Ncert Solution for Class 12 Physics Chapter 14 Semiconductor Electronics

Semiconductor Electronics ncert solutions: Ncert Class 12 Physics Chapter 14 Exercise Solutions

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Question 14.1: In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.

Solution 14.1: The correct statement is (c).

In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.

Question 14.2: Which of the statements given in Exercise 14.1 is true for p-type semiconductos.

Solution 14.2: The correct statement is (d).

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In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms.

Question 14.3: Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to \(\left (E_{g} \right )_{Si}\), \(\left (E_{g} \right )_{c}\) and \(\left (E_{g} \right )_{Ge}\) . Which of the following statements is true?
(a) (E_g)_Si < (E_g)_Ge < (E_g)_C
(b) (E_g)_C < (E_g)_Ge > (E_g)_Si
(c) (E_g)_C > (E_g)_Si > (E_g)_Ge
(d) (E_g)_C = (E_g)_Si = (E_g)_Ge

Solution 14.3: The correct statement is (c).

Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least.

The energy band gap of these elements are related as: (E_g)_C > (E_g)_Si > (E_g)_Ge

Question 14.4: In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.

Solution 14.4: The correct statement is (c).

The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

Question 14.5: When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.

Solution 14.5: The correct statement is (c).

When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced.

Question 14.6: In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency

Solution 14.6: For a half-wave rectifier, the output frequency is equal to the input frequency. In this case, the input frequency of the half-wave rectifier is 50 Hz.

On the other hand, the output frequency for a full-wave rectifier is twice the input frequency.
Therefore, the output frequency is 2 × 50 = 100 Hz.

Question 14.7: A p-n photodiode is fabricated from a semiconductor with a bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm?

Solution 14.7: No, the photodiode cannot detect the wavelength of 6000 nm because of the following reason:

The energy bandgap of the given photodiode, Eg = 2.8 eV

The wavelength is given by λ = 6000 nm = 6000 × 10−9 m

We can find the energy of the signal from the following relation:

E = hc/λ

In the equation, h is Planck’s constant = 6.626 × 10−34 J and c is the speed of light = 3 × 108 m/s

Substituting the values in the equation, we get

E = (6.626 x 10-34 x 3 x 108) / 6000 x 10-9 = 3.313 x 10-20 J

But, 1.6 × 10 −19 J = 1 eV

Therefore, E = 3.313 × 10−20 J = 3.313 x 10-20 / 1.6 x 10-19 = 0.207 eV

The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.

Question 14.8: The number of silicon atoms per m³ is 5 × 10²⁸. This is doped simultaneously with 5 × 10²² atoms per m³ of Arsenic and 5 × 10²⁰ per m³ atoms of Indium. Calculate the number of electrons and holes. Given that n_I = 1.5 × 10¹⁶m^–3. Is the material n-type or p-type?

Solution 14.8: The following values are given in the question:

Number of silicon atoms, N = 5 × 10 ²⁸ atoms/m³

Number of arsenic atoms, n_AS =5×10²²atoms/m³

Number of indium atoms, n_In=5×10²²atoms/m³

n_i=1.5×10¹⁶electrons/m³

n_e=5×10²²−1.5×10¹⁶=4.99×10²²

Let us consider the number of holes to be n_h

In the thermal equilibrium, n_en_h = n_i²

Calculating, we get

n_h=4.51×10⁹

Here, n_e>n_h, therefore, the material is an n-type semiconductor.