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1 Advancing Physics G495 June 2015 SET #1 ANSWERS Field and Particle Pictures Seeing with electrons The compound optical microscope Q1. Before attempting this question it may be helpful to review ray diagram drawing from GCSE. [Lines 10-15] The following data relate to an optical compound microscope: Focal length of objective lens (f 1 ) = 20 mm Distance from lens to object = 30 mm Focal length of eyepiece lens (f 2 ) = 60 mm Distance between objective lens and eyepiece lens = 100 mm Height of object (arrow) measured from principal axis = 10 mm DIAGRAM NOT TO SCALE L1 L2 object objective lens eyepiece lens (a) Construct a ray diagram on graph paper (full scale) for the situation shown above showing the principal axis, the positions of the two lenses and their principal foci and the position of the object. Use ray tracing to find the position of the image formed by the objective lens and the image formed by the eyepiece lens. [5] See attached at end. (b) Write down the equation that links focal length, lens-object distance and lens-image distance, identifying each of the terms. [2] 1/v = 1/u + 1/f [1] (Curvature out = Curvature in + Curvature added by lens) f = focal length, u = lens object distance, v = lens-image distance [1] (c) Use this equation to determine the position of the real image formed by the objective lens. Check your calculated answer using the ray diagram. [2] 1/v = 1/ /0.02 = = D [1] v = 1/16.67 = m = 60 mm [1] (d) Calculate the magnification of the first stage of the microscope. Check your answer using the ray diagram. [2] M = v/u = 60/-30 [1] = -2.0 [1] (e) The image formed by the objective lens acts as the object for the eyepiece lens. Use the equation to determine the position of the image formed by the second, eyepiece lens. Check your answer using the ray diagram. [2] 1/v = 1/u + 1/f = 1/ /0.060 = = D [1]
2 v = 1/-8.33 = m [1] (f) What type of image is formed by the second lens? [1] Virtual image [1] (g) Estimate the overall magnification of the system using either your ray diagram or using the appropriate equations [2] Overall magnification = -2 x 3 [1] = -6 [1] Q2. An optical microscope is connected to a digital camera and is used to record an image of part of the optic nerve of a human eye. The magnification of the part of the imaging system that focuses light from the object on the CCD array is 40x. The image from the digital camera is recorded on a CCD array of dimensions 12.8 mm x 12.8 mm (1050 x 1050 pixels). [Lines 16-25] (a) Calculate the linear dimensions of a pixel on the CCD array. [1] 12.8 mm / 1050 = mm [1] (b) Use the magnification of the system to estimate the smallest scale of features resolvable with this microscope-ccd system. [1] Scale of features resolvable = mm / 40 = mm (0.3 μm) [1] (c) The image detected by the CCD array is displayed on a monitor screen. The image dimensions on the screen are 20 cm x 20 cm. Calculate the overall magnification of the microscope imaging system. [2] Magnification of second stage (CCD to screen) = 200 / 12.8 = 15.6 [1] Overall magnification = 40 x 15.6 = 625x [1] Q3. Using ideas about diffraction, explain carefully why the resolution of an optical system is limited by the wavelength of the light being used. [Lines 20-25] [3] Wave fronts passing through optical system experiences diffraction due to passage through apertures [1] So wave not focused to a point but to a diffraction-limited spot [1] Overlap of spots places limit on resolution [1] Q4. The diffraction limit of resolution of an optical system in the mid-visible is given approximately by d = λ/2, where λ is the wavelength of light being used. Which of the following biological structures will be resolvable when illuminated with green light (500 nm)? [Lines 20-25] Cells μm Viruses 100 nm Proteins 10 nm [2] d = 500 nm / 2 = 250 nm [1] Only cells resolvable [1]
3 Improving the view Q1. Explain why the use of ultraviolet light offers the highest resolution in optical microscopy. [Lines 28-30] [1] UV light has shortest wavelength, so highest resolution. [1] Q2. The diagram below shows a magnetic lens used to focus a beam of electrons in an electron microscope. Explain why the beam follows a spiral path. Consider the magnetic field lines of the electromagnet coil to lie along the beam axis, and the beam of electrons to be travelling at an angle to the axis as shown. [Lines and Fig. 4] [4] Beam electrons have velocity components parallel and perpendicular to axis. [1] Force is exerted on electrons as a result of their component of velocity perpendicular to the axis / field direction (F=Bqv) [1] This force is directed at right angles to electron motion perpendicular to axis and field direction. [1] This causes circular motion in a plane perpendicular to axis, which combined with constant axial velocity component produces spiralling. [1] (Variation in field strength will produce focusing effect.) Q3. Suggest why the equation I= Ce -ε/kt, which includes the Boltzmann factor, gives an approximate value for the current I from the electron microscope filament. (ε is the work function of the metal, T the temperature, k is Boltzmann s constant and C a constant.) [Lines 40-44] [2] Current is determined by rate at which electrons leave the surface owtte; [1] Probability that an electron will (have energy ε to) be able to leave the surface (at temperature T) is proportional to the Boltzmann factor [1] Q4. Assuming that the work function of the filament metal is 4 ev, by what factor does the current increase when the temperature is increased from 1000 K to 2000 K? [Lines 40-44] [3] Boltzmann factor at 1000 K ε/kt = (4 x 1.6 x 10-19)/(1.4 x x 1000) = 46 f 1000 =e -46 = [1] Boltzmann factor at 2000 K ε/kt = (4 x 1.6 x )/(1.4 x x 2000) = 23 f 2000 =e -23 = [1] f 2000 /f 1000 = /10-20 = [1]
4 Q5. Why is it essential that the electron microscope is evacuated? [Lines 50-52] [1] If no vacuum, beam electrons strike air molecules, lose energy, get scattered, beam destroyed. [1] Q6. The mean free path (D) is the average distance a gas molecule travels between collisions, and is given approximately for air at 20 o C by the equation D (cm) = where P is the pressure measured in Pascals. A good vacuum in an electron microscope corresponds to 10-4 Pa. (a) Estimate the mean free path for air molecules in the electron microscope chamber and explain its significance in the context of operating the electron microscope. [Lines 50-54] [3] D (cm) = 0.7 / 10-4 = 7000 cm = 70 m [1] Mean free path long compared to dimensions of apparatus [1] So air molecules unlikely to strike each other as number density so low, and electrons unlikely to strike air molecules for same reason [1] (b) Explain why the requirement for high vacuum in the electron microscope limits its usefulness. [Lines 50-54] [2] Samples may be damaged by outgassing as pressure is lowered [1] Biological samples can be difficult to study as water in them evaporates [1] Any other valid points Q7. The interaction between high-speed electrons and the sample could give rise to the production of X rays. [Lines 55-57] (a) Explain how x rays could be produced in this situation. [3] Energetic beam electron ejects electron from deep-lying energy level in atom when inelastic scattering takes place [1] Electron from higher energy level drops down into vacancy [1] Emitting x ray photon in doing so [1] Any other valid mechanism (b) Estimate the shortest-wavelength X ray that could be emitted by an atom that absorbs a 50 kv electron. State any assumptions made. [3] Assume that all of the electron energy is used in ejecting an electron from an energy level lying 50 kev below ionisation limit. [1] Energy lost by outer electron dropping down 50 kev to fill vacated level = x 1.6 x = 8 x J [1] λ =hc/e = 6.6 x x 3 x 10 8 / 8 x = 2.48 x m (24.8 pm) [1] Q8. In this question, you will use the de Broglie relationship to estimate the resolution of an electron microscope. [Lines 60-64]
5 (a) Eliminate v from the equations Kinetic Energy E K = ½ mv 2 and momentum = mv to obtain an expression for the momentum of an accelerated electron in terms of its (non-relativistic) mass and its kinetic energy. [2] v = p/m substitute for v in KE equation E K = ½ m (p/m) 2 = p 2 /2m [1] Rearranging and solving for p gives p = 2mE K [1] (b) Use the equation to calculate the momentum (in kg ms -1 ) of an electron accelerated through a potential difference of 50kV. [2] E K = x 1.6 x = 8 x J [1] p = (2 x 9.1 x x 8 x ) 1/2 = 1.21 x kg ms -1 [1] (c) Use the de Broglie equation to show that the de Broglie wavelength of a 50 kv electron is approximately 5 pm. [1] λ = h/p = 6.6 x / 1.21 x = 5.45 x m [1] (d) Use the relevant equations for relativistic momentum and energy relationships to calculate the relativistic momentum of a 50 kv electron. Compare with the nonrelativistic value obtained earlier and comment. [4] Use p =γm 0 v and γ = E total /E rest and v 2 /c 2 = 1-1/ γ 2 γ = E total /E rest = (E K + E rest )/E rest = (50 kev MeV)/0.512 MeV = [1] v 2 /c 2 = 1-1/ γ 2 = 1-1/ = = So v/c = 0.911, and v = 0.413c = 1.24 x 10 8 ms -1 [1] p =γm 0 v = x 9.1 x x 1.24 x 10 8 = 1.24 x kg ms -1 [1] Non relativistic value calculated earlier = 1.21 x kg ms -1 Discrepancy very small (as kinetic energy is small compared to rest energy) [1] (Note: problem can be solved using E 2 = p 2 c 2 +m 0 2 c 4 instead) (e) By what factor is the theoretical resolution of a 50kV electron microscope better than an optical microscope using 500 nm illumination? [2] Assuming that resolution of electron microscope given approximately by de Broglie wavelength, and using the answer from Q4 in the section on the compound optical microscope 5 pm / 250 nm = 5 x / 250 x 10-9 = 1/50000, [1] so electron microscope resolution is better in theory [1]
6 Q9. A de Broglie wavelength of 5 pm for 50 kv electrons suggests that the resolution of an electron microscope is considerably better than the best optical microscopes. What factors may limit the resolution of an electron microscope in practice? [Lines 60-65] [2] Electrons all have negative charge so repel one another, limiting minimum spot size [1] Electrons may cause local heating of sample that leads to secondary electrons [1] Any other valid points Q10. TEM specimens are required to be at most hundreds of nanometres thick, as unlike neutron or X ray radiation the electron beam interacts readily with the sample, an effect that increases roughly with atomic number squared (Z 2 ). Source: (a) Why is it important that the sample in a Transmission Electron Microscope is thin? [Lines 55-58] [1] If sample is too thick, then electron beam will not penetrate through appreciably. [1] (b) In a certain TEM measurement, the electron beam is attenuated by 50 % for every 25 nm depth. What percentage of the incident beam intensity is transmitted by a sample 75 nm thick? [Lines 55-58] [2] 75 nm = 3 half thicknesses [1] 1/2 3 = = 12.5 % [1] (c) Why do electrons interact more strongly with sample atoms than neutrons or X rays? [Lines 55-58] [2] Electrons are charged, neutrons and X rays are uncharged [1] So electrons can interact with nuclei and atomic electrons via the electrostatic force [1] (d) Why does the strength of interaction increase with atomic number? [Lines 55-58] [2] The larger the atomic number, the greater the nuclear charge / number of atomic electrons for sample atoms [1] so stronger electrostatic interactions with beam electrons can occur [1] (e) Why are samples coated with a thin layer of a metal such as lead or gold prior to TEM investigations? [Lines 55-58] [2] Lead and gold have very large atomic numbers [1] Therefore interaction with beam electrons will be strong [1] Q11. Draw a diagram to show how the planes of atoms in crystalline material may act as a diffraction grating and as a result scatter some of the beam electrons. [Line 58] [2]
7 Maximum in scattered intensity when path difference is a whole number of de Broglie wavelengths. This implies that 2d sin θ = nλ [1] [1] Q12. A scanning electron microscope with a beam energy of 25 kev ejects secondary electrons with a range of kinetic energies from 0 to 300 ev from a sample. The electrons are ejected from a range of energy levels in atoms of the sample. [Lines 65-68] (a) Why is a beam electron able to eject many secondary electrons from the sample material? [1] Beam electrons have much higher energy than binding energy of electrons being ejected. [1] (b) The quantised nature of atomic energy levels implies that the secondary electrons should be ejected with specific energies. Why in practice are secondary electrons emitted with a range of energies? [2] Electrons lose energy as they collide with atoms when migrating towards the surface [1] This gives secondary electrons a range of kinetic energies [1] Q13. Draw a diagram to show how electric fields could be used to raster-scan the electron beam across the sample surface. [Lines 70-72] [2] [1] Voltages applied to x plates and y plates are varied with time to deflect beam horizontally and vertically across the sample. [1] Q14. Estimate the magnification of the image of the flea, assuming that the flea is 2 mm long. [Lines and Fig.5a] [2] Length of image of flea = 64 mm [1] Magnification = 64 mm / 2 mm = 32x [1] Q15. Estimate the size of a single lens element in the fly s eye shown in Figure 5b. Take the magnification as 1000x. [Lines and Fig.5a] [2]
8 Size of one element on image = 3 mm [1] Size on fly = 3 x 10-3 / 1000 = 3 x 10-6 m (3 μm) [1] Scrutinising the future Q1. What advantages does atomic force microscopy offer over SEM or TEM? [Lines 80-82] [3] Don t need vacuum to investigate samples [1] Sample preparation time is less [1] Samples not damaged by beam [1] Any other valid points
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