BACHELOR OF SCIENCE (B.Sc.) Term-End Examination December, 2016

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1 No. of Printed Pages : 8 I PHE-13 I BACHELOR OF SCIENCE (B.Sc.) Term-End Examination December, PHYSICS PHE-13 : PHYSICS OF SOLIDS Time : 2 hours Maximum Marks : 50 Note : Attempt all questions. However, internal choices are given. The marks for each question are given against it. You may use log tables or non-programmable calculators. Symbols have their usual meanings. The values of physical constants are given at the end. 1. Attempt any five parts : 5x3=15 (a) (c) With the help of an appropriate diagram explain all the symmetry elements of a methane molecule. Show that nearly half the volume of the unit cell of a simple cubic lattice is occupied by its atoms. Explain three types of Van der Waals bonding. (d) Draw a labelled energy band diagram for a p-type semiconductor. PHE-13 1 P.T.O.

2 (e) The potential energy function is given by a f3 U(r) = + r6 r 12 Obtain the inter-molecular distance for which the potential energy is minimum. (f) (g) (h) Fermi energy of an intrinsic semiconductor is 0.7 ev. The low lying energy levels in the conduction band is 0.2 ev above the Fermi level. Calculate the probability of occupation of this level by an electron, at room temperature. Explain with the help of appropriate examples the difference between a piezoelectric and a pyroelectric crystal. Explain the application of thin films as interference filters. 2. Answer any two parts : 2x5=10 (a) What is rotational symmetry in a lattice? Show that 5-fold rotational symmetry is not possible in a 2-D lattice. 1+4 (c) Describe the Laue's method of crystal structure determination. Write down its limitations. 4+1 Prove that the reciprocal lattice of a bcc is an fcc structure. PHE-13 2

3 3. Answer any one part : 1x5=5 (a) 9N 1 Cv = (kbtd )3 T State the basic difference between the Einstein's and Debye's theories of specific heat capacity. The expression for heat capacity according to Debye's theory is given by (kbtd)4 71) 1] [exp( kbtd f 443d4 1 0 {exp( 4 ) kbt where symbols have their usual meanings. Obtain the T3 law. 1+4 Derive the expression for dispersion relation for a linear array of the same type of atoms Answer any two parts : 2x5=10 (a) Show that the effective mass of an electron in a crystalline solid is given by h 2 m = (d 2 Eidk 2 ) 5 (i) Explain how a superconductor can be used for magnetic shielding. (ii) Calculate the limiting value of the magnetic field for which Nb will act as a superconductor at 4 K. Take B ac(0) as 1970 Oe and Te for Nb to be 9.25 K. 2 (c) Derive an expression for Hall coefficient. What are its applications? 4+1 PHE-13 3 P.T.O.

4 5. Answer any two parts : 2x5=10 (a) (c) Distinguish between dia-, para- and ferro-magnetism. Give two examples of each category. 5 Classify the defects in crystals. Explain any one of these classes in detail with suitable diagrams. 2+3 What is a "Transducer"? Explain the working of any one transducer. 1+4 Physical Constants : h = 6.62 x Jo 71 =1.05x10-34 Js NA = 6.02 x 1023 morl e = 1.6 x C kb = 1.38 x J K me = 9.1 x kg PHE-13 4

5 lif1.rw.t =Ird4 (*.R/1 111.) forrall, 2016 Oft 01.1 tft.ka.f.-13 :TR amarr OR* R-497: 2 EPS' 3#0*-Ār 37.* ": 50 gl# eff / cg, che41 77' / J/4"*" NV* dff4 faii4 7/k7 / 377ET tv/tl~id/ 31274T 3751k11477 eleggriej itt 3/* W1- We} 1 / Nee ff/ / OW- Petaio7 7477,3k#R77R- ff/ tV I * 5x3=15 ("W) sitiv rift 3T-44-41*t maw *PAR I Ri4 *tftr f Aloct..* tictw 4F wript artit atrzrffq t4. yt1i J t I (IT) At4 Wit * must girt' 31T 0t0 *I WIRT '407 (4) p-awk i eW 4114 at1111 Ilan l PHE-13 5 P.T.O.

6 (19 dwi odiach 14HiCi Rgo : a r3 U(r) = + T 6 r 12 fld-q Triq t4s *t "PiMs 4 ai-du-atur Tra- *tl* N) 7-* 47 atti-4tfw tsiff.i,711 "Mq 0.7 ev t I milzfri.ftk th-rff tic t 0.2 ev 'W:R t I QtR5 (T) WI qt Iv 'RR 4 * trrr 7r4 '4t mincbdi *VA I () z-q-sffi tr will kiiiww i* TA' I %I, att f4-tri% atk eicm (7) -ff3 (tr-t-41.) 14)04 * oeacbtui -1**-4-* * vcr ardsrzhir wow I 2. f+-i \irk tf* : 2x5=10 Alo.r. ' vii i kii1i 1? f,q *F1* i* 2-D Alcich 5-citii 44T4 t I 1+4 f4ttkerf A C113 f4fit Ws au *rnr 4tieR tiliii* -FAR I 4+1 (TT) #T4 *If4R cophii Algicb t 1T-414 (bcc) 4414 (fcc)tth PHE-13 6

7 3. v ITITT ZW( tra7 : 1x5=5 () -1:41tz SIT tuftm ' 3T14R:ETF c ti -ft-o-rt fiat * 419. wita. 3TK Gici14 I it-0 * atfr wan vrftur 04,3m 9N CV = (kbtd)3 PHIZiRgi»lnl : D kbtd (kbt )4 ff 4 3 d4 [exprn 11 0 [exp( 4 T I 1 J TOt 3ltr4 31 I titl T3 Ptiii srrrrfa I 1+4 (N) i * 1R **t kricb (sitifff ) * tfftweruf 144 VF c t) 41* I k--41 lurrl 39 tflr : 2x5 = 10 (W) RIZ *rf* DAtzflii ZIET ra-vq 04,31,4) f417-4fft : h 2 m = (d2 E/dk 2 ) (i) aedw-ew v-61 cift(kiut 3441) )(1 M 41,1t 6411 M' tithcll t I 2 (ii) 4 K Nb 'AO atd-- r-ew el-1r ttg * W RR kiiket)rem *ii* #tf* Bac(0) = 1970 Oe 3k Nb RiL Tc = 9-25 K I 3 (TO tr-0'-±jutict) 04,31q) c4jc I 4ti argaritt cif? 4+1 PHE-13 7 P.T.O.

8 5. 14,41 ITA 39 tp4r : 2x5=10 (") mr.et) i dqwui tf. A. I 5 (174) -I*R- 1 A.Tiffff *tf=4r 34Tr) atiw t cr tr *mem 1). iwt *IPA7 I 2+3 (TT) "p-iis-oot" TEIT? walzpi tii-miw I 1+4. irlf4w HtIffieh : h = 6.62 x Js = 1.05 x Js NA = 6.02 x 1023 mo1-1 e = 1.6 x C kb = 1.38 x J K me = 9.1 x kg PHE ,500

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No. of Printed Pages : 8 BP11E-106/PHE-06 BACHELOR OF SCIENCE (B.Sc.) Term-End Examination June, 2018 PHYSICS BPHE-106/PHE-06 : THERMODYNAMICS AND STATISTICAL MECHANICS Time : 2 hours Maximum Marks : 50