81. In the energy band diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is
    

1. 1. an insulator ​

   
   
   

   2. a metal

   
   
   

   3. an n-type semiconductor ​

   
   
   

   4. ​a p-type semiconductor

   
   
   

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1. View Hint View Answer Discuss in Forum

   For a p-type semiconductor, the acceptor energy level, as shown in the diagram, is slightly above the top E_v of the valence band. With very small supply of energy an electron from the valence band can jump to the level E_A and ionise acceptor negatively.

   Correct Option: D

   For a p-type semiconductor, the acceptor energy level, as shown in the diagram, is slightly above the top E_v of the valence band. With very small supply of energy an electron from the valence band can jump to the level E_A and ionise acceptor negatively.

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82. For a cubic crystal structure which one of the following relations indicating indicating the cell characteristics is correct?​​

1. 1. a ≠ b ≠ c and α = β = γ = 90° ​

   
   
   

   2. a = b = c and α ≠  β ≠ γ = 90°

   
   
   

   3. a = b = c and α =  β = γ = 90° ​

   
   
   

   4. a ≠ b ≠ c and α ≠  β and γ ≠ 90°

   
   
   

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1. View Hint View Answer Discuss in Forum

   For a cubic crystal, ​
   a = b = c and α = β = γ = 90°

   Correct Option: C

   For a cubic crystal, ​
   a = b = c and α = β = γ = 90°

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83. A p-n photo diode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly

1. 1. 10 × 10¹⁴ Hz ​

   
   
   

   2. ​5 ×10¹⁴ Hz ​

   
   
   

   3. 1 × 10¹⁴ Hz

   
   
   

   4. ​20 × 10¹⁴ Hz

   
   
   

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1. View Hint View Answer Discuss in Forum

   E_g = 2.0 eV = 2 × 1.6 × 10^–19 J ​
   E_g = hv
   

   ∴ v =	Eg	=	2 × 1.6 × 10-19 J
   	h		6.62 × 10-34 Js

   
   = 0.4833 × 10¹⁵ s^-1 = 4.833 × 10¹⁴ Hz
   ≃ 5 × 10¹⁴

   Correct Option: B

   E_g = 2.0 eV = 2 × 1.6 × 10^–19 J ​
   E_g = hv
   

   ∴ v =	Eg	=	2 × 1.6 × 10-19 J
   	h		6.62 × 10-34 Js

   
   = 0.4833 × 10¹⁵ s^-1 = 4.833 × 10¹⁴ Hz
   ≃ 5 × 10¹⁴

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84. Sodium has body centred packing. Distance between two nearest atoms is 3.7 Å. The lattice parameter is ​

1. 1. 4.3 Å ​

   
   
   

   2. ​3.0 Å

   
   
   

   3. 8.6 Å

   
   
   

   4. 6.8 Å

   
   
   

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1. View Hint View Answer Discuss in Forum

   d =	√3	a
   	2

   
   

   3.7 =	√3	a
   	2

   
   

   a =	2 × 3.7	= 4.3 Å
   	√3

   Correct Option: A

   d =	√3	a
   	2

   
   

   3.7 =	√3	a
   	2

   
   

   a =	2 × 3.7	= 4.3 Å
   	√3

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85. A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength

1. 1. 4000 nm ​

   
   
   

   2. 6000 nm ​

   
   
   

   3. 4000 Å

   
   
   

   4. ​6000 Å

   
   
   

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1. View Hint View Answer Discuss in Forum

   λmax =	hc
   	E

   
   

   =	6.6 × 10-34 × 3 × 108	≃ 5000 Å
   	2.5 × 1.6 × 10-19

   
   The wavelength detected by photodiode should be less than λ_max .
   Hence it can detect a signal of wavelength 4000 Å.

   Correct Option: C

   λmax =	hc
   	E

   
   

   =	6.6 × 10-34 × 3 × 108	≃ 5000 Å
   	2.5 × 1.6 × 10-19

   
   The wavelength detected by photodiode should be less than λ_max .
   Hence it can detect a signal of wavelength 4000 Å.

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