(b) State the characteristics of photon. 2 Solution:

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1 Max Marks 40 ENGINEERING PHYSICS / VRK/KS/14/6549 Faculty of Engineering and Technology RTMNU SUMMER PAPER SOLUTION First Semester B.E. (C.B.S.)Examination ENGINEERING PHYSICS Q. 1. (a) What is Compton effect? On the basis of quantum theory, explain the existence of modified and unmodified component in Compton scattering. 1+3 Compton Effect : When a beam of monochromatic X-ray, is scattered by a scattering material (like graphite), the scattered radiation contains two components of X-rays. One of the components having the same wavelength as that of the incident beam called as Primary or Unmodified component denoted by λi or λ and the other component having a longer wavelength than that of the incident x-rays is called as Secondary or Modified Component λ f or λ. This phenomenon of change in wavelength due to scattering is known as Compton Effect. The change in the wavelength (difference) λ= λ - λ is called the Compton shift. In Compton scattering, the scattered radiation consists of two component wavelengths namely the modified and unmodified wavelength. The Unmodified component (Wavelength) arises when photons are scattered by tightly bound electrons. In this case, the whole atom of scattering substance is involved in the collision and hence the value of mass of electron (m 0 ) in the equation λ = h (1-cos θ) (i) m 0 c is to be replaced by the mass of the atom ( M). Since the mass of the atom is many times larger than that of an electron, the Compton shift is negligible and correspondingly there is no change in wavelength. i.e. λ = h (1-cos θ) M c As M>> m 0 λ = λ f - λ i = 0 λ f = λ i i.e. Unmodified component Hence the intensity of unmodified component is large. In case of higher atomic number elements (Heavy elements), number of tightly bound electrons are more and hence Unmodified component is stronger (intense) in Higher atomic number elements. The modified wavelength arises due to the collision of X-ray photon with either free or loosely bound electrons. For high energy photons, most of the atomic electrons appear free and hence large fraction of incident X-rays undergoes a wavelength shift. Therefore in case of low atomic number elements (Light elements) the modified component will be more intense. (b) State the characteristics of photon. 1. The energy of photon is determined by its frequency and is given by E=h ν. Velocity : Photon always travels with the velocity c. 3. Mass : The rest mass of photon is zero since a photon can never be at rest. Thus m 0 =0 4. Relativistic mass Prepared by Department of Applied Physics, RGCER, Nagpur Page No 1

2 As photons travels with velocity c, it will have relativistic mass. It is given by mass energy equivalence relation as mm = EE = h ν cc cc (c) An X-ray photon of wavelength 0.5A 0 is scattered through an angle 60 0 by a loosely bound electron. Calculate recoil energy and velocity of an electron. 4 Sol: Given: λ =0.5 Å; θ =60 0 ; E =?, v e =? Compton Shift Δ λ = (h/ m o c)(1- cosθ) λ λ = 0.04 (1- cos60 0 ) λ = (1-0.5) λ = 0.515A 0 K.E e = hc [ 1 1 ] = λ λ 1 1 ] 0.5 x x x 10 x 3 x 10 [ K.E = 9.498X J = x x ev = ev V e = (K.E./m) 1/ x = ( ) x 10 1 ev =3.61 x m/s OR Q.. (a) Explain the concept of matter waves. Illustrate it with the help of Davisson and Germer experiment. 4 In 194,Louis de-broglie a French Physicist suggested that the wave-particle dualism need not be special feature of light alone but the material particles must also exhibit such dual behavior. He reasoned out that since nature exhibits a great amount of symmetry, light as well as material particles exhibits properties of both particles and waves.thus any moving particle is associated with a wave. The waves associated with moving particles are known as de Broglie waves or matter waves. and the wavelength of matter waves expressed by equation λ = h / mc is called as de-broglie wavelength. Waves exhibit diffraction. If the de-broglie hypothesis is valid, then the matter waves should also exhibit diffraction effects. In 197, Davisson and Germer observed the diffraction of an electron beam incident on a nickel crystal. This experiment provided a convincing proof of wave nature of matter (electron). Experimental arrangement: Prepared by Department of Applied Physics, RGCER, Nagpur Page No

3 The experimental set up of Davisson and Germer is shown in fig. It consists of an electron gun, where the electrons are produced from a heated filament and are accelerated by an anode A which is connected to a variable voltage source. The electron beam is collimated by suitable slits to obtain a fine narrow beam. The electron beam is allowed to fall normally on a single crystal of nickel C. These electrons are scattered by nickel crystal. The number of electrons scattered by the crystal in different directions is studied with the help of a detector D, which can be moved on a circular scale. Analysis: During their experiment Davisson and Germer moved the detector on the circular scale to various positions and the current was measured. The detector current is a measure of intensity of the diffracted beam. A polar graph was then plotted between the detector current (Intensity) and the angle between incident beam and diffracted beam. Such polar curves were obtained for electrons accelerated through different voltages. It was found that a hump appears in the polar curve when electrons accelerated with V=44 volt were incident on Nickel crystal. The hump grow in size as the accelerating voltage is increased and it was found that when electrons are accelerated through a potential difference of V=54 volts and incident normally on the crystal face, the scattering of electrons is more pronounced at Ф =50 0,as shown in fig. The maximum is an indication that electrons are being diffracted. From this, we can interpret that the rows of atoms in the crystal act like the rulings of a grating producing the diffracted beam in the direction Ф =50 0. We can extend the Bragg s law in such a case to electron diffraction and calculate the wavelength of electrons. From the fig. it is seen that the glancing angle θ=65 0.The interplanar spacing for nickel crystal is obtained by x-ray diffraction method which is equal to 0.91Å. Applying Bragg s equation, for first order diffraction, dsin θ= λ Prepared by Department of Applied Physics, RGCER, Nagpur Page No 3 3

4 x 0.91x sin(65) = λ λ = 1.65 Å The wavelength of electron can be calculated using de-broglie hypothesis. λ = = h emv 6.63x10-34 [ x9.1x10-31 x1.6x10-19 x54] 1/ = 1.66 Å It is seen that the values obtained experimentally using Bragg s equation and de-broglie equation agreed well. Therefore, Davisson-Germer experiment gave conclusive evidence that electrons exhibit diffraction property. (b) Show how the quantization of angular momentum follows the concept of matter waves. 3 According to Bohr model of an atom, the angular momentum of an electron in a stationary orbits is a integral multiple of h/π. i.e. = h/π The de-broglie concept of matter waves justifies the above hypothesis. According to de-broglie hypothesis, the electron revolving round the nucleus in an atom is associated with matter waves that propagate along the circumference again and again. In order that electron energy should not be lost, the wave should produce a stationary wave profile in the orbit. This can happen only when an integral number of wavelengths fit in the circumference of the orbit as shown in fig.(b). Hence the condition for the formation of stationary waves in the orbit is, πr = nλ where n= 1,, (1) As per de-broglie hypothesis, the wavelength of matter waves associated with electron is given by, λ = h/mv () Prepared by Department of Applied Physics, RGCER, Nagpur Page No 4 4

5 Substituting this value in eqn(1), we get πr = nh/mv mvr = nh/π but mvr = L n = angular momentum of the electron in the n th orbit. Hence, L n = nh/π (3) or L n =n ħ Eqn. (3) is Bohr s quantization condition. De Broglie thus demonstrated that the quantization of angular momentum is thus direct consequence of wave nature of electrons. (c) An electron and a particle of mass 10 gm both are moving with velocity 300 m/s. Calculate the wavelengths associated with them and interpret the results. 3 Given: M = 10g = kg; m= kg & v =300m/s Wavelength associated with the partical:- 34 h λp = = = Mv Wavelength associated with the electron:- m 34 h λe = = = mv m Since the wavelength of the partical is of the order of 10-4 A 0, it is very very small than the size of the particle & it can not be detectable. Therefore wave nature cannot be revealed through diffraction effect. Prepared by Department of Applied Physics, RGCER, Nagpur Page No 5 5

6 Q. 3. (a) What is Heisenberg s Uncertainty principle? Describe a thought experiment to arrive at this principle. 4 Heisenberg s uncertainty principle states that It is not possible to make simultaneous measurements of the position and momentum of a particle to an unlimited accuracy. If x is the uncertainty in position of the particle and p x is the uncertainty in momentum of particle, then, the product of uncertainty in position and uncertainty in momentum is greater than or equal to h/π (or ħ) Thus, x. p x ħ or more precisely x. p x h/π or x. p x ħ The corresponding relations for the other components of position and momentum are y. p y ħ z. p z ħ Experimental proof i.e Thought Experiment (Diffraction of electron at single slit) Suppose a stream of monoenergetic microparticles is moving horizontally along the x- axis as shown in figure Each particles in the stream carries a finite momentum px in x direction and momentum along y axis is zero. However there is uncertainty in the location of the particle in y direction y=. To find ourt the location of particle, let a slit of width d be placed in its pathat x=0. As the particles emerges from the slit at x=0,the particles must have been localized in the y direction within an uncertainty y=d. This uncertainty in the y direction of the particle may be reduced by making the slit narrower and narrower (i.e d is made smaller). This reduction in width of slit causes a diffraction of particle-waves on the screen by making an angular separation θ between two minima flanking the central maximum. From the diffraction condition dsinθ=nλ sinθ=λ/d (n=1, for first order) As θ is very small sinθ= θ =λ/d =λ/ y (i) ( as d= y) Because of the diffraction effect at the slit, the particles acquires a small component of momentum. p y in y direction. From figure tan θ=θ. p y / p x (ii) Prepared by Department of Applied Physics, RGCER, Nagpur Page No 6 6

7 From (i) and (ii) p y λ = px y y. p = λ. p ( iii) From de Broglie λ= h / p x, we get y. p y =h y x From this relation if y increases, then. p y decrease and vice versa. (b) Define phase velocity and group velocity. Derive the relation between phase velocity and group velocity. 3 Phase Velocity : The velocity with which the individual wave forming wave packet propagates is called phase velocity or wave velocity v p.phase velocity is given by Group Velocity : v p = w/k The velocity with which the wave group advances or propagates in the medium is called as the group velocity v g.group velocity is given by v g = dw/dk Relation between phase velocity (v p ) and group velocity (v g ) The velocity of the individual component wave of waver packet (phase velocity )is given by, v p = νλ Using v=w/π and λ= π/k into the above equation, we get w π w v p = = π k k w = kv p The group velocity is given by the relation as, dw d vg = = ( kv dk dk π As k = λ π dk = dλ λ dv p vg = v p λ dλ p ) = v and p k dk dv + k dk = Group velocity will be same as the phase velocity, if all the constituent waves travels with the same velocity i. e. in non dispersive medium v g =v p Prepared by Department of Applied Physics, RGCER, Nagpur Page No 7 7 p λ dλ

8 (c) A proton is confined to a nucleus of radius m. Calculate the minimum uncertainty in velocity of the proton. 3 Given: m p = 1.67 X 10-7 kg ; x =x10-14 m ; v =? x = = p m v v = = 7 14 m x = m s / Q. 4. (a) Show that wave function for a particle confined in an infinite one-dimensional nπx potential well of length L is given by Ψn = sin.hence, discuss the energy L L levels and their discreteness. 4 A potential well is a potential energy function that has a minimum. Consider a potential well of length L and of infinite depth Consider a particle confined to the region 0<x<L. The electron can move inside the well horizontally in the x direction but can not escape from the potential well. It means that V=0 in the region 0<x<Land rises to infinity V= at x=0 and x=l. This situation is called a one dimensional potential box or well. Let ψ be the wave function associated with the particle inside the potential well. The probability of finding the electron outside the well must be zero. That is ψ=0 at x 0 and x L. Prepared by Department of Applied Physics, RGCER, Nagpur Page No 8 8

9 To find out the wave function and energy of the particle, we apply Schrodinger s one dimensional time independent wave equation to the particle inside the box. Time independent Schrodinger wave equation: 8π m Ψ + ( E V ) Ψ = 0 h The particle moves along x-direction only and V=0 inside the potential well. Hence equation will be d Ψ 8π meψ + = 0 dx h d Ψ 8 me + k Ψ = 0 where k = π ( i) dx h The solution of the above equation is ikx ikx Ψ( x) = Ae + Be ( ii) are: Where A and B are constants whose value can be determined by applying boundary conditions. The particle is enclosed between two rigid walls, hence two boundary conditions Ψ = 0 at x=0 Ψ = 0 at x=l Putting Ψ = 0 at x=0 in eqn.(ii),we get 0 = A + B B = -A (iii) Putting second boundary condition in eqn.(ii),we get 0 = A e ikl + B e -ikl Using equation (iii), we get A e ikl - A e -ikl = 0 A(e ikl - e -ikl ) = (iv) Using Euler s equation 1 iθ iθ sinθ = ( e e ) i iθ iθ ( e e ) = i sinθ Hence equation (iv) may be re expressed as Ai.sin kl =0 The factor Ai can not be zero, since it is trivial solution Hence sin kl = 0 kl = nπ (where n= 1,,3,----) k=nπ/l k = n π /h (v) As k = π / λ π π L λ = = = k nπ n L Prepared by Department of Applied Physics, RGCER, Nagpur Page No 9 9

10 This implies that the wave equation has solution only when the electron wavelength is restricted to discrete values such that only a whole number of half wavelengths (λ/) are formed over the length L. It means that electron waves form standing wave pattern within the potential well. The allowed wave function of a particle is called eigen function which is given by, Ψ = ia.sinkx Or Ψ = Asin (nπ/l) x Applying the normalization condition, we can determine the value of constant A. L ψψ dx = 1 L 0 (ia) sin (nπ/l)x.dx = 1 0 (i A) (1 -cos (nπ/l)x dx = 1 L (ia) = 1 ia = L Hence, using the value of ia in equation of wave function, Ψ ( x) = nπx sin L L (b) Explain the phenomenon of tunneling when a beam of electron is incident on potential barrier of finite width. 3 When an electromagnetic wave strikes at the interface of the two media, it is partly reflected and partly transmitted through the interface and enters the second medium. Similarly, the wave nature of micro particles makes it possible to get partly reflected from the boundary of the potential barrier and partly penetrate through the barrier. The penetration of a barrier by a quantum particle is called tunneling. A potential barrier is the opposite of potential well; it is a potential energy function with a maximum. Let us consider a potential barrier of height h and thickness L. The potential energy is zero for x<0 and x>l has a value V for 0 <, x < L. An electron of total energy E approaches the barrier from the left. From the view point of classical physics, the electron would be reflected from the barrier because its energy E is less than V. For the particle to overcome the potential barrier, it must have an energy equal to or greater than V. Quantum mechanics leads to entirely new result. It shows that there is a finite chance for the electron to leak to the other side of the barrier. We say that the electron tunneled through the potential barrier and hence in quantum mechanics, the phenomenon is called quantum tunneling. Prepared by Department of Applied Physics, RGCER, Nagpur Page No 10 10

11 (c) An electron is trapped in one dimensional potential well of infinite depth and width 1mm. Calculate first three lowest energy values. 3 Given: L e = 1 x 10-3 m; m e =9.1 x kg; First three energy levels for electron For n =1, For n =, For n =3, h ( ) 8mL (10 ) E = = = J = 3.76x10 ev 31 3 h E = n ( ) = n E 1 = x10 ev = 15.05x10 ev 8mL h E3 = n ( ) = n E 1 = x10 ev = x10 ev 8mL Q. 5. (a) Define the following: (i) Unit cell (ii) Packing fraction (iii) Miller Indices of a plane. 3 Unit Cell:- Unit Cell may be defined as the smallest geometrical unit,which when repeated in space indefinitely, generates the space lattice and carries a full description of entire lattice. Atomic packing fraction (APF): The fraction of the space occupied by atoms in the unit cell is known as atomic packing fraction. It is defined as the ratio of the volume of effective number of atoms in the unit cell to the total volume of the unit cell. Thus, AAAAAA = (No. of atoms /unit cccccccc)(volume of each atom) Volume of the unit cell Prepared by Department of Applied Physics, RGCER, Nagpur Page No 11 11

12 Therefore, AAAAAA = ZZZZ VV The atomic packing fraction denotes the efficiency with which the available space in a unit cell is utilized. Miller Indices Miller Indices are the reciprocals of the intercepts, made by the plane on the crystallographic axes, when reduced to smallest integers. (b) Obtain the values of voids in case of FCC and BCC structures and also show that BCC has maximum voids. 4 Void space: The void space in the unit cell is the vacant space left unutilized in the cell. It is equal to (1- APF). It is often expressed as %.. It is commonly known as interstitial space. Thus, %Void space= (1 APF) x 100 APF for BCC is = 0.68 BCC Void Space = [1- APF] x 100 BCC Void Space = [1-0.68] x 100 = 3% APF for FCC is = 0.74 FCC Void space = (1 APF) x 100 = (1-0.74) x100 = 6% The result shows that BCC has maximum voids. (c) Aluminum has FCC structure and density 700 kg/m 3. Calculate unit cell dimension and atomic diameter. Atomic weight of Aluminum is Density ρ = 700 kg / m 3 Molecular weight M = 6.98; For FCC Z = 4. ZM ρ = Na a 3 A 3 ZM = = = 4.04A 6 N ρ x 700 A ( ) a 4.04 r = = = 1.48A Atomic diameter = r = x 1.48 =.856 A 0 Prepared by Department of Applied Physics, RGCER, Nagpur Page No 1 1

13 Q. 6. (a) Derive the relation between inter planer spacing, lattice constant and Miller indices of the planes for a cubic crystal. 4 The distance between successive members of a series of a parallel planes is known as a interplaner distance. Let d hkl represents the distance between two adjacent parallel planes having Miller Indices (hkl). Let the plane ABC be one of the planes that has intercepts OA=a/h, OB=b/k and OC=c/l. The origin of the coordinates O is in next plane of set parallel to ABC. The perpendicular OD in fig. from the origin to the plane is equal to d hkl. Let the direction cosines of OD be cosα, cosβ, cosγ. The intercepts of the plane on the coordinate axes are OA= a/h; OB=a/k; OC=a/l (As a=b=c= Unit length of edge a) Where a is the length of the cube edge. Denoting OD=d Cos α = OD OA = hd a ; Cos β = OD OB = kd a Cos γ = OD OC = ld a Since cos α+ cos β+ cos γ=1 hhhh dd aa + kk dd aa + ll dd aa = 11 dd aa (hhhh + kk + ll ) = 11 dd = aa (hhhh +kk +ll ) aa dd = (hhhh + kk + ll ) aa dd hhkkkk = (hhhh + kk + ll ) (b) Draw the crystal planes for (00) and ( 100). Z Z Y Y X ( 0 0) x Prepared by Department of Applied Physics, RGCER, Nagpur Page No 13 13

14 (c) X-ray of unknown wavelength give first order Bragg reflection at glancing angle 0 0 with (1) planes of copper having FCC structure. Find the wavelength of x-rays. If the lattice constant for copper is 3.615A 0. 4 Given : Angle = θ = 0 0 ; a = A 0 ; m=1 dsin d1 = a h + k + l = = 1.05A θ = mλ Bragg s Law 1.05 sin 0 = 1λ λ = 0.843A Q. 7. (a) What is Fermi Dirac Distribution Function? Show that Fermi energy level lies midway between conduction band and valence band in intrinsic semi conductor. 4 Fermi-dirac distribution functions: The expression that governs the distribution of electrons among the energy levels as a function of temperature is known as Fermi-Dirac distribution functions in honour of Fermi and Dirac who derived it independently in 196. The probability of occupation of an energy level E by an electron is given by fermi-dirac distribution function f(e). 1 f ( E) = 1+ exp[( E E ) / kt ]. Prepared by Department of Applied Physics, RGCER, Nagpur Page No F where, T is absolute temperature and k is called Boltzman s constant. Fermi level is defined as the highest filled energy level in a conductor at 0 K. In terms of energy band diagram, a conduction band and a valence band is separately by a smaller energy gap,energy band gap E g.. At temperature above absolute zero in semiconductor two types of charge carriers electrons and holes exist ; (electrons in the conduction band and holes in the valence band). Free electrons and holes are produced due to breaking of covalent bonds therefore the number of free electrons is equal to the number of holes in an intrinsic semiconductor.

15 The electron concentration in the conduction band is given by n = N C e (E C E F ) kt The hole concentration in the valence band is given by p = N V e (E F E V ) kt In an intrinsic semiconductor the electron and hole concentrations are equal. Thus, n = p N C e (E C E F ) kt = N V e (E F E V ) kt Taking logarithm on both sides, we get (E C E F ) = ln N V (E F E V ) kt N C kt EE CC + EE FF = kkkk llll NN VV EE NN FF + EE VV CC E F = (E C + E V ) + kt ln N V N C E F = (E C +E V ) + 1 kt ln N V (1) N C But N C = πm e 3 kt h and NV = πm h 3 kt h N V N C = m 3 h m e ln N V = 3 N C ln m h m e Substituting the value of ln N V N C in Equation (1) E F = E C +E V + 3 kt ln 4 m h m e Effective mass of a free electron is assumed to be equal to the effective mass of a hole, i.e., m h = m e ln m h = log 1 = 0 m e E F = E C +E V We can write E F = E C E V + E V E F = E g + E V If we denote the top of the valence band EV as zero level, EV =0 Prepared by Department of Applied Physics, RGCER, Nagpur Page No 15 15

16 E F = E g This shows that in an intrinsic semiconductor the Fermi level lies in the middle of the forbidden gap. (b) What is drift and diffusion current? Drift current When an electric field is applied across the semiconductor material, the charge carriers attain a certain drift velocity. Holes move towards the negative terminal of the battery and electrons move towards the positive terminal of the battery. This combined effect of movement of the charge carriers constitutes a current known as the drift current.thus the drift current is defined as the flow of electric current due to the motion of the charge carriers under the influence of an external electric field. Diffusion current It is possible for an electric current to flow in a semiconductor even in the absence of the applied voltage provided a concentration gradient exists in the material. A concentration gradient exists if the number of charge carriers is greater in one region of a semiconductor as compared to the rest of the Region. In a semiconductor material the charge carriers have the tendency to move from the region of higher concentration to that of lower concentration of the same type of charge carriers. Thus the movement of charge carriers takes place resulting in a current called diffusion current. (c) A Sample of intrinsic silicon at room temperature has carrier concentration of 1.5 x m 3 If a donar impurity is added to the extent of 1 donar atom per 10 8 atoms of silicon. If the concentration of silicon atom is 5 x 108 atom/m3 determine the resistivity of the material. (Given: µ e = m /V.s and µ h = m /V.s. ) 4 Given: Intrinsic carrier Density, ni = 1.5 x10 16 /m 3 Electron Mobility µ e = m /v.s Hole Mobility µ h = m /v.s σσ ii = eenn ii (μμ ee + μμ hh ) σσ ii = ( ) σσ ii = mm VV 11 ss 11 σσ ii = mmmmmm/mm ρρ ii = = σσ ii ρρ ii = oooooo mm Prepared by Department of Applied Physics, RGCER, Nagpur Page No 16 16

17 OR Q. 8. (a) Draw neat energy band diagram for symmetrically doped PN junction diode. When it is : i) unbiased ii) Forward biased. 3 Energy band diagram of a p-n junction in unbiased condition Energy Band Diagram of PN junction diode in forward biased : (b) Explain the induction of Hall voltage in a semiconductor carrying current and placed in transverse magnetic field. Also obtain an expression of Hall coefficient. 4 If a metal or a semiconductor carrying a current I is placed in a transverse magnetic field B, a potential difference V H is produced in the direction normal to both the current and magnetic field directions.let us consider a bar of P-type semiconductor crystal and assume the charge carriers to be holes having a charge +e. Let an electric field E x be applied to the bar which produced a current I in the x-direction in the crystal. The Current through the semiconductor wafer is given by I = peav d (1) Where p is the hole concentration A is the area of cross section of the end face of semiconductor wafer e is electrical charge associated with hole v d is the drift velocity of holes The current density is given by J x = I / A J x = pev d () Prepared by Department of Applied Physics, RGCER, Nagpur Page No 17 17

18 Let a magnetic field B act in the Z-direction. As holes moves in the bar with a velocity, say v, they experience a Lorentz force F L due to the transverse magnetic field B.As a result they deviate sidewise towards the front face of the bar. Because of the deflection due to magnetic field, the holes is case of p-type crystal accumulate on front face and make it positively charged while the rare face becomes negatively charged with respect to the front face. Hence a potential V H called the Hall Voltage appears between the front and rare faces. The potential builds up in such a way that the electric field E H due to it discourages the further building up of the charges on the faces. The Process of accumulation of charges on front face lasts till the electric field just balances the Lorentz force F L. Once the balance is established the charge accumulation stops and the crystal attains equilibrium states where holes moves parallel to the faces once again. In the equilibrium state F E = F L ee H = Bev d E H = V H /w where, w is the dist between the front and rare faces of the crystal. ( V H /w ) = Bv d (3) From Eq. () v d = Jx / pe Eq. (3) becomes ( V H /w ) = BJx / pe V H = wbjx / pe = wbi / pea If t is the thickness of semiconductor plate A =wt V H = BI /pet Hall coefficient R H : Hall coefficient R H is defined as Hall electric field per unit current density per unit magnetic induction. R H = E H / J x B Prepared by Department of Applied Physics, RGCER, Nagpur Page No 18 18

19 = B. J x /p.e J x B = 1/pe (4) Current is given by I = J x wt. Where w t is the area of cross section of the crystal R H = (V H /w) / (J x. B) = V H w t/wi B R H = V H t/i B R H = 1/ p.e (c) An n-type Ge sample has a donor density of 10 1 /m 3. It is arranged in Hall effect experiment having magnetic field of 0.5 T. and current density 500 A/m. Find the Hall voltage if the sample is 3 mm thick. 3 Given: n = 10 1 /m 3 ; B=0.5T; J=500 A/m ; w=3mm = 3x10-3 m; V H =? V H = BJw/ne V H = [0.5 x 500 x 3 x 10-3 ] / [10 1 x 1.60 x ] Ans V H = 4.68 x 10-3 V = 4.68 mv Prepared by Department of Applied Physics, RGCER, Nagpur Page No 19 19

20 Prepared by Department of Applied Physics, RGCER, Nagpur Page No 0 0