
induced dipole moment of 5.4×10-30 C-m. This dipole
moment is not constant, due to the thermal vibrations. Since thermal
vibrations are 2.3×10-12 m, and assuming linear variation
of dipole moment as the shape oscillates, the variation of the dipole moment
should be approximately
(5.4×10-30 C-m)
(2.3×10-12 m)/(10 nm) = 1.2×10-33 C-m.
It is this time varying part of the dipole moment that leads to the Raman scattering. The total
radiated power from a dipole is:
I assume the laser light is green light, with λ = 514.5 nm, so that
ω = 3.7×1015 rad/s. In that case, the total power
radiated from a single silicon "nanobump" is 3.6×10-20 W.
I assume that a region of the sample of size 1 μm × 1 μm is imaged by
a microscope into the spectrograph. The large "nanobumps" are approximately
10 nm in size
and 50 nm apart. So the imaged region contains 4×102
such nanobumps. The total scattered power from all of them is
1.4×10-17 W. However, due to imaging optics aperature
limitations, I would guess that only 5% of the light is received into the
spectrograph. Of that, approximately 10% makes it to the CCD.
This is 7.2×10-20 W. In other words,
only 0.2 light photons are received per second.
Assuming 50% quantum efficiency, there will be 0.1 electrons accumulated
per second. So it is possible that the
Raman scattered light is too faint to see.
Daniel Murray
Associate Professor
Math, Stats & Physics Unit
University of British Columbia - Okanagan
Kelowna, BC, Canada
daniel "dot" murray "at" ubc "dot" ca
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