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### homework7sol

Course: ECE 407, Fall 2009
School: Cornell
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Word Count: 1168

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407: ECE Homework 7 Solutions (By Farhan Rana) Problem 7.1 a) The answer follows from elementary vector calculus result that that the gradient of any function is perpendicular to the surface of constant value of the function. In the present case, the velocity is related to the gradient of the energy function, rr 1rr vc k = k E k h And therefore the velocity must be perpendicular to the surfaces of constant...

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407: ECE Homework 7 Solutions (By Farhan Rana) Problem 7.1 a) The answer follows from elementary vector calculus result that that the gradient of any function is perpendicular to the surface of constant value of the function. In the present case, the velocity is related to the gradient of the energy function, rr 1rr vc k = k E k h And therefore the velocity must be perpendicular to the surfaces of constant energy. () () Problem 7.2 rr r B.k a) First note that: k|| = r B r r rr r r d hk B. = e B . v c k B dt r r d hk B. =0 dt rr d B.k =0 dt r d k|| =0 dt Now take the dot product on both sides of the crystal momentum equation with the magnetic field to get: [ () ] r r r r rr dE c k (t ) 1 r d h k (t ) r r = k E c k (t ) . = v c k (t ) . e v c k (t ) B b) dt dt h () () ( ) [ [ ( ) ] ]= 0 c) The complete argument follows in two steps: i) Since the energy of the electron remains unchanged and equal to its initial value E o , the motion in k-space of the electron must be confined to a constant energy surface corresponding to the initial energy E o of the electron. Since the component of its crystal momentum parallel to the magnetic field also remains r unchanged and equal to its initial value k|| , the motion in k-space of the electron must also be confined to the plane in k-space on which the components of all crystal momenta in the r direction of the magnetic field is k|| (convince yourself that this later condition defines a plane in k-space that is perpendicular to the magnetic field). Since both (i) and (ii) above must be true, the desired result follows. ii) r d) One can always write the vector r (t ) in terms of its components in two orthogonal directions: parallel and perpendicular to the magnetic field: 1 r r r r (t ) = r (t ) + r|| (t ) But: r r r r (t ) . B r r|| (t ) = r B 2 B Therefore: r r r r r (t ) . B r r (t ) = r (t ) r B 2 B e) Start from, r rr r d hk = e v c k B dt Take thercross product on both sides with the magnetic field: r r d hk rr r = e B v c k B B dt r r d hk r r rrr r B = e v c k B B . v c k dt r r r r d hk dr (t ) r 2 r r dr (t ) Note : B B B . B = e dt dt dt r r r d hk r 2 dr (t ) B = e B dt dt Now integrate both sides from time 0 to t to get: r r r hrr r (t ) r (t = 0 ) = r B k (t ) k (t = 0 ) 2 eB [ () ] [ () ] [ () ( ( ))] rrr rrr rrr A B C = B A . C C A . B ( ) ( )( ) [ ] f) We have from lecture handouts: rr rr r rr d vc k = e v c k B o z = e B o v c k z M. dt rr rr d vc k or : = e B o M 1 . v c k z dt The above equation shows that the equations for the x- and y-components of the velocity depend only on the x- and y-components of the velocity and not to the z-component of the velocity. Consequently, we can write the equations for only the x- and y-components as follows: 1 m xx 1 m xy v cy d v cx e = B o dt v cy 1 m yx 1 m yy v cx () [ () ] [ () ] () [ () ] We can now solve this simpler equation for the cyclotron frequency. The above are two coupled linear differential equations. One can take the derivative of the first and use the second to get a second order differential equation: 2 1 d 2 v cx 1 1 1 2 = e 2 Bo v cx dt 2 m xx m yy m xy m xy 1 1 1 1 2 = e 2 Bo v cx m xx m yy m xy m yx This implies the cyclotron frequency is: The term in the square brackets is the determinant of the upper-left 2x2 submatrix of the inverse effective mass tensor. 2 c = e 2 Bo 1 1 1 1 m xx m yy m xy m yx Problem 7.3 2 T The time period T is the time taken by the electron moving in k-space in the negative x-direction to traverse the FBZ (from the X-point (2 a ,0,0 ) to the X-point ( 2 a ,0,0 ) covering a total distance of 4 a in k-space. This will take time equal to: 4 h a e E x . 2 a e E x B = = 2h T a) B = b) We have, 2 a e E x B = = = 2.14 1013 rad/s T 2h T = 294 fs. Therefore, electrons will not complete even a single period before they are scattered. Problem 7.4 (Conductivity tensor of germanium) For the pocket at ( a , a , a ) we have: M 1 1 3m l + 2 3mt = 1 3m l 1 3m t 1 3m l 1 3m t 1 3m l + 2 3m t 1 3m l 1 3m t 1 3m l 1 3m t 1 3m l 1 3 m t 1 3m l 1 3 m t 1 3m l + 2 3m t This pockets inverse effective mass tensor will serve as the reference. Now, for the pocket at ( a , a , a ) it was shown that: M 1 1 3m l 1 3 m t (1 3m l 1 3mt ) 1 3m l + 2 3 m t 1 3m l + 2 3m t = 1 3m l 1 3m t (1 3m l 1 3mt ) (1 3m l 1 3mt ) (1 3m l 1 3mt ) 1 3m l + 2 3mt 3 Now we look at the pocket at ( a , a , a ) . If we let E x become...

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