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Therefore
…………………… 1.96Therefore
……………………………………………………………1.97Therefore in semiconductors the mean free path will be the product of the thermal velocity and the mean free time. Mean free time is calculated from mobility of the mobile carriers which is determined experimentally.
From Table(1.10) we obtain the mobility values. In Table (1.11) the mobility, mean free time, thermal velocity and mean free paths are tabulated for Ge , Si and GaAs.
Table(1.11) Mobilities, Mean Free Times, Thermal Velocities and Mean Free Paths of Ge, Si and GaAs.
Semiconductor | μ n (cm^ 2 / (V-sec)) | τ (femtosec) | v e (m/sec) | L* (A°) |
Ge | 3900 | 2217 | 0.95×10^ 5 | 2106 |
Si | 1350 | 767.6 | 0.95×10^ 5 | 729 |
GaAs | 8600 | 4890 | 0.95×10^ 5 | 4645.5 |
As we see electron has much larger mobility in semiconductors as compared to that in metals. This implies that the mean free path of electrons is greater by one order of magnitude in semiconductor as compared to that in metal. But why is the scattering less in semiconductors as compared to that in metal.? This answer is obtained by determining the de Broglie wavelength of electron and by using wave optics.
We will determine the velocity of a conducting electron in Electron Microscope, in metal and in semiconductor. In these three cases the conducting electron gains Kinetic Energy equal to the Potential Energy it loses while falling through a potential difference of 10kV in case of Electron Microscope(because 10kV is the accelerating voltage in Electron Microscope), through a potential difference of 4V in case of metal(because average kinetic energy associated with conducting electron is (3/5)E _{F} and E _{F} is 7eV in copper) and through a potential difference 0.025V in case of semiconductor ( since thermal voltage at 300K Room Temperature is kT/q= 0.025V). From the kinetic velocity the de Broglie wavelength is determined. The set of equations are: Kinetic Energy gained =
Therefore momentum gained
;Therefore de Broglie wavelength:
;In Table (1.12) the de Broglie wavelengths are tabulated:
Table1.12. de Broglie wavelengths of conducting electron in Electron Microscope, Metal and Semiconductor.
V acc | v e (m/sec) | λ(m) | Implications | |
Electron Microscope | 10kV | 59×10^ 6 | 10^ -11 m = (1/50)(5A°) | λ<<a (lattice constant) |
100kV | 187.6×10^ 6 | 4×10^ -12 m | ||
Metal | 4V | 10^ 6 | 6×10^ -10 m = (5A°) | λ~ a (lattice constant) |
Semiconductor | 0.025V | 10^ 5 | 7.75×10^ -9 m = (78A°) | λ>>a (lattice constant) |
As seen from Table(1.12), we see that de Borglie wavelength is much less than the lattice constant in case of Electron Microscope. For 100kV , theoretically the resolution should be (1/100)(4A°) This is like Sunlight falling through a broad aperture. Sun-ray will pass in a straight line and shadow of the aperture should fall on the screen behind the aperture. Hence in an Electron Microscope, a regular lattice array does not scatter an electron beam. The shadow of the crystal lattice should be imaged. But this theoretical resolution is never achieved since we are using magnetostatic focusing. Only 1A° is the resolution actually achieved. In case of 10kV, though the theoretical resolution (1/50)(5A°) but in practice only 10A° resolution is achieved. The electron beam can penetrate through a thin specimen and produce the image of its broad features without being influenced by the atomic details.
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