DATENAMEEVENT 1897J. J. ThompsonDiscovers the electron 1924Louis deBroglieIdentifies a wavelength to moving electrons λ=h/mv where λ = wavelength h = Planck's constant (6.626 x 10 -34 J s) m = mass v = velocity (For an electron at 60kV λ = 0.005 nm) 1926H. BuschMagnetic or electric fields act as lenses for electrons 1929E. RuskaPh.D thesis on magnetic lenses 1931Knoll & RuskaFirst electron microscope built 1931Davisson & CalbrickProperties of electrostatic lenses 1934Driest & MullerSurpass resolution of the LM 1938von Borries & RuskaFirst practical EM (Siemens) - 10 nm resolution 1940RCACommercial EM with 2.4 nm resolution 1945 1.0 nm resolution p = momentum Viewing the nanoscale
Abberation Spherical aberration occurs when electrons travelling further from the optic axis, which are more likely to be deviated, are focussed at different points along the optic axis. Each point in the object is therefore imaged a disc of radius, Δr s Spherical aberration constant usually C s 1-2mm (1.1)
Abberation Chromatic aberration arises because of a spread in energy acting on the electrons, leading to a small divergence of the electron beam, Δr c Astigmatism occurs when electrons go through different focal lengths, depending on the plane of the ray paths, Δr A
Electron waves λ = h / m.v de Broglie An electron’s momentum is usually defined by falling through a potential difference: eV = ½ m.v 2 e is the electronic charge in volts (1 electron volt = 1.602 x 10 -19 J), therefore: λ = h / (2.m.eV) ½ When V is large electrons approach speed of light, therefore electron mass must be accounted for through the relativistic accelerating voltage V r, but usually when V ≥ 10 5 volts: V r = V.[1 + eV / 2.m 0.c 2 ] So, when accelerating voltage is 100 kV, the wavelength is 0.0037nm and the optimum resolving power is 0.23nm, less than lattice spacing of graphite!
EM imaging Contrast effects are generated by small phase variations among the electrons passing through the sample. By combining all or most of the beams transmitted and scattered by the sample in the formation of the image, a phase contrast image is produced. A model of a nanotube interacting with an electron beam is shown above. Part of the electron beam scatters at the top and bottom parts of the tube (point H). The contrast of this scatter is relatively weak, due to a low number of atoms in this (hk0) plane aligned to the electron beam, but structural detail here is often observed on precisely tuned TEMs. Part of the electron beam also scatters at the sides of the tube (point V). The contrast is relatively stronger as there are a greater number of atoms in the (00l) plane aligned to the electron beam. This gives rise to the appearance of the nanotube “walls”.
When the electron beam passes through a sample, a diffraction pattern is formed at the back focal plane of the objective lens, which contains all of the information present in the specimen. The diffracted beams forming this pattern can be recombined with the transmitted electrons in the image plane to form a lattice image. By collecting the transmitted electrons only, achieved by placing an aperture to obstruct the diffracted electrons, a bright field image is obtained. Conversely, by only collecting the diffracted beams and obstructing the transmitted beams, a dark field image can be obtained. EM imaging
Electron diffraction (TEM cont.) The crystal lattice of a sample causes diffraction of the electron beam. When the crystal planes are aligned with an hkl plane at an angle θ to the beam axis, commonly scattered beams interfere constructively. Since the Bragg angle for high-velocity electron beams is very small, sin θ approximates to θ, and for n = 1 the Bragg equation (2d hkl sin θ = nλ) is simplified: In the electron microscope, the film is placed at a distance L (the camera length) from the sample. The distance r from the centre spot O to the diffracted spot H is expressed: Bragg Equation Standard x-ray diffraction
When the electron beam strikes the sample, an incident electron may collide with an electron in the K shell and ejects it, resulting in ionisation of the atom. Another electron, from say a higher shell, can fall into this vacancy and emits its excess energy as an X-ray photon. Such a transition from the L to K shell results in a Kα line and from the M to K shell gives a Kβ line, etc. These X- rays are detected by energy dispersive techniques. Energy Dispersive X-ray (TEM cont.)
Electron Energy Loss Spectroscopy (TEM cont.) EELS is concerned with the study of the electron excitation processes, each of which results in a fast electron losing a characteristic amount of energy. After interacting with the sample, the transmitted electrons are directed into a high- resolution electron spectrometer which separates the electrons according to their kinetic energy and produces an EELS spectrum showing the scattered intensity as a function of the decrease in kinetic energy of the fast electron: Inelastic scattering from outer-shell electrons is visible as a peak in the 5-50 eV region of the spectrum. Inner-shell excitations appear in the form of edges, which represent the ionisation threshold. Since inner-shell binding energies depend on the atomic number of the scattering atom, the ionisation edges present in an EELS spectra indicate which elements occur within the sample. By monitoring particular energy losses within a selected range, it is possible to produce energy-filtered images that reveal the elemental mappings of desired elements.
J-P. Salvetat, et al., Phys. Rev. Lett., 82, 944 (1999) Individual arc-MWCNT, Young’s modulus = 270 to 950 Gpa Individual cvd-MWCNT = 15- 50 Gpa Carbon nanotube strength + AFM “The superposition principle implies that the total deflection, σ, is the sum of the deflections due to bending, σ B, and to shear, σ S. If we use the unit-load method for a concentrated load F, the deflection at the middle becomes σ = σ B + σ S = F L 3 /192.E.I + f s.F.L /4.G.A where L is the suspended length, E is the elastic modulus, f s is the shape factor (equal to 10/9 for a cylindrical beam), G is the shear modulus, I is the second moment of area of the beam (I = πD 4 /64 for a filled cylinder), and A is the cross-sectional area. The ratio σ B / σ S increases with the ratio of beam length to diameter.”
Questions on microscopy 1.Why does electron microscopy have a better resolution than light microscopy? 2.Briefly describe the three aberrations that exist within electron microscopy. 3.What is the difference between bright field and dark field imaging? 4.What are the differences between transmission electron microscopy and scanning electron microscopy? 5.What are the differences between atomic force microscopy and scanning tunnelling microscopy?
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