A pure semiconductor like Ge or Si, when doped with

Chapter 14- Semiconductor Electronics

Case Based Question:- A pure semiconductor like Ge or Si, when doped with a small amount of suitable impurity, becomes an extrinsic semiconductor. In thermal equilibrium, the electron and hole concentration in it are related to the concentration of intrinsic charge carries. A p-type or n-type semiconductor can be converted into p-n junction by doping it with suitable impurity. Two process, diffusion and drift take place during formation of a p-n junction. A semiconductor diode is basically a p-n junction with metallic contacts provide at the ends for the application of an external voltage. A p-n junction diode allows currents to pass only in one direction when it is forward biased. Due to this property, a diode is widely used to rectify alternating voltage, in half-wave of full wave configuration.                     4×1 = 4        [CBSE  2024]

(i) When Ge is doped with pentavalent impurity, the energy required to free the weakly bound electron from the dopant is about

(A) 0.001 eV                   (B) 0.01 eV

(C) 0.72 eV                     (D) 1.1 eV

(ii) At a given temperature, the number of intrinsic charge carries in a semiconductor is 2.0 \times 10^{10}  cm^{-3}. It is doped with pentavalent impurity atoms. As a result, the number of holes in it becomes 8 \times 10^3  cm^{-3}. The number of electrons in the semiconductor is

(A) 2 \times 10^{24}  m^{-3}

(B) 4 \times 10^{23}  m^{-3}

(C) 1 \times 10^{22}  m^{-3}

(D) 5 \times 10^{22}  m^{-3}

(iii) (a) During the formation of a p-n junction-

(A) electrons diffuse from p-region into n-region and holes diffuse from n-region into p-region.

(B) both electrons and holes diffuse from n-region into p-region.

(C) electrons diffuse from n-region into p-region and holes diffuse from p-region into n-region.

(D) both electrons and holes diffuse from p-region into n-region.

               OR

(iii) (b) Initially during the formation of a p-n junction-

(A) diffusion current is large and drift current is small.

(B) diffusion current is small and drift current is large.

(C) both the diffusion and the drift currents are large.

(D) both the diffusion and the drift currents are small.

(iv) An ac voltage V = 0.5 sin (100 πt) volt is applied, in turn, across a half-wave rectifier and a full-wave rectifier. The frequency of the output voltage across them respectively will be

(A) 25 Hz, 50 Hz             (B) 25 Hz, 100 Hz

(C) 50 Hz, 50 Hz             (D) 50 Hz, 100 Hz

Solution:-

(i) Answer (C) E = 0.72 eV

(ii) Answer (D) 5 \times 10^{22}  m^{-3}

(iii) (a) Answer (C) both the diffusion and the drift currents are large.

            OR

(iii) (b) Answer (A) diffusion current is large and drift current is small.

(iv) Answer (D) 50 Hz, 100 Hz


 

Chapter 9- Ray Optics and Optical Instruments

Case Based question:- A lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical. Applying the formula of image formation by a single spherical surface successively at the two surfaces of a thin lens, a formula known as lens maker’s formula and hence the basic lens formula can be obtained. The focal length (or power) of a lens depends on the radii of its surfaces and the refractive index of its material withrespect to the surrounding medium. the refractive index of a material depends on the wavelength of light used. Combination of lenses helps us to obtain diverging or converging lenses of desired power and magnification.                                         4×1 = 4            [CBSE  2024]

(i) A thin converging lens of focal length 20 cm and a thin diverging lens of focal length 15 cm are placed coaxially in contact. The power of the combination is

(A) -5/6 D                 (B) -5/3 D

(C) 4/3 D                   (D) 3/2 D

(ii) The radii of curvature of two surfaces of a convex lens are R and 2R. If the focal length of this lens is (4/3) R, the refractive index of the material of the lens is :

(A) 5/3                     (B) 4/3

(C) 3/2                     (D) 7/5

(iii) the focal length of an equiconvex lens

(A) Increases when the lense is dipped in water.

(B) Increases when the wavelength of incident light decreases.

(C) Increases with decrease in radius of curvature of its surface.

(D) Decreases when the lens is cut into two identical parts along its principal axis.

(iv) (a) A thin convex lens L of focal length 10 cm and a concave mirror M of focal length 15 cm are placed coaxially 40 cm apart as shown in figure. A beam of light coming parallel to the principal axis is incident on the lens. The image will be formed at a distance of

(A) 10 cm, left of lens         (B) 10 cm, right of lens

(C) 20 cm, left of lens          (D) 20 cm, right of lens

        OR

(iv) (b) A beam of light coming parallel to the principal axis of a convex lens L_1 of focal length 16 cm is incident on it. Another convex lens L_2 of focal length 12 cm is placed coaxially at a distance 40 cm from L_1. The nature and distance of the final image from L_2 will be

(A) real, 24 cm           (B) virtual, 12 cm

(C) real, 32 cm           (D) virtual, 18 cm

Solution:- See the solution

Some other Case study Question

1.The lens Maker’s formula is usefull to design lenses

2. An oscillating charge is an example of accelerating charge

3.Case Study Chapter 2: Electrostatic Potential and Capacitance :

4.Case Study Physics Chapter 3 – Current Electricity

5.Case Study Chapter 5 : Magnetism and Matter

Case Study

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