Structure factor of diamond. The crystal structure of diamond is described in Chapter 1. The basis consists of eight atoms if the cell is taken as the conventional cube. (a) Find the structure factor S of this basis. (b) Find the zeros of S and show that the allowed reflections of the diamond structure satisfy v, + U2+ Uz = 4n, where all indices are even and n is any integer, or else all indices are odd (Fig. 18). (Notice that h, k, l may be written for v, v2, Uz and this is often done.)
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- For a coefficient of 1.8 × 10^15 m−1 J^-0.5, at what wavelength is the absorption coefficient equal to 1000 cm-1 for GaAs at 300K.J 3Tungsten (W) crystallizes in cubic structure. The edge length of the unit cell of this crystal structure is a = 3.1648 Å. When the X-ray diffraction experiment is performed, scattering occurs from the following planes: (110), (200), (211), (220), (310), (222), (321), (400), (411), (420), (332), (431) Which of the x-rays scattered from the (110) and (200) planes has the greatest intensity? Hint: The intensity of the x-ray scattered from any atom decreases as the scattering angle increases. Also note that λ / 2d = sinθ.
- Q4: Find interpalanar space of the direction ( 2 2 0 ) for ( FCC) crystal plane its redii ( 1.414 A).The skin reflects most visible and IR-A (near-infrared) radiation. The epidermis is highly absorbing at UV-B and UV-C wavelengths and at IR-B and IR-C wavelengths. True or FalseThe maximum number of bits which a noisy channel can carry = B log2 (1 + S/N) with B = channel bandwidth in Hz; S/N = Signal-to-Noise ratio; S = the signal power; and N = the noise power. %3D %3D %3D In addition, when the Signal-to-Noise ratio (SNR) is given in decibels (DB), (we will represent this as SNRDB), meaning that it is expressed on a log scale as the quantity 10*log10 S/N. That is, SNRDB = 10* log10 S/N The Asymmetric Digital Subscriber Line (ASDL) which provides Internet service from our local library to the phone exchange has SNRDB of 40DB and a bandwidth of 1MHZ. What is the maximum number of bits the ASDL can carry in a second? 16,800,000bps 14,190,000bps 13,290,000bps 11,208,000bps
- A GaAs DH injection laser (refractive index: 3.6) has an optical cavity of length 50 μm and width 20 μm. At normal operating temperature (300 K) the loss coefficient is 10 cm-1, the mirror reflectivity at each end of the optical cavity is 0.3 and the gain factor for the device is 3.76x10-2 cm/A. The device emits at a wavelength of 870 nm. 1.1 Calculate the frequency separation of the longitudinal modes and the number of longitudinal modes emitted by the device. 1.2 Determine the current threshold for the device at 300 K. 1.3 Determine the required threshold current when the diode is operated at a temperature of 60˚C.Consider an asymmetric slab waveguide, with a core index ng= 1.5, a cladding index ne= 1.0 (air-clad), and a substrate index n₁=1.47. a) What is the maximum allowed value for kh? (K is the transverse wavevector) b) Using the graphical method, determine the allowed values for K and consequently for ẞ for a waveguide height of 5 μm and 7 μm, respectively. Assume an excitation wavelength of 1 μm. One application of slab waveguides is in sensing. By having an air-cladding, it is possible to disturb the waveguide transmission through any surface contamination. To achieve this, we need to know the ratio of the power in the evanescent tail vs. the core power. c) Find the ratio of the power-per-unit-length-in-the-y-direction in the cladding to the power-per- unit-length-in-the-y-direction in the core.Find V.E for E = pcos op – psin OÓ + pz2 i.e. F, = pcos ø, F = -p sin ø, F, = pz %3D %3D