Quetions From Emw Waves Semiconductors and Atom Molecule Nuclei

THE EASY WAYS ( A GOOD INSTITUTE NEAR YOU)

From The Desk Of Gaurav Sharma

QUES 1. Discuss Davisson and Germer experiment to verify de-Broglie waves. Ans. It consists of a filament F of tungsten coated with barium oxide, which on heating emits electrons. It acts as cathode. A is a cylinder with fine hole along its axis. It is kep at a positive potential w.r.t. cathode and is called anode. The cathode and anode form an electron gun, by which a fine beam of electrons can be obtained under different accelerating potentials. N is a nickel crystal cut along cubical diagonal. D is an electron detector which can be rotated on a circular scale. A sensitive galvanometer is connected to it to record the current. A fine beam of accelerated electrons is made to fall normally on the nickel crystal. The incident electrons are scattered in different directions by the atoms of the crystal. The intensity of the scattered electrons in a direction is found by the use of detector. By rotating the detector on a circular scale at different positions, the intensity of the scattered beam is measured for different values off latitude angle (or angle of scattering), the angle between the incident and the scattered electron beam. Graphs are plotted between angle and the intensity of scattered beam. www.theeasyways.com Ques. 2 Draw a schematic diagram of Hertz's experimental set-up to produce electromagnetic waves. Give its principle. Ans. Hertz experiment was based on the fact that an oscillating electric charge radiates electromagnetic waves. Two metal sheets of copper or zinc, placed about 60cm apart are connected to two metallic spheres through thick copper wires. A high potential difference of several thousand volts is applied across the spheres using induction coil. Due to high potential difference the air between the spheres gets ionized and provides a path for discharge of plates. Due to it, a spark is produced between spheres and electromagnetic waves of high frequency are radiated. The two plates act as a capacitor having a small capacitance C and the connecting wires provides the low inductance L. The high frequency of oscillations of charges on the plates is given by v = 1/2LC. www.theeasyways.com



QUES.3 A)Name the different layers of earth's atmosphere. How does earth's atmosphere help in the propagation of radiowaves? Ans. The atmosphere has been divided into the following four layers: (i) Troposphere, (ii) Stratosphere, (iii) Mesosphere and (iv) Ionosphere. The ionosphere plays an important role in transmission of radio waves to different parts of earth's surface. It reflects the radio waves back to earth's surface. Thus, it helps in long distance radio communication. B).Why short wave communication over long distance is not possible via ground wave? Ans. In ground wave propagation, loss of energy due to intersection with objects on the earth's surface increases with increase in frequency (or decrease in wave length). Radio waves of frequencies higher than 1500 kHz (or wavelength less than 200m) get heavily damped. Therefore, short wave communication via ground wave is not possible. www.theeasyways.com QUES.4 Why are microwaves preferred over other radiowaves to send signals in particular direction? Briefly describe their uses in radar and telecommunication. Ans. The radio waves having frequencies more than that of television signals are called microwaves. Because of their extremely high frequency they travel as intense beam in a particular direction and cannot be diffracted or bent by objects of normal dimensions. Uses of Microwaves: www.theeasyways.com In radar. Microwaves are used in radars to locate flying objects and to estimate their distance. As microwaves travel in free space with speed of light, by recording the time delay in transmitting and receiving the reflected signal, the distance can be estimated. In telecommunication. With the help of geostationary satellites, microwaves can be transmitted from one place on earth's surface to practically any other place on the earth. In recent years this technique has been extensively used in telecommunication.



QUES.5 A Plane electromagnetic wave travels in vaccum along z-direction. What can you about the directions of its electric and negative field’s vectors? If the frequency of the wave is 30 MHz, what is its wavelength? Ans. Since the electric and magnetic field vectors are always perpendicular to the direction of propagation, E and B are in x-y plane and are mutually perpendicular. Wavelength = c/v = 3 108/3 107 m = 10m. QUES.6 A radio can tune to any station in the 7.5 MHz and 12MHz band. What is the corresponding wavelength band? Ans. Radio waves, irrespective of their frequencies, travel wih same speed. 1 = c/v1 = 3 108 m/s/ 7.5 106 s-1 = 40m 2 = c/v2 = 3 108 ms-1/12 106 s-1 = 25m Thus, the corresponding wavelength band is 40m to 25m. QUES.7 Answer the following questions: Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why? Some scientists have predicted that a global nuclear war on the earth would be followed by a severe nuclear winter with a devastating effect on life on earth what might be the basis of this prediction? Ans. (a) In Optical and radio telescopes visible and radio waves respectively are used which can pass through the atmosphere and therefore these telescopes are built on the grounds. X-rays are absorbed the atmosphere and also it is not desirable to pass X-rays through atmosphere. For these reasons X-ray astronomy is possible only from satellites orbiting the earth. (b) The prediction is based on the fact that the clouds produced by global nuclear war would cover the substantial parts of the sky preventing scalar light from reaching the earth. The absence of solar radiation would cause a 'Winter' which would naturally have a devastating effect on all forms of life on the earth. QUES.8. Distinguish between conductors (or metals), insulators and semiconductors on the basis of their energy bands. Ans. Conductors (Metals). In metallic conductors either the conduction band is partially filled as in case of Na, Li, K, etc. or there is partial overlapping of conduction band and valence band as in case of Zn, Mg, Be, etc. In both the



situation large number of free electrons are available in the conduction band. Therefore, on applying even a small electric field, the metals conduct electricity. Thus, metals are good conductors of electricity. Insulators. In insulators the unfilled conduction band is separated from the filled valence band by a forbidden gap of energy 3eV or more. This means that a minimum of 3eV energy is required by an electron to jump to the conduction band. When electric field is applied across such a solid, the electrons find it difficult to acquire such a large amount of energy and so the conduction band continues to be empty and hence no current flows through it. Semiconductors. The energy band structure of the semiconductors is similar to the insulators but with a smaller forbidden gap energy of about 1eV. The forbidden gap energy is 0.785 eV for germanium and 1.21eV for silicon. At 0K, electrons are not able to cross even this small forbidden gap and hence the conduction band remains totally empty. Therefore, at 0K, semiconductors behave as insulators . QUES.9 Differentiate between n-type and p-type semiconductors on the basis of energy band diagrams. Explain the process of conduction in both types of materials. Ans. Conduction in n-type semiconductors. When an electric field is applied across an n-type semiconductor , a large number of electrons (majority carriers) present in conduction band (at room temperature) start drifting towards the positive terminal and a small number of holes (minority carriers) drift towards the negative terminal. The movement of electrons and holes together constitute the electric current. Conduction in p-type semiconductors. When an electric field is applied across a p-type semiconductor, large number of holes (majority carriers) start drifting towards the negative terminal and a small number of electrons (minority carriers) drift towards the positive terminal. The movement of electrons and holes together constitute the electric current. www.theeasyways.com



QUES 10. Define current gain of a transistor in (i) common base and (ii) common emitter configurations. Obtain a relation between them. Ans. (i) Current gain (a.c. or simply ) of a transistor in common base configuration is defined as the ratio of small change in collector current (dIc) to the small change in emitter current (dIe), at constant collector voltage. = dIc/dIe ………..(i) (ii) Current gain (a.c. or simply ) of a transistor in common emitter configuration is defined as the ratio of the small change in collector current (dIc) to the small change in base current (dIb), at constant collector voltage. = dIc/dIb ………….(ii) Relation between and . In a transistor, Ie = Ib + Ic dIe = dIb + dIc Dividing both sides, by dIc, we get dIe/dIc = dIb/dIc + 1 As dIc/dIe = and dIc = 1/ = 1/ + 1 Or = /1 - . www.theeasyways.com QUES.11 Give a circuit diagram for common base characteristics of p-n-p transistor. How are input and output characteristics drawn? Explain them. Ans. (1) Input or emitter characteristics. The curves drawn between emitter-base (input) voltage VEB and emitter (input) current IE at various fixed collector-base voltage VCB are known as input characteristics. To obtain input characteristics, collector-base voltage is fixed at a constant value. Emitter current is noted at various emitter-base voltages. The process is repeated at various values of VCB. Graphs are plotted between VEB and IE at various VCB. A few typical input characteristic curves. These input characteristic curves have following important features: For a given collector-base voltage, the emitter current increases rapidly with increasing values of emitter- base voltage. This means that input resistance is very small. At higher negative collector-base voltages, emitter current rises more rapidly with the emitter-base voltage.



(2) Output or collector characteristics. The curves drawn between collector-base voltage VCB and collector (output) current IC at various fixed emitter current IE are known as output characteristics. A suitable constant value of emitter current is adjusted and various values of collector current is noted by varying collector-base voltage. The process is repeated at various constant emitter currents. Graphs are plotted between VCB and IC at various IE. A few typical output characteristic curves. These curves have following important features: (i) For a given value of emitter current, the collector current is not zero when collector base voltage is zero. (ii) For a given emitter current, there is a rapid increase in the collector current for an increase in low negative collector-base voltage. In this region, collector has low resistance. (iii) For a given emitter current, the collector current becomes saturated beyond a certain value of VCB. In this region collector has high resistance. QUES.12 Give the circuit diagram for obtaining the characteristics of an n-p-n transistor in common emitter configuration. Draw the characteristic curves and explain them. Ans. Common emitter characteristics of a n-p-n transistor. The circuit diagram for drawing common emitter characteristics of a n-p-n transistor. (1) Input characteristics. The curves drawn between base-emitter (input) voltage VBE and base (input) current IB at various constant collector-emitter voltage VCE are known as input characteristics. To obtain input characteristics, collector-emitter voltage is fixed at a particular value and variation of IB with changing VBE is noted down. The process is repeated at various values of VCE. Graphs are plotted between. VBE and IB at various constant values of BCE. The input characteristics have the following important features. As long as VBE is less than barrier voltage, base current is very small. When VBE exceeds the barrier voltage, base current increases first slowly and then very rapidly as in the case of a forward biased diode. (2) Output characteristics. The curves drawn between collector-emitter (output) voltage VCE and collector (Output) current IC at different constant values of base current are known as output characteristics: To obtain output characteristics, a suitable base current is adjusted and the variation of IC with changing VCE is recorded. The process is repeated at various-



constant base currents. The output characteristics have the following important features: The collector current increases very rapidly with increasing VCE initially but soon it gets saturated and becomes almost independent of collector-emitter voltage. For a given value of VCE, IC is large for larger value of IB. QUES.13 Draw the circuit diagram of a common-emitter amplifier, with appropriate biasing. What is the phase difference between the input and output signals? State two reasons why a common-emitter amplifier is preferred to a common base amplifier. Ans. A CE amplifier is preferred to a CB amplifier because. The voltage gain and power gain of a CE amplifier is larger than that of a CB amplifier. The difference between input impedance and output impedance is lower in a CE amplifier than that in CB amplifier. QUES.14 Explain with the help of a circuit diagram, why the output voltage is out of phase with the input voltage in a transistor common emitter amplifier. Ans. The circuit diagram of a common emitter amplifier is shown. The output voltage of this amplifier is given by V0 = VC – VCC = ICRL During the positive half—cycle of input (a.c.) signal, the forward biasing of emitter base junction becomes higher. Due to which, emitter current (Ie) and hence collector current (IC) increases. As a result, the collector voltage (as evident from Eq.(i)) decreases. Since the collector is connected to the positive terminal of VCC, therefore, decreases in collector voltage means that the collector will become less positive i.e. negative w.r.t. initial value. This implies that during positive half cycle of input voltage, the negative half cycle of the output is obtained. During the negative half cycle of input (a.c.)signal, the opposite process take place. The base emitter junction becomes less forward biased which result in a decreased IC. Consequently the output voltage V0 increases (Eq (i)) i.e., output voltage becomes more positive. Thus, during the negative half cycle of input



signal, the positive half cycle of output signal is obtained. Thus, we see that the output voltage is out of phase with the input voltage in a CE amplifier. QUES.15 Two independent light sources cannot produce sustained interference pattern. Explain. Ans. Two independent light sources cannot produce sustained interference pattern because the light waves emitted by two independent sources do not have same phase or a constant phase difference. It is because, in a source there are billions of atoms and they emit light while returning to lower energy state after staying for nearly 10-8 second in excited state. The condition of constant phase difference cannot be maintained by these billions of waves from the two independent sources. QUES 16.A slit, S is illuminated by a monochromatic source of light to give two coherent sources P1 and P2. These give bright and dark bands on a screen. At a point R, on the screen, there is a dark fringe. What relation must exist between the lengths P1R and P2R? Ans. The condition for minima, i.e. dark fringe is path difference, P = P2R – P1R = (2n + 1) . /2 Where is the wavelength of light and n = 0, 1, 2, 3,….. QUES 17. What is the effect on the interference fringes in a Young’s double slit experiment due to each of the following operations? Give reason for your answer. (i) Separation between the slits is increased. (ii) Monochromatic source is replaced by a source of white light. Ans. In a Young’s double slit experiment the fringe width () in the interference pattern is given by = D/d Where D = distance between slits and screen = wavelength of light used D = separation between slit (i) We see, 1/d



Therefore, if separation between the slits (d) is increased, the fringe width will decrease i.e. fringes will come closer to each other. White light consists seven colours (VIBGYOR) in the wavelength range 4000 Ao to 7500 Ao. The interference pattern due to different colours will overlap and nothing can be clearly seen. However, at the centre of the screen the path difference is zero for all components and so central fringe is bright and white. QUES 18. Distinguish between magnifying power and resolving power of a telescope. Ans. The magnifying power of a telescope, in normal adjustment, is defined as the ratio of the angle subtended by the image at the eye as seen through the telescope to the angle subtended by the object seen directly, when both the object and the image lie at infinity. It is given by M = f0/fe Resolving power of a telescope is defined as the reciprocal of the smallest angular separation between two distant objects whose images are distinctly separated by the telescope. It is given by Resolving power = 1/d = a/1.22 Where is the wavelength of the light used, and a is the diameter of the telescope objective also called aperture. QUES 19. What is the affect of increasing the diameter of the objective on its (i) magnifying power and (ii) resolving power? Give reasons. Ans. (i) Magnifying power (= f0/fe) is independent of the diameter of the objective. (ii) Resolving power (= a/1.22, a – diameter of objective) will increase on increasing the diameter of the objective as resolving power is directly proportional to diameter of the objective. QUES 20. Why is diffraction of sound waves easier to observe than diffraction of light waves? What two main changes in diffraction pattern of a single slit will you observe when the monochromatic source of light is replaced by a source of white light? Ans. Diffraction of sound waves are easier to observe than diffraction of light waves because the wavelength of sound waves is much large and comparable to the size of commonly found aperture or obstacle (e.g., a window) and, therefore



sound waves undergo diffraction. While the wavelength of light is very small and aperture of comparable size is not commonly found and, therefore, light waves ordinarily does not show diffraction. When source of light is monochromatic, the diffraction pattern, due to a single slit, consists of alternate bright and dark bands of unequal width. The central bright fringe has maximum intensity. The intensity of secondary maxima falls off rapidly. If replaced by a source of white light, the diffraction pattern becomes colourd, however, the central maximum is white. Since band width (n = n/a) is proportional to wavelength, red band is wider than other colours. The first order secondary maximum corresponding to each seven constituent colours are visible. Thereafter, over lapping begins and the clarity is lost. QUES 21. In a single slit experiment, how is the angular width of central bright maximum changed, when (i) the slit width is decreased (ii) the distance between the slit and the screen is increased (iii) light of smaller wavelength is used? Ans. In single slit diffraction pattern, the width of central bright maximum is equal to 2D/a. Therefore, when slit width a is decreased, the width of central bright maximum increases; When the distance between the slit and the screen D is increased, the width of central bright maximum increases; When light of smaller wavelength is used, the width of central bright maximum decreases.

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Quetions From Emw Waves Semiconductors and Atom Molecule Nuclei