MANIPAL INSTITUTE OF TECHNOLOGY


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1 MANIPAL INSTITUTE OF TECHNOLOGY MANIPAL UNIVERSITY, MANIPAL SECOND SEMESTER B.Tech. ENDSEMESTER EXAMINATION  MAY 013 SUBJECT: ENGINEERING PHYSICS (PHY101/10) Time: 3 Hrs. Max. Marks: 50 Nte: Answer any FIVE FULL questins. Each questin carries 10 marks Answer all the sub questins f a main questin in a cntinuus sequence. Write specific and precise answers. Any missing data may suitably be assumed. Write questin number n the margin nly. Draw neat sketches wherever necessary. Physical Cnstants: Speed f light in vacuum = 3.00 X 10 8 m/s Electrn charge = 1.60 X C Electrn mass = 9.11 X kg atmic mass unit (u) = 1.66 X 10 7 kg Bltzmann cnstant = 1.38 X 10 3 J/ K Planck s cnstant = 6.63 X J.s 1A. Discuss the diffractin due t singleslit. Obtain the lcatins f the minima and maxima qualitatively. Draw the necessary diagrams wherever necessary. [5] 1B. Shw that the grup speed f a wavepacket is equal t the particle speed. [3] 1C. The ttal ptential energy (U) f an inic crystal is where α = Madelung cnstant, k e = electrical cnstant, r = nearest neighbr distance, e = electrn charge, m = small integer. Obtain the cnstant B when U is minimum (=U ) at r = r. And, hence shw that the inic chesive energy f the slid is. [] A. The wave functins fr Hatm in 1sstate and sstate, respectively, are and [a = Bhr radius]. Obtain the expressins fr the radial prbability density f Hatm in these states. Sketch schematically the plt f these prbability densities vs. radial distance. [5] B. Sketch the diagram f a MOSFET and name the parts in it. [3] C. Explain the law f Malus with a diagram. []. 3A. Sketch schematically the plt f resistance f a supercnducting material vs. temperature, near the critical temperature. Explain briefly the BCS thery f supercnductivity in metals. [5] 3B. Obtain an expressin fr dispersin by a diffractin grating. [3] 3C. Explain Einstein s phtelectric equatin with a schematic graph f the kinetic energy f mst energetic phtelectrn vs. frequency f incident radiatin. [] 1 / 14
2 4A. Use phasr methd t find the resultant E(t) f the fllwing wave disturbances. E 1 = E sin t E = E sin ( t + 15 ) E 3 = E sin ( t + 30 ) E 4 = E sin ( t + 45 ) [5] 4B. The radius f ur Sun is 6.96 x 10 8 m, and its ttal pwer utput s 3.77 x 10 6 W. Assuming that the Sun s surface emits as a black bdy, calculate its surface temperature. StefanBltzmann cnstant = x 10 8 W/m K 4. [3] 4C. Cnsider a system f electrns cnfined t a threedimensinal bx. Calculate the rati f the number f allwed energy levels at 8.50 ev t the number at 7.00 ev. [] 5A. The wavelengths f the K and K Xrays frm mlybdenum target are 63 pm and 71 pm respectively. Frm these values calculate the wavelength f L Xrays frm mlybdenum target. [5] 5B. Obtain the value f the Madelung cnstant (α) in the ptential energy expressin U(r) = α k e e / r, fr a nedimensinal chain f alternating psitive and negative ins shwn in the figure, where e = electrn charge, k e = electrical cnstant. [3] ln Use the series expansin: x x 3 x 4 1 x x C. When red light in vacuum is incident at the plarizing angle n a certain glass slab, the angle f refractin is What are the index f refractin f the glass and the plarizing angle? [] 6A. A H mlecule is in its vibratinal and rtatinal grund states. It absrbs a phtn f wavelength.11 μm and jumps t the v = 1, J = 1 energy level. It then drps t the v = 0, J = energy level, while emitting a phtn f wavelength.4054 μm. Calculate the mment f inertia (I) f the H mlecule abut an axis thrugh its centre f mass and perpendicular t the HH bnd. The energy (E vj ) f a diatmic mlecule is E vj = (v +½) hf + (h / 8 I) J (J+1), where f = vibratinal frequency. [5] 6B. In a duble slit experiment, the distance f the screen frm the slits is 5 cm, the wavelength is 480 nm, slit separatin is 0.1 mm and the slit width is 0.05 mm. What is the spacing between adjacent fringes? What is the distance frm the central maximum t the first minimum f the fringe envelpe? [3] 6C. An electrn is cnfined between tw impenetrable walls 0.0 nm apart. Determine the energy f the electrn in the grund state in ev. []
3 1A. Discuss the diffractin due t singleslit. Obtain the lcatins f the minima and maxima qualitatively. Draw the necessary diagrams wherever necessary. [5] Diffractin pattern due t single slit has central maximum flanked by the minima and secndary maxima, as shwn in the figure. All the diffracted rays arriving at P are inphase. Hence they interfere cnstructively and prduce maximum (central maximum) f intensity I at P. At pint P 1, path difference between r 1 and r is (a/) sin θ. S the cnditin fr the first minimum is (a/) sin θ = λ/. r a sin θ = λ. This is satisfied fr every pair f rays, ne f which is frm upper half f the slit and the ther is a crrespnding ray frm lwer half f the slit. At pint P, path difference between r 1 and r is (a/4) sin. S the cnditin fr the secnd minimum is (a/4) sin θ = λ/. r a sin θ = λ. This is satisfied fr every pair f rays, separated by a distance a/4. In general, the cnditin fr the m TH minima is a sin θ = m λ, m = ±1, ±, ±3,
4 1B. Shw that the grup speed f a wavepacket is equal t the particle speed. [3] Wavelength f a quantum particle is λ = h / p, p = mmentum f the particle.. Frequency f the quantum particle is f = E / h. E = ttal energy f the particle.. Grup speed f the quantum particle is. Fr a classical particle f mass m mving with speed u, the kinetic energy E is given by and r., particle speed. 1C. The ttal ptential energy (U) f an inic crystal is where α = Madelung cnstant, k e = electrical cnstant, r = nearest neighbr distance, e = electrn charge, m = small integer. Obtain the cnstant B when U is minimum (=U ) at r = r. And, hence shw that the inic chesive energy f the slid is. [] At r = r, U is minimum. ( ). ( ). 4 14
5 A. The wave functins fr Hatm in 1sstate and sstate, respectively, are and [a = Bhr radius]. Obtain the expressins fr the radial prbability density f Hatm in these states. Sketch schematically the plt f these prbability densities vs. radial distance. [5]. Radial prbability density: P(r) = 4 r ( ) ( ) ( ) fr indexing the axes r, P(r) prperly (grids & graduatins need nt be shwn) fr the plt f P 1s (r) with prper shape & peak shwn at r = a fr the plt f P s (r) with prper shape & majr peak shwn at r = 5a a 5a r 5 14
6 B. Sketch the diagram f a MOSFET and name the parts in it. [3] fr prper diagram fr shwing the nchannel, the depletin regin, the ptype substrate fr shwing the surce, the gate, the drain, the silicn dixide C. Explain the law f Malus with a diagram. [] fr prper diagram If E is the magnitude f electric vectr f the incident light (alng xdirectin), nly E cs θ (y cmpnent) passes thrugh the plarid. θ is the angle between E and the plarizing directin. Transmitted intensity: I = I m cs θ where [Law f Malus] I m maximum intensity (ie θ = 0 r 180 ) fr prper explanatin 6 14
7 3A. Sketch schematically the plt f resistance f a supercnducting material vs. temperature, near the critical temperature. Explain briefly the BCS thery f supercnductivity in metals. [5] R fr the plt with prper shape (fr the curve) and axes prperly indexed 0 T T C BCS thery f supercnductivity in metals: Accrding t this thery, tw electrns can interact via distrtins in the array f lattice ins s that there is a net attractive frce between the electrns. As a result, the tw electrns, are bund int an entity called a Cper pair. The Cper pair behaves like a bsn (= particle with integral spin that d nt bey the Pauli exclusin principle). At very lw temperature, it is pssible fr all bsns in a cllectin f such particles t be in the lwest quantum state. As a result, the entire cllectin f Cper pairs in the metal is described by a single wave functin. Abve the energy level assciated with this wave functin is an energy gap equal t the binding energy f a Cper pair. Under the actin f an applied electric field, the Cper pairs experience an electric frce and mve thrugh the metal. A randm scattering event f a Cper pair frm a lattice in wuld represent resistance t the electric current. Such a cllisin wuld change the energy f the Cper pair because sme energy wuld be transferred t the lattice in. But there are n available energy levels belw that f the Cper pair (it is already in the lwest state) and nne available abve, because f the energy gap. As a result, cllisins d nt ccur and there is n resistance t the mvement f Cper pairs. 7 14
8 3B. Obtain an expressin fr dispersin by a diffractin grating. [3] Angularseparatin betweenspectral lines Dispersin Difference betweenwavelengthf spectrallines Dispersin: Diffractin grating equatin: Differentiating the equatin d sin = m d cs = m Dispersin: m = rder f diffractin, d = grating space, θ = angle f diffractin. 3C. Explain Einstein s phtelectric equatin with a schematic graph f the kinetic energy f mst energetic phtelectrn vs. frequency f incident radiatin. [] Einstein s phtelectric equatin: K MAX = hf  = wrk functin f the metal ( = minimum energy with which an electrn is bund in the metal) K MAX = kinetic energy f mst energetic phtelectrn hf = energy f the incident phtn f = frequency f the incident phtn K MAX f SLOPE = h f φ 8 14
9 4A. Use phasr methd t find the resultant E(t) f the fllwing wave disturbances. E 1 = E sin t E = E sin ( t + 15 ) E 3 = E sin ( t + 30 ) E 4 = E sin ( t + 45 ) [HRKSP41.] [5] Phasrs at t = 0 1 ST METHOD E 1 = 0 E = E sin 15 = E (0.588) E 3 = E sin 30 = E (0.5000) E 4 = E sin 45 = E (0.7071) Resultant: E = E 1 + E + E 3 + E 4 = E θ sin = E (1.466) 1 E 1X = E E X = E cs 15 = E (0.9659) E 3X = E cs 30 = E (0.8660) E 4X = E cs 45 = E (0.7071) E 4 E E 3 E E X = E 1X + E X + E 3X + E 4X E θ E 1X E X E 3X E 4X tan = E / E X = E X = E 1X + E X + E 3X + E 4X = E (3.539) 1 = tan 1 (0.414) =.5 E(ωt) = E θ sin ( t + ) = 3.83 E sin ( t +.5 ) ND METHOD [text bk methd] SUM OF INTERIOR ANGLES IN POLYGON OF 5 SIDES (5 ) x 180 = + 3(180 φ) =.5 1 C φ D E E = OD = OA + A B + B C + C D θ OD = OA + AB + B C + C D OD = OA cs + AB cs ( φ) + BC cs ( φ) + CD cs E θ = E cs.5 + E cs E cs E cs.5 E = 3.83 E θ E B E θ B E B φ C E E(t) = E sin (ωt + ) θ A φ E(t) = 3.83 E sin (ωt +.5 ) E A ωt O 9 14
10 4B. The radius f ur Sun is 6.96 x 10 8 m, and its ttal pwer utput s 3.77 x 10 6 W. Assuming that the Sun s surface emits as a black bdy, calculate its surface temperature. StefanBltzmann cnstant = x 10 8 W/m K 4. [3] [SJP40.5] Perfect black bdy: e = 1 σ = x 10 8 W/m K 4 r = 6.96 x 10 8 m P = 3.77 x 10 6 W A = 4 r = x m StefanBltzmann law: P = σ A e T 4 T = 5750 K 4C. Cnsider a system f electrns cnfined t a threedimensinal bx. Calculate the rati f the number f allwed energy states at 8.50 ev t the number at 7.00 ev. [SJP43.35(a)] [] Number f allwed energy states at energy E: g(e) 4 m 3 h 3 E 1 g(8.5 ev) g(7.0 ev) OR (8.5) (7.0) 1 1 g(e) = [CONSTANT] E ½
11 5A. The wavelengths f the K and K Xrays frm mlybdenum target are 63 pm and 71 pm respectively. Frm these values calculate the wavelength f L Xrays frm mlybdenum target. [HRKE48.1(a) Extensin][5] E M E L hc λ α Frm the figure hc λ α hc λ β OR E K λ K = 63 pm, h = 6.63 x J.s, λ Kα = 71 pm c = 3.00 x 10 8 m/s = x J =.801 x J = 3.56 x J OR = 5.60 x m = 560 pm = 560 pm 11 14
12 5B. Obtain the value f the Madelung cnstant (α) in the ptential energy expressin U(r) = α k e e / r, fr a nedimensinal chain f alternating psitive and negative ins shwn in the figure, where e = electrn charge, k e = electrical cnstant. r = interatmic distance. [SJP43.5] [3] ln Use the series expansin: x x 3 x 4 1 x x ( ), putting x = 1 in the expansin f ln (1+x) α = ln 5C. When red light in vacuum is incident at the plarizing angle n a certain glass slab, the angle f refractin is What are the index f refractin f the glass and the plarizing angle? [HRKE44.1] [] r = 31.8 i + r = 90 Plarising angle: θ P = i = 90 r = 58. Index f refractin: n = tan θ P =
13 λ 1 =.11 μm λ =.4054 μm MU  MIT I BTE C H SECOND SEMESTER END EXAMINATION MAY 013 ENGINEERING PHYSICS 6A. A H mlecule is in its vibratinal and rtatinal grund states. It absrbs a phtn f wavelength.11 μm and jumps t the v = 1, J = 1 energy level. It then drps t the v = 0, J = energy level, while emitting a phtn f wavelength.4054 μm. Calculate the mment f inertia (I) f the H mlecule abut an axis thrugh its centre f mass and perpendicular t the HH bnd. The energy (E vj ) f a diatmic mlecule is E vj = (v +½) hf + (h / 8 I) J (J+1), where f = vibratinal frequency. [SJP43.19] [5] ( )h [ ] V=1 E v,j E 11 J= J=1 J=0 V=0 E 0 J= J=1 h c ( ) E 00 J=0 ( ) I = 4.60 x kg.m 13 14
14 6B. In a duble slit experiment, the distance f the screen frm the slits is 5 cm, the wavelength is 480 nm, slit separatin is 0.1 mm and the slit width is 0.05 mm. What is the spacing between adjacent fringes? What is the distance frm the central maximum t the first minimum f the fringe envelpe? [HRKSP4.6] [3] D = 0.5 m, λ = 480 nm, d = 0.1 mm, a = 0.05 mm Spacing between adjacent fringes: Δy = λ D / d = (480 x 10 9 m) (0.5 m) / m) = m =.1 mm 1 a sin θ = 1 λ fr 1 ST diffractin minimum at θ, sin θ = y 1 / D a y 1 / D = λ y 1 = λ D / a = m = 10 mm 1 6C. An electrn is cnfined between tw impenetrable walls 0.0 nm apart. Determine the energy f the electrn in the grund state in ev. [SJEx41.1 ] [] L = 0.0 nm, m = 9.11 x kg, h = 6.63 x J.s E 1 = h / 8 m L = 1.5 x J 1 = 9.4 ev
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