Section 6

Question 1

If E > V(±∞)

Answer 1

we have a scattering state

Question 2

Scattering formalism

Incoming plane wave with t-dependence hidden:

Answer 2

Ψ(x) = Aexp[ikx]

Complex so non-zero flux

J = (ℏk)/m |A|^2

Question 3

Reflection coefficient

Answer 3

R = J(R)/J(I) = |B|^2/|A|^2

Question 4

Transmission coefficient

Answer 4

T = J(T)/J(I) = k2/k1 |C|^2/|A|^2

Question 5

Flux conservation

Answer 5

J(I) = k2|C|^2 = k1(|A|^2-|B|^2)

Question 6

Scattering from allowed potential step

R = |B|^2 /|A|^2 =

Answer 6

(k1-k2)^2/(k1+k2)^2

Question 7

Scattering from allowed potential step

T = k2/k1 |C|^2 /|A|^2 =

Answer 7

1 - R = 4k1k2/(k1+k2)^2

Question 8

Scattering from forbidden potential step

R =|B|^2 /|A|^2 =

Answer 8

|ik1+k2|^2 / |ik1-k2|^2 = 1

Question 9

Scattering from forbidden potential step

T = 1 - R =

Answer 9

0

Question 10

In forbidden potential steps barrier penetration occurs, but

Answer 10

can’t pass through, i.e. no flux

Question 11

skin depth

Answer 11

d = 1/2k2

Question 12

Scattering from potential well

4klA =

Answer 12

[ (k+l)^2 exp[-2ila] - (k-l)^2 exp[2ila] ] F exp[2ika]

Question 13

Resonant scattering from potential well

Limit 1: E &laquo_space;|V| => k &laquo_space;l

Answer 13

4klA ≈ l^2 (exp[-2ila] - exp[2ila]) F exp[2ika]

= -2iFl^2 sin(2la) exp[2ika]

T = |F|^2 / |A|^2 = 0

100% transmission

Question 14

Resonant scattering from potential well

Limit 2: E&raquo_space; |V| => k ≈ l

Answer 14

4k^2A ≈ (2k)^2 exp[-2ila] F exp[2ika]

=> T = | exp[2ila] exp[-2ika] |^2
= 1

Perfect transmission

Question 15

Resonant transmission when

Answer 15

2la = nπ

Question 16

Ramsauer-Townsend effect

Answer 16

energy of electrons scattering from noble gases can be tuned to values with no scattering: no classical explanation

Question 17

Scanning tunnelling microscope (STM) =

Answer 17

T = [ 1+ (V^2sinh^2(2la)) / (4E(V-E) ]^(-1)

≈ 16E(V-E) / V^2 exp[-2Ll]

Question 18

For large barrier opacity ,

Answer 18

2la = Ll&raquo_space; 1

Question 19

Alpha decay

Answer 19

is emission of He nuclei (2n2p) from atomic nuclei

Question 20

In alpha decay the nucleus is bound by

Answer 20

strong nuclear force vs coulomb repulsion of protons

strong wins at small radii, Coulomb at large.

Question 21

alpha is very tightly bound:

Answer 21

can spontaneously form by nuclear disintegration, but potential barrier to emission

Question 22

Gamow modelled barrier as a series of barriers

Answer 22

T = Πi Ti(Vi,ai)