Tuesday, July 11, 2023

H C Verma solutions, SEMICONDUCTORS AND SEMICONDUCTOR DEVICES, Chapter-45, OBJECTIVE - I, Concepts of Physics, Part-II

OBJECTIVE - I


       1. Electric conduction in a semiconductor takes place due to 
(a) electrons only  
(b) holes only 
(c) both electrons and holes
(d) neither electrons nor holes.  

ANSWER: (c).      

EXPLANATION: Both electrons and holes. When a valence electron in the semiconductor moves to the conduction band, the vacancy created by it is called a hole. When an electric field is applied across the semiconductor, the conduction electrons drift opposite to the field and holes drift along the electric field. So electric conduction takes place in a semiconductor due to both electrons and holes. Option (c) is correct.                







       2. An electric field is applied to a semiconductor. Let the number of charge carriers be n and the average drift speed v. If the temperature is increased, 
(a) both n and v will increase
(b) n will increase and v will decrease
(c) v will increase but n will decrease
(d) both n and v will decrease.    

ANSWER: (b).      

EXPLANATION: Due to the increase in temperature, the average energy exchanged in a collision increases which results in more valance electrons crossing the gap and the number of electron-hole pairs increases i.e. n increases.   
          Due to the increase in thermal collisions, the average drift speed v decreases. 
Option (b) is correct.                






       3. Let nₚ and nₑ be the number of holes and conduction electrons in an intrinsic semiconductor.
(a) nₚ > nₑ
(b) nₚ = nₑ
(c) nₚ < nₑ
(d) nₚ ≠ nₑ.  

ANSWER: (b).      

EXPLANATION: In an intrinsic semiconductor, a hole is created due to a shift of a valance electron to the conduction band, they are in pairs, and the number of holes equals the number of conduction electrons.   
         Option (b) is correct.   






       4. Let nₚ and nₑ be the number of holes and conduction electrons in an extrinsic semiconductor.
(a) nₚ > nₑ
(b) nₚ = nₑ
(c) nₚ < nₑ
(d) nₚ ≠ nₑ.  

ANSWER: (d).      

EXPLANATION: Extrinsic semiconductors are made by doping the pure (intrinsic) semiconductors with impurities having five or three valance electrons. These impurities do not create electron-hole pairs. Either they make available extra conduction electrons or extra holes in the valance band depending upon the type of the impurity. So the number of conduction electrons and holes in an extrinsic semiconductor is not equal.  
   Option (d) is correct.               






       5. A p-type semiconductor is 
(a) positively charged 
(b) negatively charged
(c) uncharged
(d) uncharged at 0 K but charged at higher temperatures.     

ANSWER: (c).      

EXPLANATION: In a pure semiconductor such as silicon, each atom's four valance electrons make four covalent bonds with neighboring silicon atoms. When an impurity element having three valance electrons is added in a very low quantity it replaces silicon atoms at different places. Now one of the neighboring silicon atoms has four covalent bonds but the one with the impurity atom is incomplete because only silicon shares the valance electron. So a hole is created and this extrinsic semiconductor has abundant holes so it is called a p-type semiconductor. Neither the silicon nor the added impurity is electrically charged and the resulting p-type semiconductor is also uncharged.   
       Option (c) is correct.  






       6. When an impurity is doped into an intrinsic semiconductor, the conductivity of the semiconductor 
(a) increases 
(b) decreases
(c) remains the sane
(d) becomes zero.     

ANSWER: (a).      

EXPLANATION: When an impurity is doped into an intrinsic semiconductor, the number of charged carriers increases. Hence the conductivity of the semiconductor increases. 
        Option (a) is correct.                






       7. If the two ends of the p-n junction are joined by a wire 
(a) there will not be a steady current in the circuit
(b) there will be a steady current from the n-side to the p-side 
(c) there will be a steady current from the p-side to the n-side 
(d) there may or may not be a current depending upon the resistance of the connecting wire. 

ANSWER: (a).      

EXPLANATION: When the two ends of a p-n junction are joined by a wire, there will be diffusion and drift currents that will cancel each other. So there will not be a steady current in the circuit.   
   Option (a) is correct.                






       8. The drift current in a p-n junction is 
(a) from the n-side to the p-side
(b) from the p-side to the n-side
(c) from the n-side to the p-side if the junction is forward-biased and in the opposite direction if it is reverse-biased. 
(d) from the p-side to the n-side if the junction is forward-biased and in the opposite direction if it is reverse-biased. 

ANSWER: (a).      

EXPLANATION: If an electron-hole pair is created in the depletion region, the electron is quickly pushed by the electric field toward the n-side and the hole toward the p-side. Due to this continuous process, there is a regular flow of electrons toward the n-side and the holes toward the p-side. This makes a current from the n-side to the p-side. This current is called the drift current.    
   Option (a) is current.          






       9. The diffusion current in a p-n junction is 
(a) from the n-side to the p-side   
(b) from the p-side to the n-side
(c) from the n-side to the p-side if the junction is forward-biased and in the opposite direction if it is reverse-biased. 
(d) from the p-side to the n-side if the junction is forward-biased and in the opposite direction if it is reverse-biased. 

ANSWER: (b).      

EXPLANATION: Due to the presence of an electric field in the depletion region, holes in the p-side and conduction electrons in the n-side find it difficult to diffuse to the other side against the field. Only a few energetic holes and conduction electrons are able to overcome the electric field in the depletion region and the holes from the p-side crossover to the n-side while electrons from the n-side crossover to the p-side. Thus there is a diffusion current from the p-side to the n-side.  
        Option (b) is correct.             






       10. Diffusion current in a p-n junction is greater than the drift current in magnitude 
(a) if the junction is forward-biased
(b) if the junction is reverse-biased
(c) if the junction is unbiased
(d) in no case.     

ANSWER: (a).      

EXPLANATION: When no battery is connected, the potential of the n-side is higher than the p-side. If the p-n junction is connected in forward bias, the potential of the p-side is raised and hence the height of the potential barrier decreases. The width of the depletion region is also reduced in forward bias. So more diffusion is allowed and the diffusion current increases. The drift current is unchanged because the rate of formation of new electron-hole pairs is fairly independent of the electric field. Thus, in this case, the diffusion current is more than the drift current.    
    If the p-n junction is unbiased, the diffusion current and drift current are in equal and opposite directions.  
     If the junction is reverse-biased, the diffusion current will be less than the drift current.
     Hence option (a) is correct.   






       11. Two identical p-n junctions may be connected in series with a battery in three ways (figure 45-Q1). The potential difference across the two p-n junctions is equal in  
(a) circuit 1 and Circuit 2 
(b) circuit 2 and Circuit 3
(c) circuit 3 and Circuit 1 
(d) circuit 1 only.   
The figure for Q-11


ANSWER: (b).      

EXPLANATION: In Circuit 1, one junction is in forward bias thus offering almost no resistance. So it can be replaced with a wire with no potential difference across it. Another junction is in reverse bias which offers maximum resistance and behaves like a cut in the circuit. So there is the maximum potential difference across it. Thus both junctions have unequal potential differences. 
      In Circuit 2, both junctions are in forward bias, thus offering almost no resistance. So the potential difference across each of them is equal to almost zero. 
      In circuit 3, both the junctions are connected in reverse bias thus offering maximum resistance and almost no current across them. Hence the potential difference across each of them is equal and maximum. 
     So in Circuit 2 and Circuit 3, the potential difference across the two p-n junctions are equal. Option (b) is correct.              






       12. Two identical capacitors A and B are charged to the same potential V and are connected in two circuits at t = 0 as shown in Figure (45-Q2). The charges on the capacitors at time t =CR are respectively, 
(a) VC, VC 
(b) VC/e, VC
(c) VC, VC/e
(d) VC/e, VC/e.     
The figure for Q-12


ANSWER: (b).      

EXPLANATION: Before connection, the charge on the capacitor is equal to 
Q =VC. 
    Since B is offering reverse bias to the p-n junction, there will be no current flow in the circuit and the charge on it will remain the same (=VC) even at time t =CR.  
    Capacitor A is offering forward bias to the p-n junction which will allow the current through it with almost no resistance. The p-n junction may be replaced as a conducting wire allowing the capacitor to start discharging. The remaining charge on the capacitor at time t is given as, 
 q =Qe(-t/CR)  
For t =CR,
q =Qe⁻¹ =VC/e.

   So option (b) is correct.      






       13. A hole diffuses from the p-side to the n-side in a p-n junction. This means that 
(a) a bond is broken on the n-side and the electron freed from the bond jumps to the conduction band 
(b) a conduction electron on the p-side jumps to a broken bond to complete it
(c) a bond is broken on the n-side and the electron freed from the bond jumps to a broken bond on the p-side to complete it
(d) a bond is broken on the p-side and the electron freed from the bond jumps to a broken bond on the n-side to complete it.

ANSWER: (c).      

EXPLANATION: A hole diffuses from the p-side to the n-side in a p-n junction, which means a broken bond is created in the n-side and a broken bond disappears in the p-side by accepting a conduction electron which is a result of the broken bond on the n-side. Thus a hole in the p-side is said to be diffused to the n-side.  
   So option (c) is correct.             






       14. In a transistor, 
(a) the emitter has the least concentration of impurity.    
(b) the collector has the least concentration of impurity. 
(c) the base has the least concentration of impurity
(d) all the three regions have equal concentrations of impurity.
 
ANSWER: (c).      

EXPLANATION: The emitter is heavily doped which means that it has the highest concentration of impurity. Option (a) is incorrect.  
  The collector is moderately doped and the middle layer called the base is very lightly doped. This means that the base has the least concentration of impurities and the collector's impurity concentration is in between the base and emitter. 
   Clearly, only option (c) is correct.            






       15. An incomplete sentence about transistors is given below. The emitter- .......... junction is __ and the collector- ......... junction is __· The appropriate words for the dotted empty positions are, respectively,    
(a) 'collector' and 'base'. 
(b) 'base' and 'emitter' 
(c) collector' and 'emitter'
(d) 'base' and 'base'.   

ANSWER: (d).      

EXPLANATION: There is no emitter-collector junction in a transistor, hence options (a), (b), and (c) are incorrect. 
     A transistor has an emitter-base junction and a collector-base junction. Thus only option (d) is correct.       

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Links to the Chapters


CHAPTER- 45- Semiconductors and Semiconductor Devices




CHAPTER- 34- Magnetic Field

CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics




CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion









CHAPTER- 11 - Gravitation





CHAPTER- 10 - Rotational Mechanics






CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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Click here for → OBJECTIVE-II

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Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

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Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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CHAPTER- 6 - Friction

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Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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