sp01mod_sol

sp01mod_sol - Modern Physics and Quantum Mechanics Spring...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 = 8 . 99 10 9 Nm 2 C 2 = 8 . 85 10- 12 C 2 Nm 2 1. Consider a one-dimensional 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 time-independent Schr odinger equation to determine the form of the particles 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 1 dx 2 = E 1 , (1) and 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 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 2 dx 2 + V 2 = E 2 .- h 2 2 m d 2 2 dx 2 = ( E- V ) 2 . and 2 has the form of a plane wave moving to the right (the transmitted wave), so that 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 * h im 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 !...
View Full Document

Page1 / 8

sp01mod_sol - Modern Physics and Quantum Mechanics Spring...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online