Interpretation: The group in the periodic table that contains elements which are semiconductors at room temperature should 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 have small energy gap between valence and conduction band hence its electrical conductivity lies between conductor and insulator.
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
Periodic Table: The available chemical elements are arranged considering their
In periodic table the horizontal rows are called periods and the vertical column are called group.
There are seven periods and 18 groups present in the table and some of those groups are given special name as follows,
Atomic Number: Atomic number of the element is equal to the number of protons present in the nucleus of the element which is denoted by symbol Z. The superscript presents on the left side of the

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Chapter 24 Solutions
CHEMISTRY: ATOMS FIRST VOL 1 W/CONNECT
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- What are the retrosynthesis and forward synthesis of these reactions?arrow_forwardWhich of the given reactions would form meso product? H₂O, H2SO4 III m CH3 CH₂ONa CH3OH || H₂O, H2SO4 CH3 1. LiAlH4, THF 2. H₂O CH3 IVarrow_forwardWhat is the major product of the following reaction? O IV III HCI D = III ა IVarrow_forward
- The reaction of what nucleophile and substrate is represented by the following transition state? CH3 CH3O -Br อ δ CH3 Methanol with 2-bromopropane Methanol with 1-bromopropane Methoxide with 1-bromopropane Methoxide with 2-bromopropanearrow_forwardWhat is the stepwise mechanism for this reaction?arrow_forward32. Consider a two-state system in which the low energy level is 300 J mol 1 and the higher energy level is 800 J mol 1, and the temperature is 300 K. Find the population of each level. Hint: Pay attention to your units. A. What is the partition function for this system? B. What are the populations of each level? Now instead, consider a system with energy levels of 0 J mol C. Now what is the partition function? D. And what are the populations of the two levels? E. Finally, repeat the second calculation at 500 K. and 500 J mol 1 at 300 K. F. What do you notice about the populations as you increase the temperature? At what temperature would you expect the states to have equal populations?arrow_forward
- 30. We will derive the forms of the molecular partition functions for atoms and molecules shortly in class, but the partition function that describes the translational and rotational motion of a homonuclear diatomic molecule is given by Itrans (V,T) = = 2πmkBT h² V grot (T) 4π²IKBT h² Where h is Planck's constant and I is molecular moment of inertia. The overall partition function is qmolec Qtrans qrot. Find the energy, enthalpy, entropy, and Helmholtz free energy for the translational and rotational modes of 1 mole of oxygen molecules and 1 mole of iodine molecules at 50 K and at 300 K and with a volume of 1 m³. Here is some useful data: Moment of inertia: I2 I 7.46 x 10- 45 kg m² 2 O2 I 1.91 x 101 -46 kg m²arrow_forwardK for each reaction step. Be sure to account for all bond-breaking and bond-making steps. HI HaC Drawing Arrows! H3C OCH3 H 4 59°F Mostly sunny H CH3 HO O CH3 'C' CH3 Select to Add Arrows CH3 1 L H&C. OCH3 H H H H Select to Add Arrows Q Search Problem 30 of 20 H. H3C + :0: H CH3 CH3 20 H2C Undo Reset Done DELLarrow_forwardDraw the principal organic product of the following reaction.arrow_forward
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