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Unformatted text preview: Modern Physics and Quantum Mechanics Spring 2001 Proficiency Useful constants: • e = 1 . 60 × 10 19 C • hc = 1240 eV · nm • c = 3 . 00 × 10 8 m s • m e = 0 . 511 MeV c 2 • k = 1 4 π = 8 . 99 × 10 9 Nm 2 C 2 • = 8 . 85 × 10 12 C 2 Nm 2 1. Consider a onedimensional step potential of the form V ( x ) = ( x < V x ≥ , where V > . A particle with total energy E > V and mass m is incident on the step potential “from the left” (in other words: the particle starts at negative values of x and travels toward positive values of x ). (a) Use the timeindependent Schr¨ odinger equation to determine the form of the particle’s wave function in the two regions x < and x ≥ . In the x < 0 region (call this region 1) the potential is zero so the Schr¨ odinger equation has the form ¯ h 2 2 m d 2 Ψ 1 dx 2 = E Ψ 1 , (1) and Ψ 1 has the form of a plane wave moving to the right (the incident wave) and another moving to the left (the reflected wave), so that Ψ 1 = Ae ik 1 x + Be ik 1 x . (2) Substituting this into the Schr¨ odinger equation we see that it is a solution if ¯ h 2 2 m ( k 2 1 ) = E k 1 = s 2 mE ¯ h 2 = √ 2 mE ¯ h . In the x > 0 region (call this region 2) we have ¯ h 2 2 m d 2 Ψ 2 dx 2 + V Ψ 2 = E Ψ 2 . ¯ h 2 2 m d 2 Ψ 2 dx 2 = ( E V )Ψ 2 . and Ψ 2 has the form of a plane wave moving to the right (the transmitted wave), so that Ψ 2 = Ce ik 2 x , (3) with k 2 = q 2 m ( E V ) ¯ h . (4) (b) Derive expressions for the probabilities that the particle is (i) reflected ( R ), and (ii) transmitted ( T ). Hint: Recall that the probability density current is given by j ( x ) = Re Ψ * ¯ h im ∂ Ψ ∂x ! , and that R and T are ratios of probability density currents. We have to derive the values of A , B , and C using the boundary conditions, which are that the wavefunction and its derivative must be continuous at the boundary ( x = 0). Continuity of the wavefunction gives that A + B = C, (5) while that of the derivative at x = 0 gives that k 1 A k 1 B = k 2 C, (6) since taking the derivative of the plane wave just brings down a factor of ik , and we have divided by i . For that reason the incident probability current is simply j in = Re " Ae k 1 x ¯ h im ik 1 Ae ik 1 x # = ¯ h m k 1 A 2 , (7) and similarly j refl = ¯ h m k 1 B 2 j trans = ¯ h m k 2 C 2 . Putting these together we get that the reflection probability is R = j refl j in = B 2 A 2 , (8) and we can find B/A by combining (1) and (2) above, k 1 ( A B ) = k 2 ( A + B ) , (9) so that B A = k 1 k 2 k 1 + k 2 , (10) and R = k 1 k 2 k 1 + k 2 !...
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 mechanics, Energy, Fundamental physics concepts, Total Energy

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