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(a)
Interpretation:
The type of semiconductor formed when silicon combine with the given set of elements has to be determined.
Concept Introduction:
Semiconductors are substances that conduct electricity either by addition of an impurity or by the effects of temperature on it. Semiconductors electrically conductivity lies between conductor and insulator. Semiconductors have small energy gap between valence band and conduction band.
Addition of impurity to a semiconductor is termed as doping. Doping alters the conductivity of a semiconductor. The addition of an element having either more or less number of valence electrons than the natural semiconductor decides the combination as the following two types of semiconductor.
- n- type semiconductor: (conduction is due to movement of extra electrons)
The element added will have more valence electron than the natural semiconductor. Therefore, the extra electron from the added element resides in conduction band and increase the conductivity.
Example: Silicon (natural semiconductor) and Phosphorus
- p-type semiconductor: (conduction is due to movement of holes)
The element added will have less valence electron than the natural semiconductor. Here, instead of extra electron, there will be “holes” at the places, where a semiconductor is replaced by added element. A p-type semiconductor increases conductivity because the holes (effective positive charge; lies at valence band) move through the natural semiconductor rather than electrons.
Example: Silicon (natural semiconductor) and Gallium
(b)
Interpretation:
The type of semiconductor formed when silicon combine with the given set of elements has to be determined.
Concept Introduction:
Semiconductors are substances that conduct electricity either by addition of an impurity or by the effects of temperature on it. Semiconductors electrically conductivity lies between conductor and insulator. Semiconductors have small energy gap between valence band and conduction band.
Addition of impurity to a semiconductor is termed as doping. Doping alters the conductivity of a semiconductor. The addition of an element having either more or less number of valence electrons than the natural semiconductor decides the combination as the following two types of semiconductor.
- n- type semiconductor: (conduction is due to movement of extra electrons)
The element added will have more valence electron than the natural semiconductor. Therefore, the extra electron from the added element resides in conduction band and increase the conductivity.
Example: Silicon (natural semiconductor) and Phosphorus
- p-type semiconductor: (conduction is due to movement of holes)
The element added will have less valence electron than the natural semiconductor. Here, instead of extra electron, there will be “holes” at the places, where a semiconductor is replaced by added element. A p-type semiconductor increases conductivity because the holes (effective positive charge; lies at valence band) move through the natural semiconductor rather than electrons.
Example: Silicon (natural semiconductor) and Gallium
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Chapter 24 Solutions
GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST
- Part I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward• Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forwardCould you redraw these and also explain how to solve them for me pleasarrow_forward
- In the image, the light blue sphere represents a mole of hydrogen atoms, the purple or teal spheres represent a mole of a conjugate base. A light blue sphere by itself is H+. Assuming there is 2.00 L of solution, answer the following: The Ka of the left & right solution is? The pH of the left & right solution is? The acid on the left & right is what kind of acid?arrow_forwardNonearrow_forwardNonearrow_forward
- Nonearrow_forwardWhat spectral features allow you to differentiate the product from the starting material? Use four separate paragraphs for each set of comparisons. You should have one paragraph each devoted to MS, HNMR, CNMR and IR. 2) For MS, the differing masses of molecular ions are a popular starting point. Including a unique fragmentation is important, too. 3) For HNMR, CNMR and IR state the peaks that are different and what makes them different (usually the presence or absence of certain groups). See if you can find two differences (in each set of IR, HNMR and CNMR spectra) due to the presence or absence of a functional group. Include peak locations. Alternatively, you can state a shift of a peak due to a change near a given functional group. Including peak locations for shifted peaks, as well as what these peaks are due to. Ideally, your focus should be on not just identifying the differences but explaining them in terms of functional group changes.arrow_forwardQuestion 6 What is the major product of the following Diels-Alder reaction? ? Aldy by day of A. H о B. C. D. E. OB OD Oc OE OAarrow_forward
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