10. Appendix

Solution to Problem 8.1
Solution to Problem 8.1 659
There are several units of pressure in use nowadays. When we try to inflate
our tires we find that the unit for pressure most commonly used in the US is
pounds per square inch (PSI). When the meteorologists report their weather
forecast they quote the atmospheric pressure in terms of inches of mercury
(Hg). The SI unit for pressure is the Pascal (Pa). However, when one is dealing
with vacuum systems and pumps the most common unit for pressure is the
torr. The conversion from torr to Pa and the cgs units (which we use here)
can be found in the inside back cover table.
Let us recall that a torr is the pressure exerted by a column of Hg 1 mm
high. Since the density of Hg 13.6 g/cm 3 , and the acceleration due to gravity
is 981 cm/sec 2 , 1 torr 1.33 × 10 3 dyn/cm 2 in cgs units. Thus, a pressure of
10 6 torr corresponds to 1.33 × 10 3 dyn/cm 2 .
Next we determine the density of oxygen molecules at this pressure and a
temperature of 300 K using the equation of state of ideal gases:
p NkBT,
where p is the pressure (in dyn/cm 2 ), N is the number of molecules per cm 3 ,
kB is the Boltzmann’s constant (1.38×10 16 erg/K, from the inside back cover)
and T the temperature in Kelvin. From this equation we obtain N 3.22 ×
10 10 molecules/cm 3 for the pressure and temperature under consideration.
Using the kinetic theory of gases (see e.g., F.J. Blatt, Principles of Physics,
(Allyn and Bacon, 1983), p. 262) we find that the pressure p is related to the
average velocity v of one molecule by the equation:
v 2 3p/NM,
where M is the mass of the molecule (we take M to be the oxygen molecule
mass: 2 × 16 × 1.67 × 10 24 gm 5.34 × 10 23 gm). Hence the average
oxygen molecule velocity v 4.82 × 10 4 cm/sec. The number of molecules
impinging on a surface area of 1 cm 2 per second is obtained by multiplying v
by N. However, one must take into account that the velocity can point along
six different directions: x, y, z, x, y, z. If the surface is perpendicular to say
the z-direction, then only the molecules with velocities along the z direction
will contribute to collisions with the surface. Hence, we have to divide N by 6
in order to obtain the total number of molecules colliding with 1 cm 2 of the
surface per second. The result is:
(1/6)v.N 2.62 × 10 14 /cm 2 sec.
Let us now consider the (001) silicon surface. Since there are two atoms per
unit lattice square on this surface, the surface density of Si atoms is: 2/a 2 0 ,
where a0, the lattice constant, is equal to 0.543 nm. The surface density of
atoms is, therefore, 6.78×10 14 atoms/cm 2 . Assuming that the oxygen molecules
split into two oxygen atoms upon collision with the surface, then after 1 Langmuir
(L) of exposure the Si surface should be 77.3% covered if the sticking