Question

2). Rough estimation of the diameter of an aluminum atom: Given aluminum density as p-2.7*10 kg/m', atomic weight is 27 (i.e. the mass of 1 mole of aluminum atoms is 27g. Imole 6.02x10 ). Assume atoms of aluminum metal are spherical with spaces in between the aluminum atoms take up 74.1% of the volume of the material. What is the diameter of an aluminum atom? (Note: Volume of a sphere with radius ris: V-(4/3)xr) 4. (20") (1). In E-beam lithography, the equivalent DeBroglie wavelength of electron beam can he calculated by 1- 2mV4l+eV/2me)) where his plank constant: -6.63x10 mkg's. m is the electron mass: m=9.1x10 kg. e is the electron charge: -1.6x10C, V is the acceleration voltage. c is the speed of light:c-3x10m/s. If the acceleration voltage V=100kV, assume e-Ve=(1/2)m-v", find out the velocity v of an electron after accelerated by given voltage V=100KV. What is the ratio of this velocity to speed of light in vacuum? Should we consider relativity theory? Find out the corresponding De Broglie (2). An electron jumps from higher energy level to lower energy level with energy difference of AE-1.8V (leV-1.6x10")). This energy difference is emitted as light (a photon). What is the wavelength () and frequency of the light? (Note: AE=h-fhol, where is speed of light in vacuum, c^3*10'm/s, his Plank's constant: -6.63x10 mkg/s). (3). The color of UB logo is purple. For purple light, its wavelength is around 400nm. What is the corresponding energy (in unit of eV) of a photon of purple light? 5. (15"XI). List the three working modes of Focused Ion Beam Lithography (FIBL). FIBL has the error correction capability during the patterning process. How? Please explain. (2). What is microcontact printing technique? Briefly explain the steps of microcontact printing technique. Sketch some diagrams to help explain the steps. What's the difference between microcontact printing and nanoimprinting technique? (3). What is Dip Pen Nanolithography? Explain the working principle with a figure. 6. (15") (1). What is SAM? Briefly explain the process to form a SAM layer with a figure. How can we guarantee that the formed SAM layer is exactly one single molecule thick? (2). Explain an example process for fabricating polymer sphere nano-capsules (with crystalline molecules inside) based on LBL (layer-by-layer) self-assembly technique. Please show the procedure step by step and roughly sketch the diagram for each step 7. (15) 1). What is the difference between top-down and bottom-up nanofabrication strategies? Which strategy is more efficient (i.e. generating less material waste)? 2). DNA-induced nanoparticle assembly: Assume we attach two batches of nanoparticles (A,B) with different single stranded DNA sequences. The DNA strands attached to nanoparticle A is A-T-G-C-T-G-A, and the DNA strands attached to nanoparticle is: G-C-T-A-C-G. Design the DNA linker (show the DNA sequence) which can be used to link the nanoparticles A and B together into AB-AB periodic structure. DNA linker A-G-T-C-G-TA A-T-G-C-T-G-A G-C Figure 2. DNA-induced nanoparticle assembly

Answer 1

Here is the solution provided below in the hand written mode. Hope it will help you.

Solutions (2) Here given, aluminium density, P = 2.7X103 kg / m3 Atomic weight, M = 27 Atoms of Aluminium metal is assumed to be spherical up 76.1% Here given Aluminium atoms take of the material . of the volm I fee lattice . So, these are the number of aluminium atoms per mit cell will be a Z=4 lattice constant. lu let be the a so so we know, exax NA Z 23 2.7 X103 x 23 x 6.02X10 3 11 27 = ) a3 26 6.64 X10 2 a = 4.05 :10-9 m Again FCC lattice, we know that fore 4.05X10-9 atomic radius, r 25 22 جا 9 1.438 10 So, diameter of aluminium atom, d=28= 2.86 x 109 m ml

47 20) (0) Equivallent de Broglie - Broglie wavelength of electron beam is given by, h 22 (0) zm ve [ it ev. 2mc n= 6.63x10-34 mikg/s m=9.1X10-31 31.kg . Again, if the acceklaration ) roltage is He then, given, e ve = t4mor fo) S01 fore. Ve = 100 kV 9 by eqn Ei 202 3 zeve 2x 1.6X10 19 X 100X10 9.1x10-31 3.52x106 v=1.87 X 108 m/s त Hence, the reatio of this velo to with speed locity will be, valulint vacuum of light in 8 1.87Xo PR 2 0.623 3 X 108 hence / since, the reatio is greater than 0.1 we should consider relativity theory.

So by equation (1) h a 11 2m Ve e [iteve 2me -34 6.63X10 2 -31x100 x103 x1-6X10 -19 1.6810-19X105 2 X 9.1X10 + - 31 14 2X9.1X10 X9X10 10 349x109 -12 22. 3.7 X10 m a 0.0037 nmi Here an election jumps from higher energy with, level to lower energy energy level 1.8ev he de = hit = we know emmitted photon, son the frequency JE fa 14 4.34 X10 19 1.8x1-6X107 6.63 X10-34 НЕ h nc Again, the wavelength , a = -7 -6-902'x10 A a= 690.2 nm

for purple light, wavelength, a=400 mm he sol the corresponding energy, E= -34 x 3x108 6.63X10 400X10-9 -19 Ez 4.969 X10 J E 2 3.102 ev This is the corresponding energy of purple light