Question 1

How many 1s energy states are present in one mole of sodium vapour? Are they all filled in normal conditions? How many 3s energy states are present in one mole of sodium vapour? Are they all filled in normal conditions?

Accepted Answer

For sodium, the atomic number is 11 . The electronic configuration of sodium is 1 s22 s22p63 s1. One sodium atom has 11 electrons. Thus, if the sodium crystals consist of N atoms, the total number of electrons will be 11 N. W know that for each atom, there are two states in the energy level 1 s. Thus, the sodium crystal will have 2N states for 1 s energy level. Similarly, the number of states in 3 s energy level will also be 2 N. 1 s  state is filled under normal condition. But the 3s state has only one electron per sodium atom, so the 3s band will be half-filled.

Question 2

There are energy bands in a solid. Do we have really continuous energy variation in a band ro do we have very closely spaced but still discrete energy levels?

Accepted Answer

A solid consists of a combination of closely spaced energy levels. These energy levels are discrete but they have very small energy gap between two consecutive levels so they are reffered as band.However, the energy levels in the band are discrete.

Question 3

The conduction band of a solid is partially filled at 0 K. Will it be a conductor, a semiconductor or an insulator?

Accepted Answer

It will be a conductor. As the elements having partially filled conduction band belong to the category of elements whose outermost subshell consists of an odd number of electrons, they are good conductors of electricity. When an electric field is applied, electrons in the partially filled band gain energy and start drifting. So, the conductor will conduct even at 0 K. A semiconductor behaves like an insulator at 0 K and an insulator conducts poorly only at very high temperatures.As the given material has free electrons to conduct even at 0 K, it is a conductor.

Question 4

In semiconductors, thermal collisions are responsible for taking a valence electron to the conduction band. Why does the number of conduction electrons not go on increasing with time as thermal collisions continuously take place?

Accepted Answer

An electron jumps from the valence band to the conduction band only when it has gained sufficient energy. The thermal collisions sometimes do not provide sufficient energy to the electron to jump. Also, energy is lost in the form of heat because of the collision of the carriers with other charge carriers and atoms. Because of all these losses, only few electrons are left with sufficient energy to jump from the valence band to the conduction band. So, the population of electron in the conduction band does not keep on increasing with time.

Question 5

When an electron goes from the valence band to the conduction band in silicon, its energy is increased by 1.1 eV. The average energy exchanged in a thermal collision is of the order of kT  which is only 0.026 eV at room temperature. How is a thermal collision able to take some to the electrons from the valence band to the conduction band?

Accepted Answer

Fermi level: it is the energy level occupied by the highest energy electron.In an extrinsic semiconductor for example in n-type semiconductor, fermi level lies close to the conduction band so it needs a very small amount of energy to excite the electron from fermi level to conduction band. This energy is comparable to the thermal excitation energy. So even at room temperature,these semiconductors can conduct.For a p-type semiconductor, fermi level lies close to valence bane because here conduction takes place majorly via holes.So by the thermal excitation,a bond is broken and an electron hole pair is created.Out of this,hole comes to the valence band for conduction or equivalently an electron goes to the conduction band. In an intrinsic semiconductor, no impurity is doped so fermi level lies at the centre of band gap. Here only few electrons get sufficient energy via repeated thermal collisions to jump from the fermi level to the conduction band.Hence the conductivity of intrinsic semiconductor is less as compared to extrinsic semiconductor.

Question 6

What is the resistance of an intrinsic semiconductor at 0 K?

Accepted Answer

At 0 K, the valence band is full and the conduction band is empty. As no electron is available for conduction in an intrinsic semiconductor, the intrinsic semiconductor at 0 K acts as an insulator and hence offers infinite resistance.

Question 7

We have valence electrons and conduction electrons in a semiconductor. Do we also have 'valence holes' and 'conduction holes'?

Accepted Answer

Holes do not exist in reality. They exist only virtually. When an electron jumps from the valence band to the conduction band, a vacancy is created at the place from where the electron had jumped. This vacancy is called a hole. So, a valence or conduction hole is a virtual concept only.

Question 8

When a p-type impurity is doped in a semiconductor, a large number of holes are created, This does not make the semiconductor charged. But when holes diffuse from the p-side to the n-side in a p-n junction, the n-side gets positively charged. Explain.

Accepted Answer

A p-type semiconductor is formed by doping a group 13 element with group 14 element (Si or Ge). As the group 13 element has only 3 electrons in its valence shell and the group 14 element has 4 electrons in its valence shell, when the group 13 element, say, Al, replaces one Si in the silicon crystal, only 3 covalent bonds are formed by it. And the fourth covalent bond is left in need of one electron. So, it creates a hole. Since the atom as a whole is electriclly neutral, the p-type semiconductor is also neutral.In a p‒n junction, when the diffusion of holes takes place across the junction because of the difference in the concentration of charge carriers from p to n sides, these holes neutralise some of the electrons on the n side. So, the atom attached with that electron becomes one electron deficient and hence positively charged. This makes the n side of the p‒n junction positively charged and the p side of the p‒n junction negatively charged.

Question 9

The drift current in a reverse-biased p-n junction is increased in magnitude if the temperature of the junction is increased. Explain this on the basis of creation of hole-electron pairs.

Accepted Answer

When the temperature of a reverse-biassed p‒n junction is increased, the breaking of bonds takes place because of the increase in the thermal energy of the charge carriers. Drift current is due to the flow of the minority carriers across the junction. So, when a p‒n junction is reverse biassed, the applied voltage supports the flow of minority charge carriers across the junction. Thus, the drift current increases with increase in temperature in a reverse-biassed p‒n junction.

Question 10

An ideal diode should pass a current freely in one direction and should stop it completely in the opposite direction. Which is closer to ideal-vacuum diode or a p-njunction diode?

Accepted Answer

It should be an ideal vacuum diode. When a pn junction diode is reverse biassed then a small current called reverse current flows across the diode.As the the p‒n junction diode allows some current in reverse biassed condition also so the given diode can not be a pn junction diode.

Question 11

Consider an amplifier circuit using a transistor. The output power is several times greater than the input power. Where does the extra power come from?

Accepted Answer

The amplifier takes this energy from the power supply. Amplifiers take the energy from the power supply and control the output to match with the input signal but with greater amplitude.

Question 12

Electric conduction in a semiconductor takes place due to:

Accepted Answer

Both electrons and holesA hole is created in a semiconductor when a valence electron moves to the conduction band. When potential difference is applied across the semiconductor, the electron drifts opposite to the electric field applied, while the hole moves along the electric field. Therefore, electric conduction takes place in a semiconductor because of both electrons and holes.

Question 13

An electric field is applied to a semiconductor. Let the number of charge carries be n and the average drift speed by v. If the temperature is increased:

Accepted Answer

n will increase but v will decrease.As we increase the temperature, additional electron-hole pairs are created in a semiconductor. As a result, the number of charge carriers increases.Now, drift velocity (vd​) is given byvd​=m−eEτ​ As the temperature increases, the relaxation time of charge carriers (τ) decreases. As a result, vd​ decreases.

Question 14

Let np​  and ne​  be the number of holes and conduction electrons in an intrinsic semiconductor.then:

Accepted Answer

np​=ne​ As the intrinsic semiconductor is free from all impurities, the number of electrons is equal to the number of holes.

Question 15

Let np  and ne  be the numbers of holes and conduction electrons in an extrinsic semiconductor.

Accepted Answer

np ≠ neExtrinsic semiconductors are formed by doping either an n-type material or a p-type material with a pure semiconductor, Thus, it may have more number of holes (if the doping material is of p type) or more number of electrons (if the doping material is of n type).

Question 16

A p−type  semiconductor is:

Accepted Answer

UnchargedA p-type semiconductor is formed by doping a pure semiconductor with a p-type material. As impurity atoms take the position of the germanium atom in a germanium crystal, three electrons of a p-type material form covalent bonds by sharing electrons with three neighbouring germanium atoms. However, the fourth covalent bond is left incomplete, with a want of one electron. This creates a hole. As the atom as a whole is neutral, the p-type material is also neutral.

Question 17

When an impurity is doped into an intrinsic semiconductor, the conductivity of the semiconductor:

Accepted Answer

IncreasesWhen an impurity (either a p-type atom or an n-type atom) is doped into an intrinsic semiconductor, it increases the number of charge carriers in the intrinsic semiconductor. As conductivity is directly related to the number of charge carriers, the conductivity of a semiconductor increases with doping.

Question 18

If the two ends of a p−n  junction are joined by a wire:

Accepted Answer

There will not be a steady current in the circuit.     In a p‒n junction, current flows only if it is connected to the battery. If two ends of a p‒n junction are joined by a wire, then there will be diffusion and drift currents in the circuit and they will cancel each other. Hence, no current will flow in the circuit.

Question 19

The drift current in a p−n  junction is:

Accepted Answer

From the n-side to the p-sideAfter the diffusion of majority charge carriers across a p‒n junction, an electric field is set up because of the accumulation of immobile ions at the junction. These further oppose the motion of majority charge carriers across the junction. As a result, electrons from the p region start moving to the n region and holes from the n region start moving to the p region. This constitutes the drift current. As the direction of the current is opposite to the direction of the motion of the electrons, the direction of the drift current is from the n side to the p side.In forward biasing, there is no movement of electrons from the p region to the n region and of holes from the n region to the p region. Hence, there is not drift current.

Question 20

The diffusion current in a p-n junction is:

Accepted Answer

From the p-side to the n-sideWhen a p‒n junction is formed then because of the difference in the concentration of charge carriers in the two regions, electrons from the n region move to the p region and holes from the p region move to the n region. Since the direction of the current is always opposite to the motion of electron, the direction of the current is from the p side to the n side. Similarly, when the junction is forward biassed, the positive terminal of the battery is connected to the p side of the p‒n junction and the negative terminal of the battery is connected to the n side of the p‒n junction. As a result, electrons in the n side of the p‒n junction are repelled by the negative terminal of the battery and they move to the p side, where the positive terminal of the battery attracts them. Similarly, holes from the p side of the p‒n junction are repelled by the positive terminal of the battery and they move to the n side, where the negative terminal of the battery attracts them. Thus, they give diffusion current from the p side to the n side across the p‒n junction.In reverse biassing, there is no flow of majority carriers across the junction; hence, there is not diffusion current. Here, the flow of majority carriers is opposed by the applied voltage.

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Solution for Class 12, Physics, Semiconductor Electronics: Materials, Devices and Simple Circuits

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