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HW1_solutions - Homework 1 Solutions 1 Consider how the de...

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Homework 1: Solutions 1. Consider how the de Broglie’s suggestion might explain some properties of the hydrogen atom. a. Show that the assumption p = = h λ and the ‘quantization condiction’ that the length of a circular orbit be an integer multiple of the length of the electron wavelentgh (that is: = 2 πr ,where r is the raduis of the orbit and n an integer) imply that only discrete orbits are allowed. b. Calculate the total energy (kinetic plus potential) of the electron in each orbit characterized by n . Hint: In part a find two equations descreibing the balance between the centrifugal and the Coulomb (centripetal) force. Solve for the radius r and for the angular velocity ω . Now, in b insert these expressions into the formulae for the kinetic and potential energy. Solution: See ECE618 Notes,Part 1. 2. Problem. Consider the tunneling problem with the potential barrier: V ( z ) = 0 for z 0 V > 0 for 0 < z < L 0 for z L . Write the wavefunction as: ψ ( z ) = Ae ikz + Be ikz for z 0 Ce κz + De κz for 0 < z < L F e ikz for z L , with k = (2 mE ) 1 / 2 / ¯ h , κ = [2 m ( V E )] 1 / 2 / ¯ h . a. Write the system of four equations expressing the continuity of the wavefunction and its derivative at z = 0 and at z = L . b. Find the transmission coefficient, T = | F | 2 / | A | 2 . There’s no need to solve the full system. Be creative. ECE609 Spring 2010 1
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Hint: Multiply the equation expressing continuity of ψ at z = 0 by ik and add and substract it from the equation expressing continuity of the derivatives at z = 0 . Do a similar thing with the other two equations (by multiplying one by κ ). Now it should be relatively easy to solve for F in terms of A alone. This gives you T . Solution . a. The equations expressing continuity of the wavefunction and its derivatives at z = 0 and z = L are: ψ (0 ) = ψ (0 + ) A + B = C + D , (1) (0 ) dz = (0 + ) dz ikA ikB = κC κD , (2) ψ ( L ) = ψ ( L + ) Ce κL + De κL = F e ikL , (3) ( L ) dz = ( L + ) dz κCe κL κDe κL = ikF e ikL . (4) b. Multiply Eq. (1) by ik ,add and substract it from Eq. (2): 2 ikA = ( ik + κ ) C + ( ik κ ) D , (5) 2 ikB = ( ik κ ) C + ( ik + κ ) D . (6) Multiply Eq. (3) by κ ,add and substract it from Eq. (4): 2 κCe κL = ( ik + κ ) F e ikL , (7) 2 κDe κL = ( ik κ ) F e ikL . (8) From Eqns. (7) and (8) we can express C and D in terms of F . Inserting these expressions into Eq. (5) we can finally express F in terms of A : 2 ikA = e ikL F 2 κ [( κ 2 k 2 ) sinh( κL ) + 2 iκk cosh( κL )] , (9) ECE609 Spring 2010 2
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so that the transmission coeficient is: T = | F | 2 | A | 2 = 4 κ 2 k 2 ( κ 2 k 2 ) 2 sinh 2 ( κL ) + 4 κ 2 k 2 cosh 2 ( κL ) . (10) 3. Problem. From the Schr¨odinger equation derive the continuity equation: ∂ρ ∂t + ∇ · S = 0 , where ρ = | Ψ | 2 is the ‘probability density’ and S = i ¯ h 2 m Ψ Ψ Ψ] is the ‘probability density current’.
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