- AThe average energy of electrons and holes.
- BThe energy of electrons in conduction band.
- CThe energy of holes in valence band.
- DThe energy of forbidden region.
Explanation:
In an intrinsic semiconductor, n=p, where, n is number of electrons and p is number of holes in intrinsic semiconductor.
This implies that there is an equal chance of finding an electron at the conduction band edge as there is of finding a hole at the valence band edge. Thus, the average energy level of electrons and holes is half of the energy band gap in intrinsic semiconductors.
Also the Fermi energy level lie exactly in the middle of energy band gap in intrinsic semiconductors. Thus, the value indicated by Fermi energy level in an intrinsic semiconductor is the average energy of electrons and holes.
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The value of internal resistance of an ideal cell is
|
(a) Zero |
(b) 0.5 Ω |
(c) 1 Ω |
(d) Infinity |
The current in the arm CD of the circuit will be

|
(a) |
(b) |
(c) |
(d) |
The energy band diagrams for three semiconductor samples of silicon are as shown. We can then assert that

|
(a) Sample X is undoped while samples Y and Z have been doped with a third group and a fifth group impurity respectively |
|
(b) Sample X is undoped while both samples Y and Z have been doped with a fifth group impurity |
|
(c) Sample X has been doped with equal amounts of third and fifth group impurities while samples Y and Z are undoped |
|
(d) Sample X is undoped while samples Y and Z have been doped with a fifth group and a third group impurity respectively |
Two rods of same material and length have their electric resistance in ratio 1 : 2. When both rods are dipped in water, the correct statement will be
|
(a) A has more loss of weight |
(b) B has more loss of weight |
|
(c) Both have same loss of weight |
(d) Loss of weight will be in the ratio 1 : 2 |
A circular loop has a radius of 5 cm and it is carrying a current of 0.1 amp. Its magnetic moment is
|
(a) 1.32 |
(b) 2.62 |
|
(c) 5.25 |
(d) 7.85 |
Which one is not the correct statement
|
(a) 1 volt × 1 coulomb = 1 joule |
(b) 1 volt × 1 ampere = 1 joule/second |
|
(c) 1 volt × 1 watt = 1 H.P. |
(d) Watt-hour can be expressed in eV |
Two cells of equal e.m.f. and of internal resistances and
are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be
|
(a) |
(b) |
(c) |
(d) |