This set of Engineering Physics Questions and Answers for Freshers focuses on “Properties of Semiconductors II”.
1. Which method can be used to distinguish between the two types of carriers?
a) Hall effect
b) Rayleigh method
c) Doppler effect
d) Fermi effect
Explanation: When a conductor carrying current is placed in a transverse magnetic field, an electric field is produced inside the conductor in a direction normal to both the current and the magnetic field. This phenomenon is known as the Hall Effect.
2. Find the resistance of an intrinsic Ge rod cm long, 1mm wide and 1mm thick at 300K.
a) 2.32 ohm
b) 5314 ohm
c) 4310 ohm
d) 431 ohm
Explanation: Conductivity of an intrinsic semiconductor = nie(μe+μh)
Conductivity = 2.32
Resistance = ρl/A = l/(conductivy ×A)
Resistance = 4310 ohm.
3. A semiconducting crystal 12mm long, 5mm wide and 1mm thick has a magnetic flux density of 0.5Wb/m2 applied from front to back perpendicular to largest faces. When a current of 20mA flows length wise through the specimen, the voltage measured across its width is found to be 37μV. What is the Hall coefficient of this semiconductor?
a) 37×10-6 m3/C
b) 3.7×10-6 m3/C
c) 3.7×106 m3/C
Explanation: Hall coefficient = (VH b)/(IH B)
Hall coefficient = 3.7×10-6 m3/C.
4. The intrinsic carrier density at room temperature in Ge is 2.37×1019/m3. If the electron and hole mobilities are 0.38 and 0.18 m2/Vs respectively. Calculate its resistivity.
a) 0.18ohm m
b) 0.460ohm m
c) 0.4587ohm m
d) 0.709ohm m
Explanation: Conductivity = nie(μe+μh)
Conductivity = 2.12352/ohm m
Resistivity = 1/Conductivity
Resistivity = 0.4709ohm m.
5. A silicon plate of thickness 1mm, breadth 10mm and length 100mm is placed in a magnetic field of 0.5 Wb/m2 acting perpendicular to its thickness. If 10-3 A current flows along its length, calculate the Hall voltage developed, if the Hall coefficient is 3.66×104 m3/Coulomb.
Explanation:VH = (RH IH B)/t
VH = 1.83×10-3 Volts.
6. The conductivity of germanium at 20°C is 2/ohm m. What is its conductivity at 40°C? Eg=0.72eV
a) 1.38×10-23/Ohm m
b) 1.0002/ Ohm m
c) 293/ Ohm m
d) 313/ Ohm m
Explanation: σ = Ce(-E/2KT)
σ1/σ2 = e(-E/2KT)/e(-E/2KT)
σ2 = 1.0002/ Ohm m.
7. What is the Fermi energy of a n-type semiconductor?
b) E(F )= (Ec+ Ev)/2
c) EF = (Ec+ Ed)/2
d) EF = (Ev+ Ea)/2
Explanation: The Fermi energy level of n-type semiconductor lies exactly between the acceptor energy level and the maximum energy level of valence band. Therefore the Fermi energy level of n-type semiconductor is EF = (Ec+ Ed)/2.
8. EF = (Ec+ Ev)/2, this represents the Fermi energy level of which of the following?
a) Extrinsic semiconductor
b) N-type semiconductor
c) P-type semiconductor
d) Intrinsic semiconductor
Explanation: The Fermi energy of an intrinsic semiconductor is EF = (Ec+ Ev)/2. That is the Fermi energy level exactly lies between the lowest energy level of conduction band and highest energy level of valence band.
9. For semiconductors, the resistivity is inversely proportional to the temperature for semiconducting materials.
Explanation: For semiconductors, the resistivity is inversely proportional to the temperature of the material, that is, it has a negative temperature coefficient. When the temperature of the semiconductor is increased, large numbers of charge carriers are produced due to the breaking of covalent bonds. These charge carriers move freely, hence conductivity increases and therefore the resistivity decreases.
Sanfoundry Global Education & Learning Series – Engineering Physics.
To practice all areas of Engineering Physics for Freshers, here is complete set of 1000+ Multiple Choice Questions and Answers.