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(a)
Interpretation:
What type of semiconductor that would be formed from the given combination of elements has to be explained.
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 lie 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:
What type of semiconductor that would be formed from the given combination of elements has to be explained
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 lie 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
- Redraw the flowchartarrow_forwardredraw the flowchart with boxes and molecules written in themarrow_forwardPart 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_forwardNonearrow_forward
- Draw the curved-arrow mechanism with the drawings of the molecules, not just abbreviations. -NO₂ Sn, HCl (aq) E D H (CH3CO)₂O -NH2 CH3arrow_forwardWhat is/are the product(s) of the following reaction? Select all that apply. * HI A B C OD OH A B OH D Carrow_forwardIn 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_forward
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