Question 1.13: PARTICLE FLUX DENSITY AND PRESSURE Consider a vacuum deposit...
PARTICLE FLUX DENSITY AND PRESSURE Consider a vacuum deposition process in which atoms will be deposited onto a substrate. We wish to calculate the rate of impingement of atmospheric molecules in the chamber on to a surface area A on the substrate as shown in Figure 1.24c. Put differently, we wish to calculate the flux of molecules arriving on the area A. Suppose that \Delta N number of molecules reach the area A in time \Delta t as shown in Figure 1.24c. The flux density \Gamma that characterizes the flow rate of such particles per unit area is generally defined by
\Gamma=\frac{\Delta N}{A \Delta t} [1.32]
It is clear in Figure 1.24c that only those molecules with a velocity component along the positive x-direction can reach A. Suppose that the average speed parallel to the x-direction is v_{x} . In a time interval \Delta t, those molecules will travel v_{x} \Delta t along x. Only those molecules that are a distance v_{x} \Delta t away from A and also within the area A can reach A as shown in Figure 1.24d. The number of these molecules in the volume Av_{x} \Delta t is n(Av_{x} \Delta t), where n is the number of molecules per unit volume. However, only half of these will be moving along +x and the other half along −x, so the actual \Delta N reaching A is \frac{1}{2} n A v_{x} \Delta t. Substituting this into Equation 1.32, the flux density along the positive +x direction is
\Gamma_{x}=\frac{1}{2} n v_{x} [1.33]
Calculate the flux density of impinging N_{2} molecules on a semiconductor substrate in a vacuum chamber maintained at 1 atm (760 torr) and 10^{−9} torr, which represents ultra-high vacuum. What is the rate at which a typical atom on the substrate surface gets bombarded by N_{2} molecules, assuming that an atom on the surface is roughly a square with a side a on the order of 0.2 nm? Assume the temperature is 300 K. What is your conclusion?
We can use the effective velocity along x for the average velocity along this direction, that is, \frac{1}{2} M \overline{v_{x}^{2}}=\frac{1}{2} k T in which M is the mass of the N_{2} molecule, given by 2 M_{ at } / N_{A}=2\left(14 g mol ^{-1}\right) / \left(6.022 \times 10^{23} mol ^{-1}\right)=4.65 \times 10^{-26} kg . Substituting T = 300 K, we find the rms velocity along x, v_{x}( rms )=298.5 m s ^{-1} .
We have already calculated the N_{2} concentration n under a pressure of 1 atm in the chamber in Example 1.12 by using Equation 1.30, that is, n=2.45 \times 10^{25} m ^{-3} .
The flux density of N_{2} molecules impinging on the substrate is then
P=n k T [1.30]
\Gamma_{x}=\frac{1}{2} n v_{x} \approx \frac{1}{2}\left(2.45 \times 10^{25} m ^{-3}\right)\left(298.5 m s ^{-1}\right)=3.65 \times 10^{27} m ^{-2} s ^{-1}
A typical size \alpha of an atom is on the order of 0.2 nm so that an atom on the surface of a substrate typically occupies an area \alpha ^{2} of 0.04 nm² or 4 \times 10^{-20} m ^{2} . A particular atom on the surface is then bombarded at a rate \alpha ^{2} \Gamma _{x} per second, that is, \left(4 \times 10^{-20} m ^{2}\right)\left(3.65 \times 10^{27} m ^{-2} s ^{-1}\right) or 146 million times every second.
If we repeat the calculations at a pressure of 10^{-9} torr \left(1.33 \times 10^{-7} Pa \right) , we would find that n=3.22 \times 10^{13} m ^{-3} \text { and } \Gamma_{x}=4.8 \times 10^{15} m ^{-2} s ^{-1} so that a particular atom on the substrate surface is hit 1.9 \times 10^{-4} times per second, or it takes 1.4 hours for this atom to be hit by an N_{2} molecule. It is obvious that at atmospheric pressure we cannot deposit the evaporant atoms onto the substrate while the substrate is bombarded at an astronomic rate. On the other hand, under suitable vacuum conditions, we can easily deposit evaporant atoms and grow the layer we need on the substrate without air and other contaminant molecules interfering with the growth.