Wednesday, July 19, 2023

H C Verma solutions, SEMICONDUCTORS AND SEMICONDUCTOR DEVICES, Chapter-45, EXERCISES, Q1 to Q10, Concepts of Physics, Part-II

EXERCISES, Q1 - Q10


       1. Calculate the number of states per cubic meter of sodium in the 3s band. The density of sodium is 1013 kg/m³. How many of them are empty?  

ANSWER: One sodium atom has two numbers of states in the 3s band. Its atomic number is 11 with an electronic configuration of 1s²2s²2p⁶3s¹. So out of two 3s states only one is filled in each atom. 
   Let us first find the number of sodium atoms in each cubic meter. The atomic mass of sodium is 23, so 23 g of sodium will have a 6.02x10²³ number of atoms. Since one cubic meter of sodium has a weight of 1013 kg, it will have a number of sodium atoms 
=1013x10³x6.02x10²³/23 
=2.65x10²⁸

Hence the number of states in the 3s band of one cubic meter of sodium is 
=2x2.65x10²⁸
=5.3x10²⁸.   

Since half of them are empty, the number of empty states in the 3s band is
=2.65x10²⁸.    






       2. In a pure semiconductor, the number of conduction electrons is 6x10¹⁹ per cubic meter. How many holes are there in a sample of size 1 cm x 1 cm x 1 mm?

ANSWER: In a pure or intrinsic semiconductor, the number of conduction electrons and holes are equal in number because they are in pairs. Hence the number of holes in the given sample  
=0.01*0.01*0.001*6x10¹⁹  
=6x10¹².   




  
       3. Indium antimonide has a band gap of 0.23 eV between the valence and the conduction band. Find the temperature at which kT equals the band gap.  

ANSWER: For the given condition, 
kT =0.23 eV, where k is the Boltzmann constant.  
→T =0.23 eV/k 
→T =(0.23 eV)/(8.62x10⁻⁵ eV/K) 
       =2668 K
       ≈2670 K.   




 
       4. The band gap for silicon is 1.1 eV. (a) Find the ratio of the band gap to kT for silicon at room temperature 300 K. (b) At what temperature does this ratio become one-tenth of the value at 300 K? (Silicon will not retain the structure at these high temperatures.)  

ANSWER: (a) The ratio of the band gap to kT at room temperature is 
=1.1/(8.62x10⁻⁵*300) 
≈43.   

(b) Let the required temperature =T'. The ratio of the band gap to kT' is 43/10 =4.3. 
So, 1.1/(kT') =4.3
→T' =1.1/(4.3k) 
    =1.1/(4.3*8.62x10⁻⁵)  
    =2967.7 K 
    ≈3000 K.     




 
       5. When a semiconducting material is doped with an impurity, new acceptor levels are created. In a particular thermal collision, a valence electron receives energy equal to 2kT and just reaches one of the acceptor levels. Assuming that the energy of the electron was at the top edge of the valance band and that the temperature T is equal to 300 K, find the energy of the acceptor levels above the valence band.  

ANSWER: The electron at the top of the valance band receives energy equal to 2kT and just reaches one of the acceptor levels. Hence the energy of the acceptor levels above the valence band is =2kT. 
i.e. =2*8.62x10⁻⁵*300 eV 
=0.05 eV 
=50 meV.      




 
       6. The band gap between the valence and the conduction bands in zinc oxide (ZnO) is 3.2 eV. Suppose an electron in the conduction band combines with a hole in the valence band and the excess energy is released in the form of electromagnetic radiation. Find the maximum wavelength that can be emitted in this process.   

ANSWER: Maximum wavelength will be emitted when minimum energy is emitted. And it will happen when the conduction electron combines with the hole in the top level of the valence band thus releasing energy equal to the band gap. So here is the released energy,
E =3.2 eV 
→hc/λ =3.2 
→λ =hc/3.2
    =4.14x10⁻¹⁵*3x10⁸/3.2 m
    =3.88x10⁻⁷ m
    =3.88x10⁻⁷*10⁹ nm
    =388 nm
    ≈390 nm.
    




 
       7. Suppose the energy liberated in the recombination of a hole-electron pair is converted into electromagnetic radiation. If the maximum wavelength emitted is 820 nm, what is the band gap?   

ANSWER: The band gap will be equal to the minimum energy liberated in the recombination. Minimum energy is with the maximum wavelength released which is here 820 nm. So the band gap =E =hc/λ
=4.14x10⁻¹⁵*3x10⁸/(820x10⁻⁹) eV
=1.5 eV.
   




 
       8. Find the maximum wavelength of electromagnetic radiation that can create a hole-electron pair in germanium. The band gap in germanium is 0.65 eV.  

ANSWER: The minimum energy of electromagnetic radiation (i.e. having the maximum wavelength) which can create a hole-electron pair in germanium will be equal to the band gap =0.65 eV. 
Hence E =hc/λ =0.65 eV
→λ =hc/0.65
      =4.14x10⁻¹⁵*3x10⁸/0.65
      =1.9x10⁻⁶ m. 
 




 
       9. In a photodiode, the conductivity increases when the material is exposed to light. It is found that the conductivity changes only if the wavelength is less than 620 nm. What is the band gap? 

ANSWER: It is clear that the conductivity increases when the light of minimum energy with wavelength 620 nm falls on the photodiode. The valance electrons in the top levels of the valence band absorb this energy amount of energy and just reach the conduction band, thus conductivity increases. So the band gap is equal to this amount of energy.  
Hence the band gap =hc/λ
    =4.14x10⁻¹⁵*3x10⁸/620x10⁻⁹ eV
    =2.0 eV.
   



 

       10. Let ΔE denote the energy gap between the valence band and the conduction band.  The population of conduction electrons (and of the holes) is roughly proportional to e-ΔE/2kT. Find the ratio of the concentration of conduction electrons in diamond to that silicon at room temperature 300 K. ΔE for silicon is 1.1 eV and for diamond is 6.0 eV. How many conduction electrons are likely to be in one cubic meter of diamond?  

ANSWER: For silicon ΔE =1.1 eV
For diamond ΔE' =6.0 eV.
T =300 K.

From the given condition, the ratio of the concentration of the conduction electrons in diamond to that of silicon
=(e-ΔE'/2kT)/(e-ΔE/2kT)
=e(ΔE-ΔE')/2kT
=e(1.1-6.0)/2kT
=e-4.9/(2*8.62x10⁻⁵*300)
=e-94.74
=7.16x10⁻⁴².

    From this ratio, we conclude that for the diamond, the population of the conduction electrons will be about 7.16x10⁻⁴² times the silicon. The number of conduction electrons per m³ for silicon at room temperature 300 K is about 7x10¹⁵. Hence the number of conduction electrons per m³ of the diamond at room temperature is 
=7.16x10⁻⁴²*7x10¹⁵ 
=5x10⁻²⁶, which is almost zero. 

---------------------------------------------------

 Buy Home Furnishing

Click here for all links → kktutor.blogspot.com 

===<<<O>>>===


My Channel on YouTube  →  SimplePhysics with KK


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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

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".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

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

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"


Posted by at
Labels: , , , ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)