Using the Greens function of the diffusion equation P r t r 1 4 \u03c0 Dt 3 2 e r r

Using the greens function of the diffusion equation p

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Using the Green’s function of the diffusion equation, P ( r , t | r 0 , 0 ) = 1 ( 4 π Dt ) 3 / 2 e -| r - r 0 | 2 / 4 Dt analyze the time dependence of the electron density distribution. Is it correct to use this Green’s function for the case of the electron creation at the surface?
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43 Solution 4.3: For the density distribution at time t one has (by definition) n ( r , t ) = Z d 3 r 1 P ( r , t | r 1 , 0 ) n ( r 1 , 0 ) . The solution is substantially simplified because P ( r , t | r 0 , 0 ) depends only on the difference ( r - r 1 ) . Then, in general, one can use the Fourier method. However, this problem is even simpler since this function depends only on e -| r - r 1 | / 4 Dt = e - ( x - x 1 ) 2 / 4 Dt · e - ( y - y 1 ) 2 / 4 Dt · e - ( z - z 1 ) 2 / 4 Dt , while n ( r 0 , 0 ) e - x 2 1 / 2 σ · e - y 2 1 / 2 σ · e - z 2 1 / 2 σ . Thus we have to analyze only one integral, I ( x ) 1 2 π 2 σ Dt Z - dx 1 e - ( x - x 1 ) 2 / 4 Dt e - x 2 1 / 2 σ = 1 p 2 π ( σ + 2 Dt ) e - x 2 / 2 ( σ + 2 Dt ) . As a result, n ( r , t ) = N [ 2 π ( σ + 2 Dt )] 3 / 2 exp - | r | 2 2 ( σ + 2 Dt ) . This expression cannot be used for the case of surface excitation since the Green’s function for that problem has different form. Problem 4.4: In the presence of electrical field, the electrons acquire a constant (drift) velocity, proportional to the applied electric field, v d = μ E . The quantity μ is called the mobility . Let N electrons be injected at the plane x = 0, t = 0, and then diffuse in the presence of the electrical field creating the drift velocity v d k x . Find the distribution of electrons at time t . Solution 4.4: The particle current in the presence of the drift is j = n ( x ) v d - D dn dx . Thus the diffusion equation acquires the form: dn dt + d j dx = 0 dn dt + v d dn dx - D d 2 n dx 2 = 0 . It is natural to introduce a new variable, ξ = x - v d t and search solution as n ( ξ , t ) . Since dn ( ξ , t ) dt = n ( ξ , t ) t - v d n ( ξ , t ) ∂ξ .
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44 CHAPTER 4. DEFECTS AND DIFFUSION On the other hand, v d dn dx = v d n ( ξ , t ) ∂ξ ∂ξ x = v d n ( ξ , t ) ∂ξ . In this way we return to the well known diffusion equation, n t = D 2 n ∂ξ 2 having the solution n ( x , t ) = n 0 4 π Dt exp - ( x - v d t ) 2 4 Dt . Here n 0 N / A is the initial surface density of injected particles. Problem 4.5: Frenkel defects: Show that the number n of interstitial atoms in equilibrium with n lattice vacancies in a crystal having N lattice points and N 0 possible interstitial positions is given by the equation E I = k B T ln £ ( N - n )( N 0 - Nn ) / n 2 / , whence for n ¿ N , N 0 , we have n ( NN 0 ) 1 / 2 exp ( - E I / 2 k B T ) . Here E I is the energy necessary to remove an atom from a lattice site to an interstitial position. Solution 4.5: The internal energy of formation of n Frenkel defects is U = nE I . The number of ways to pick n from N is N ! n ! ( N - n ) ! . The number of ways to put n atoms into N 0 sites is N 0 ! n ! ( N 0 - n ) ! . Hence, the entropy is S = k B ln N ! n ! ( N - n ) ! + ln N 0 ! n ! ( N 0 - n ) ! . Using the Stirling’s formula we get ln N !
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