# MCQ on Semiconductor Material

## MCQ on Semiconductor Material

MCQ on Semiconductor Material, Multiple Choice Questions on Semiconductor Material, Engineering Physics MCQ, Engineering MCQ, Electronics Device & Circuit MCQ, Semiconductor Physics Multiple Choice Questions

### Multiple Choice Questions

Q.1. Which of the following, has the lowest temperature coefficient of resistivity.

Alum Silver, Copper, Gold, Aluminium

• Copper
• Silver
• Gold
• Aluminium

Q.2. In a metal

• The electrical conduction is by electrons and holes
• With rise in temperature, the conductivity decreases
• The conduction band is empty
• There is a small energy gap between the two bands

Answer: With a rise in temperature, the conductivity decreases

Q.3. The forbidden energy gap in semiconductors:

• lies just below the valence band
• is the same as the valence band
• lies just above the conduction band
• lies between the valence band and the conduction band

Answer: lies between the valence band and the conduction band

Q.4. The conduction band of a semiconductor material may be

• completely filled
• partially filled
• empty
• none

Q.5. The energy band which possesses the free electrons is called:

• valence band
• conduction band
• forbidden band
• none of these

Q.6. Which of the following behaves as an insulator.

• Diamond
• Germanium
• Silicon
• Silver

Q.7. The unit of mobility of charge carriers is

• m2/volt-sec
• m/volt-sec2
• m3/volt-sec
• m/volt-sec

Q.8. A semiconductor is electrically neutral because it has:

• no majority carriers
• no minority carriers
• no free carriers
• equal number of positive and negative carriers

Answer: an equal number of positive and negative carriers

Q.9. An intrinsic semiconductor at absolute zero temperature

• has large number of holes
• has a few hole and same number of electrons
• acts as an insulator
• acts as metallic character

Q.10. When the temperature of an intrinsic semiconductor is increased

• resistance of semiconductor is also increased
• conductivity is decreased
• energy of atoms is increased
• holes are created

Answer: energy of atoms is increased

Q.11. A doner-type impurity

• is used to obtain a p-type semiconductor
• is used to obtain a N-type semiconductor
• must possess three valence electrons
• cannot be used in silicon crystals

Answer: is used to obtain an N-type semiconductor

Q.12. If a small amount of antimony is added to Ge

• the resistance is increased
• Ge will become a p-type semi-conductor
• antimony becomes an acceptor impurity
• there will be no more free electrons than holes in a semiconductor

Answer: there will be no more free electrons than holes in a semiconductor

Q.13. In an N-type semiconductor, the concentration of minority carriers mainly depends upon

• doping technique
• number of donor atoms
• temperature of material
• quality of intrinsic materials Ge or Si

Q.14. In a P-type semiconductor, majority carriers are

• holes
• electrons
• both
• none of the above

Q.15. Which of the following atoms can be used as a p-type impurity

• Boron
• Arsenic
• Antimony
• Phosphorus

Q.16. In an intrinsic semiconductors, the Fermi level lies:

• in the middle of conduction band (CB) and valence band (VB)
• near CB
• near VB
• none of these

Answer: in the middle of CB and VB

Q.17. 10-9 ohm-m is the resistivity of

• Aluminium
• Sodium
• Bismuth
• Nickel

Q.18. Due to the illumination of light, the electron and hole concentrations in a heavily doped n-type semiconductor, increase by Δn and Δp respectively. If ni is the intrinsic concentration, then

• $\Delta n < \Delta p$
• $\Delta n > \Delta p$
• $\Delta n = \Delta p$
• $\Delta n \times \Delta p = n_{i}^{2}$

Answer: $\Delta n \times \Delta p = n_{i}^{2}$

Q.19. The current density Jo of electrons through any conductor carrying current is given by:

• $J_{0}=\frac{ne\tau E}{m}$
• $J_{0}=\frac{ne\tau E^{2}}{m}$
• $J_{0}=\frac{ne^{2}\tau E}{m}$
• $J_{0}=\frac{e^{2}\tau E}{m}$

Answer: $J_{0}=\frac{ne^{2}\tau E}{m}$

Q.20. An element used in semiconductors whose atoms have three valence electrons is

• Germanium
• A donor
• An acceptor
• Silicon

Q.21. The Hall coefficient RH is defined by

• $R_{H}=\frac{V_{H}.t}{B}$
• $R_{H}=\frac{V_{H}B}{It}$
• $R_{H}=\frac{V_{H}t}{B}$
• $R_{H}=\frac{V_{H}t}{BI}$

Answer: $R_{H}=\frac{V_{H}t}{BI}$

Q.22. The probability of occupation F(E) of an energy level E by an electron is given by

• $F(E)=[1+e^{(E-E_{F})}/kT]$
• $F(E)=\frac{1}{1+e^{(E_{F}-E)}/kT}$
• $F(E)=\frac{1}{1+e^{(E-E_{F})}/kT}$
• $F(E)=\frac{1}{1-e^{(E-E_{F})}/kT}$

Answer: $F(E)=\frac{1}{1+e^{(E-E_{F})}/kT}$

Q.23. For high conductivity and large temperature coefficient

• Pure metals are chosen
• Slightly impure materials are used
• Alloys are used
• Super-conductors are used

Q.24. The potential barrier at a pn junction is established due to the charge on either side of the Junction. These charges are

• majority carriers
• minority carriers
• both a and b
• donor and acceptor ions

Q.25. In a PN junction, holes diffuse from the p-region to the n-region because

• they move across the junction by the potential diffierence
• free electron available in the n-region attract them
• the holes concentration in the p-region is greater as compared to n-region
• all of these

Answer: the holes concentration in the p-region is greater as compared to n-region

Q.26. In a crystal diode, the barrier potential offers opposition to only

• free electrons in n-region
• holes in p-region
• majority carriers in both regions
• minority carriers in both regions

Answer: majority carriers in both regions

Q.27. When a reverse bias is applied to a crystal diode, it

• raises the potential barrier
• lowers the potential barrier
• increases the majority-carrier current greatly
• none of these

Q.28. At reverse bias, the number of minority carriers crossing the junction of a diode depends primarily on the

• concentration of doping impurities
• rate of thermal generation of electron-hole pairs
• magnitude of the potential barriers
• all of these

Q.29. When a forward bias is applied to a crystal diode, it

• raises the potential barrier
• lowers the potential barrier
• reduces the majority-carrier current to zero
• none of these