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WORK HARD UNTIL LAMP LIGHT OF STUDY TABLE BECOMES SPOT LIGHT OF STAGE.

# Note for Applied Physics - Phy By SHANMUGAM S

• Applied Physics - Phy
• Note
• Anna university - gtec
• Computer Science Engineering
• 9 Topics
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(c) Relaxation time (τr) : In the absence of electric field, the conduction electrons move in random direction, and hence the probability of finding an electron moving in any given direction is zero. i.e. Vav = 0 When an external field is applied, net positive value V1av for the average velocity of the conduction electrons in a direction opposite to the direction of field; which is equal to the drift velocity i.e Vav=V1av If the field is turned off, the average velocity Vav starts reducing exponentially as shown in the figure and is according to the equation. (1) Time counted from the instant the field is turned off, If = Relaxation time equation (1) becomes Hence relaxation time is defined as “The time interval during which drift velocity of electrons reduces to 1 e times the maximum value attained by them when applied field is turned off”. (d) Mean free path (λ): The average distance travelled by the conduction electrons between two successive collisions of conduction electrons under the influence of applied electric field is called mean free path. d) Mean collision time (τ): The average time interval between two consecutive collisions of an electron with the lattice cores in a conductor under the influence of applied electric field is called mean collision time. τ = λ/vth Where „λ‟ is the mean free path, v≈vth is velocity same as combined effect of thermal & drift velocities. Note: But ½ mvth2=3/2 KT Question (4): Discuss the failures of classical free electron theory. Classical free electron theory has failed to explain specific heat, dependence of conductivity with temperature and dependence of conductivity on electron concentration as follows (a) Specific heat: 3

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As per classical free electron theory, free electrons in a metal behave as gas molecules and 3 hence the molar specific heat of electrons at constant volume is given by Cv= R 2 Where R is universal gas constant But experimentally molar specific heat of free electrons in a metal is given by CV=10-4RT This is not only less than the experimental value but also depends on temperature. Hence classical free electron theory failed to explain specific heat. (b) Dependence of electrical conductivity on Temperature: According to the assumptions of classical free electron theory √ √ ---------------------- (1) The mean collision time „τ‟ is inversely proportional to the thermal velocity. (vth=  )  i.e. √ From the expression ------ ------------------------- (2) From (1) σ= ne 2 -------------- (3) m    Substituting for from equation (3) gives 1 σα -----------------------(4) T Therefore according to classical theory, electrical conductivity is inversely proportional to the square root of absolute temperature. But experimentally σ is inversely proportional to the temperature T. i.e.  cxpt  1 T ----------- (5) Therefore classical theory failed to explain conductivity with temperature. (c) Dependence of electrical conductivity on electron concentration: According to classical theory, electrical conductivity is directly proportional to electron concentration i.e σ = ne 2 m σαn Thus as n increases conductivity should increase. But it is contrary to the observation. Consider the data from the following table. 4

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Metals Electron Con.(n) in/m3 Conductivity in Ω-1m-1 Cu 8.45×1028 5.88×107 Ag 5.85×1028 6.3×107 Al 18×1028 3.67×107 Electrical conductivity of aluminium is lesser than Copper (Cu) and Silver (Ag); even though electron concentration in Al is higher than that of Cu and Ag. Similar observations are made with Cu and Al. Therefore σ α n do not holds good. Hence the classical free electron theory failed to explain the dependence of σ on electron concentration. Problems on classical free electron theory 1. Calculate drift velocity and thermal velocity of electrons in a metal of thickness 1mm across which a potential difference of 1volt is applied at temperature 300K.Compare this value with thermal velocity of electrons. Given that mobility of electrons is 40cm2/VS. Solution: Vd=?, Vth= ?, L=1mm=1X10-3m; V=1volt; T=300K   40cm 2 /VS  40X 104 m2 /VS E V 1   1000V L 1X 103 Vd  E =40x10-4x1000=4 ms-1 is drift velocity------(1) Vth  3KT  m  3 X 1.38X 10 23 X   300   1.17X 105 mS 1   ----(2) 9.1X 1031   Vd 4   3.14 x10 5 ms 1 Vth 1.17 x10 5 2. A uniform silver wire has a resistivity of 1.54X10-8 ohm-m at room temperature. across which an electric field of 1 volt/cm is applied. Calculate (i) relaxation time (ii) drift velocity and (iii) mobility of electrons; assuming that there are 5.8X10-28 electrons per m-3of the metal. (VTU Jun 2010) Solution:   1.54X10-8 ohm-m; E=1 volt/cm=1volt/1x10-2m=100V/m; n=5.8X10-28 electrons per m-3; Vd=?  ?; =? (i) Relaxation time Consider     m ne2  m ne2  ne2 m 9.1x1031  3.985x1014 S  8  28  19 2 1.54x10 x5.8 x10 (1.6 x10 ) (ii) Drift velocity 19  eE  1.6 x10 X 100 Vd     x3.98 x10 14 ms 1  0.699.ms 1 31 9.1x10 m 5

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(iii) Mobility of electrons Vd  E  V  0.699   d   6.99 X 10 3 m 2 / VS  E  100 3. Calculate mobility of electrons in Copper assuming that each atom contributes one free electron for conduction. Given resistivity of copper=1.7x10-8 Ω m, atomic weight=63.54,density=8.96X103kg/m3 and Avogadro number=6.023x1023/mol. 8 Solution:     ?,   1.7 X 10 m ;M=63.54; NA=6.023X1023/g mol; D=8.96X103Kgm-3; number of electrons per atom =n=1 1 No. of conduction electrons/m3=No.of electrons per atom X NAX D AtomicWt n A XN A XD n= AtomicWt 1x6.025x1023 x8.96 x103 n  8.50 x1025 kgm  3 63.54     ? Using   ne we get   ne  1 1.7 x108 x8.5x1025 x1.6 x1019  4.325m2V 1S 1 4. Calculate free electron concentration and hence mobility of electrons in a aluminium metal assuming that each atom contributes three free electron for conduction. Given resistivity of Al=2.7x10-8 Ω m, atomic weight=26.98,density=2.7X103kg/m3 and Avogadro No.=6.023x1023/mol. 8 Solution: n=?     ?,   2.7 X 10 m ;M=26.98; NA=6.023X10-23/gmol; D=2.7X103Kgm-3; number of electrons per atom =n=3 (a) (b) n= n A XN A XD AtomicWt 3x6.025x1023 x 2.7 x103 n  1.81x1026 kgm  3 26.98     ? Using   ne we get   ne  1 8 26 2.7 x10 x1.81x10 x1.6 x10 6 19  1.278m2V 1S 1