HC Verma Solutions Class 12 Chapter 23 Semiconductor and Semiconductor Devices can be used as a source of background information on topics where student’s might lack proficiency such as finding the number of conduction electrons, holes, unoccupied states in semiconductors etc. The solutions are mainly designed for class 12 students and cover the complete JEE PhysicsÂ topics as well as the exercises discussed in the chapter. Some of the common questions asked in this unit are;

- There are questions related to finding the energy band between the acceptor band and valence band.
- Questions that deal with finding the energy needed to overcome the band gap.
- Students will solve questions based on unbiased, reverse and forward biasing of current and many more.

The whole idea behind preparing and offering the solutions is to help students get familiar with the question-solving pattern and integrate them into their study activities. With the solutions, students will further get an insight into important topics and understand them more clearly in the process. Students can use the solutions to also prepare well for competitive exams like JEE.

**Download This Solution As PDF: **HC Verma Solutions Chapter 23 PDF

## Key Topics Related To Semiconductor In Physics

- Energy Bands in Solids
- p-type and n-type Semiconductors
- The Density of Charge Carriers and Conductivity
- p-n Junction
- p-n Junction Diode
- p-n Junction as a Rectifier
- Junction Transistors
- Logic Gates

## Class 12 Important Questions In Chapter 23

**1.** If the conduction band of a solid is partially filled at 0 K then which category will it belong to. Will it be a semiconductor, a conductor or an insulator?

**2.** When we increase the temperature of the junction the drift current in a reverse-biased p-n junction also increases in magnitude. On the basis of creation of hole-electron pairs can you explain why and how this occurs?

**3.** In a semiconductor, you will normally find valence electrons and conduction electrons. Does a semiconductor also have ‘conduction holes’ and ‘valence holes’?

**4.** What condition will you find a p-type semiconductor always in?

(a) It is positively charged (b) has a negative charge (c) remains uncharged at 0 K but gets charged at high temperatures (d) no charge exists

**5.** When will the diffusion current in a p-n junction be greater than the drift current in magnitude?

(a) when the junction is unbiased (b) if the junction is forward-biased (c) if the junction is reverse-biased (d) such criteria do not exist

## HC Verma Solutions Vol 2 Semiconductor and Semiconductor Devices Chapter 23

**Question 1:** Calculate the number of states per cubic metre of sodium is 3s band. The density of sodium is 1013 kg m^{-3}. How many of them are empty?

**Solution: **

Given: Mass of the sodium in 1 m^{3} = 1013 kgm^{-3}

Atomic mass of sodium = 23

Total number of atoms = N = [mass x avogadro’s number]/[atomic mass]

= [1013×10^{3}x6.022×10^{23}]/[23]

= 265.22 x 10^{26} atoms

In N atoms of sodium there will be 2N states possible states and N states will be empty.

Total number of states = 2 x N = 2 x 265.22 x 10^{26} = 5.30 x 10^{28} states

As atomic number of sodium = 11

Its electronic configuration = 1s^{2}, 2s^{2}, 2p^{6}, 3s^{1}

=> The 3s band is half-filed in case of sodium, the total number of unoccupied states is 2.65 x 10^{28}.

**Question 2: **In a pure semiconductor, the number of conduction electrons is 6 Ã— 10^{19} per cubic metre. How many holes are there in a sample of size 1 cm Ã— 1cm Ã— 1 mm?

**Solution: **

Since in pure semiconductor the number of holes = number of conduction electron

=> Number of holes = 6 Ã— 10^{19} m^{-3}

Therefore, Number of holes in 1 cm Ã— 1cm Ã— 1 mm = 6 Ã— 10^{19} x 10^{-7} = 6 x 10^{12} holes

^{-7}m

^{3}]

**Question 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.

**Solution: **

We know, kT = 0.23 eV

Where, k = Boltzmann constant and t = temperature

=> T = 0.23 eV/k = 0.23 eV/[8.62×10^{-5}eVK^{-1}] = 2670 K (approx)

**Question 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 of 300 K? (Silicon will not retain its structure at these high temperatures)

**Solution: **

The band gap for silicon is 1.1 eV.

kT at room temp = 0.026 eV

(a) Ratio of band gap to kT = 1.1/0.026 = 43 (approx)

(b) Consider new ratio is 1/10^{th} of the Part (a) result.

Now, Band gap/kT = 4.3

=> 1.1/[8.62×10^{-5}xT] = 4.3

=> T = 3000 K(approx)

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

**Solution: **

The electron was at the top of the edge of valance band and reaches acceptor level on receiving the energy given.

Energy gap between the acceptor level and valence band must be 2kT.

Where, k = Boltzmann constant and t = temperature

=> E = 2 x 8.62 x 10^{-5} x 300 = 51.72 meV

**Question 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.

**Solution: **

For the electron to move from the conduction band to valance band will have to lose energy same as the energy band gap of ZnO OR The max energy released in this process will be equal to the band gap of the material.

=> E = 3.2 eV

=> 3.2 = [1242eV-nm]/Î»

=> Î» = 388.1 nm

**Question 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?

**Solution: **

The min energy release in the recombination of the conduction band electron with valence bond hole = band gap of the material

We know, E = hc/Î»

=> E = [1242eV-nm]/820nm

[given Î» = 820nm]=> E = 1.5 eV

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

**Solution: **

The band gap in germanium is 0.65 eV.

The energy required to produce the hole-electron pair = energy band gap

We know, E = hc/Î»

=> Î» = hc/E = [4.14 x10^{-15}x3x10^{8}]/0.65

= 1.9 x 10^{-6} m

Which is required wavelength.

**Question 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?

**Solution: **

Maximum wavelength at which the conduction starts = 620 nm

and this wavelength provides minimum energy to the electron to move to conduction band.

we know, Energy of this radiation = Energy band gap

=> E = [4.14×10^{-15}x3x10^{8}]/[620×10^{-9}] = 2.003 eV = 2 eV (approx)

Which is the required band gap.

**Question 10: **Let Î”E 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^{â€“Î”E/2kT}. Find the ratio of the concentration of conduction electrons in diamond to that in silicon at room temperature 300 K. Î”E 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?

**Solution: **

Let Î”E_{1} and Î”E_{2} be the energy band gap for silicon and diamond.

=> Î”E_{1} = 1.1 eV and Î”E_{2} = 6.0 eV (Given)

Also, we are given that, population of conduction electrons (and of the holes) is roughly proportional to e^{â€“Î”E/2kT}.

Therefore,

Because of more band gap, there will be almost zero conduction electrons will be there in the diamond.

**Question 11:** The conductivity of a pure semiconductor is roughly proportional to T^{3/2} e^{â€“Î”E/2kT} where Î”E is the band gap. The band gap for germanium is 0.74 eV at 4 K and 0.67 eV at 300 K. By what factor does the conductivity of pure germanium increase as the temperature is raised from 4 K to 300 K.

**Solution: **

Let E_{1} = 0.74 eV and E_{2} = 0.67 eV

Let Ïƒ_{1} and Ïƒ_{2} be the conductivity at 4 K and 300K.

Therefore,

or Ïƒ_{2} = 0.148 x 10^{463} x Ïƒ_{1} = 10^{463} Ïƒ_{1} (approx)

Which shows that conductivity increases by the factor 10^{463}.

**Question 12:** Estimate the proportion of boron impurity which will increase the conductivity of a pure silicon sample by a factor of 100. Assume that each boron atom creates a hole and the concentration of holes in pure silicon at the same temperature is 7 Ã— 10^{15} holes per cubic meter. Density of silicon is 5 Ã— 10^{28} atoms per cubic meter.

**Solution: **

Initially, the number of charge carriers = 7×10^{15} + 7×10^{15} = 14 x 10^{15}

Total charge carriers after adding the impurity to the pure silicon = 14 x 10^{15} x 100 = 14 x 10^{17}

let “y” be holes after adding the impurity and (14 x 10^{17} – y) will be the electrons.

Now,

Product of the total conduction electrons and holes remains nearly constant.

=> 7×10^{15} x 7×10^{15} = y x (14 x 10^{17} – y)

Solving above equation for y, we get

y = 13.982 x 10^{15} or y = 1.8 x 10^{15}

As, initial number of holes must be smaller than the number of holes after doping, so neglecting 2nd value.

Thus, the number of boron atoms added = 13.982 x 10^{17} – 7×10^{15}

= 1398.2 x 10^{15} – 7×10^{15}

= 13.912 x 10^{17}

13.912 x 10^{17} atoms are added per 5 Ã— 10^{28} atoms of Silicon per cubic meter.

So, 1 atom of boron is added per [5 Ã— 10^{28}]/[13.912 x 10^{17}] = 3.59 x 10^{10} atoms of Silicon per cubic meter.

Therefore, 3.59 x 10^{10} is the proportion of born impurity.

**Question 13:** The product of the hole concentration and the conduction electron concentration turns out to be independent of the amount of any impurity doped. The concentration of conduction electrons in germanium is 6 Ã— 10^{19} per cubic meter. When some phosphorus impurity is doped into a germanium sample, the concentration of conduction electrons increases to 2 Ã— 10^{23} per cubic meter. Find the concentration of the holes in the doped germanium.

**Solution: **

Concentration of holes = 6 x 10^{19}

Product of concentration of conduction electron and that of holes remains constant.

Let n_{h} be the Concentration of holes.

=>(6 x 10^{19}) x (6 x 10^{19}) = (2 x 10^{23}) x n_{h}

As per question, 2×10^{23 }be the concentration of conduction electron after doping. And

6 x 10^{19} be the concentration of conduction electron before doping.

=> n_{h} = [36×10^{38}]/[2×10^{23}]

=> n_{h} = 18 x 10^{15}

Thus, number of holes after doping is 18 x 10^{15} per cubic metre.

**Question 14:** The conductivity of an intrinsic semiconductor depends on temperature as Ïƒ = Ïƒ_{0} e^{â€“Î”E/2kT}, where Ïƒ_{0} is a constant. Find the temperature at which the conductivity of an intrinsic germanium semiconductor will be double of its value at T = 300 K. Assume that the gap for germanium is 0.650 eV and remains constant as the temperature is increased.

**Solution: **

Î”E (Band gap) = 0.650 eV

Given: Ïƒ = Ïƒ_{0} e^{â€“Î”E/2kT} …(1)

Let the conductivities at temperatures T_{1} and T_{2} be Ïƒ_{1} and Ïƒ_{2} respectively.

Given T_{1} = 300K and T_{2} = temperature at which the conductivity is double of its value at T_{1}.

=> 2Ïƒ_{1} = Ïƒ_{2}

(1)=>

**Question 15: **A semiconducting material has a band gap of 1 eV. Acceptor impurities are doped into it which create acceptor levels 1 meV above the valence band. Assumed that the transition from one energy level to the other is almost forbidden if kT is less than 1/50 of the energy gap. Also, if kT is more than twice the gap, the upper levels have maximum population. The temperature of the semiconductor is increased from 0K. The concentration of the holes increases with temperature and after a certain temperature it becomes approximately constant. As the temperature is further increased, the hole concentration again starts increasing at a certain temperature. Find the order of the temperature range in which the hole concentration remains approximately constant.

**Solution: **

A semiconducting material has a band gap of 1 eV. Acceptor impurities are doped into it which create acceptor levels 1 meV above the valence band.

Band gap after doping = 1eV -1 meV = (1 – 0.001) = 0.999 eV

Given Condition 1: Any transition from one energy level to the other is almost forbidden if kT is less than 1/50 of the energy gap.

=> kT_{1} = 0.999/50

Where, k = Boltzmann constant

T_{1} be the upper limit for temperature over which transition becomes a forbidden transition.

=> T_{1} = 0.999/[50×8.62×10^{-5}] = 232.8 K (approx)

Given Condition 2: If kT is more than twice the gap, the upper levels have maximum population.

For max limit

kT_{2} = 2 x 10^{-3}

=> T_{2} = [2 x 10^{-3}]/[8.62×10^{-5}] = 23.2 K

Required temperature range = (23.2K – 232.8K)

**Question 16:** In a p-n junction, the depletion region is 400 nm wide and an electric field of 5 Ã— 10^{5} V m^{â€“1} exists in it.

(a) Find the height of the potential barrier (b) What should be the minimum kinetic energy of a conduction electron which can diffuse from the n-side to the p-side.

**Solution: **

(a) We know the relation between electric field and potential,

E = V/d

or V = Ed = 5Ã—10^{5} x 4Ã—10^{-7} = 0.2 volts

Given E = 5Ã—10^{5} Vm^{-1} and d = 4Ã—10^{-7} m

(b) Minimum K.E. = Potential barrier x charge on an electron

= 0.2 x 1.6 x 10^{-19}

= 0.2 eV

**Question 17:** The potential barrier existing across an unbiased p-n junction is 0.2 volt. What minimum kinetic energy a hole should have to diffuse from the p-side to the n-side if

(a) the junction is unbiased,

(b) the junction is forward biased at 0.1 volt and

(c) the junction is reverse biased at 0.1 volt?

**Solution: **

Minimum K.E. = [Potential barrier – biasing voltage] x charge on hole

(a) Biasing voltage = 0 V

Minimum K.E. = 0.2V x e = 0.2 eV

(b) Biasing Voltage for forward biased = 0.1V

Minimum K.E. = (0.2-0.1) Ã— e = 0.1

(c) Biasing Voltage for reverse biased = -0.1

Minimum K.E. = (0.2 + 0.1) Ã— e = 0.3 eV

**Question 18:** In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction

(a) from the p-side and (b) from the n-side

**Solution: **

(a) A junction will act like forward biasing when the hole approaches the junction from p-side

K.E. of the hole will decrease.

Final kinetic energy of the hole = (300-250) = 50 meV

(b) A junction will act like reverse biasing, when the hole approaches the junction from n-side

K.E. of the hole in this case will increase.

Final kinetic energy of the hole = (300+250)= 550 meV

**Question 19:** When a p-n junction is reverse biased, the current becomes almost constant at 25 Î¼A. When it is forward biased at 200 mV, a current of 75 Î¼A is obtained. Find the magnitude of diffusion current when the diode is

(a) unbiased (b) reverse biased at 200 mV (c) forward biased at 200 mV.

**Solution: **

When a p-n junction is reverse biased, the current becomes almost constant at 25 Î¼A. When it is forward biased at 200 mV, a current of 75 Î¼A is obtained.

(a) When the diode is unbiased,

Diffusion current = drift current = 25Î¼A

(b) When the diode is reverse biased, then diffusion current = 0

(c) When the diode is forward biased at 200 mV, then

Diffusion current – Drift current = Forward biasing current

Diffusion current = (75 + 25) Î¼A = 100 Î¼A

**Question 20: **The drift current in a p-n junction is 20.0 Î¼A. Estimate the number of electrons crossing a cross section per second in the depletion region.

**Solution: **

The drift current (i_{d}) in a p-n junction is 20.0 Î¼A = 20.0×10^{-6}A

we know, i_{d} = (N_{e} + N_{h}) x e

where, N_{h} = Number of holes crossing a cross section per second

N_{e}= Number of electrons crossing a cross section per second

e = magnitude of charge on a hole = 1.6Ã—10^{-19} C

Without any biasing applied on the junction, N_{h} = N_{e} = N

=> i_{d} = 2N x e

or N = i_{d}/2e = [20×10^{-6}]/[2×1.6×10^{-19}] = 6.25 x 10^{13}.