We’ve learned about this situation: the finite potential well… …but what if we “turn it upside down”? This is a finite potential barrier. When we solved.

We’ve learned about this situation: the finite potential well…

…but what if we “turn it upside down”?

This is a finite potential barrier.

When we solved this problem, our solutions looked like this…

I II IIIU

-L/2 L/2

E

What would you expect based on your knowledge of the finite box?

(in actuality the light field in the optically dense space is evanescent, i.e. exponentially decaying)

Below, the thick curves show the reflectance as the thickness of the low-index layer (air) changes from 10 to 900 nm. Note that as the layer thickness

increases, the reflectance becomes closer to total at 41 degrees. That is, FTR gives way to TIR.

Qualitatively:

(pure momentum states)

)()( tkxitkxiI BeAe ωω −−− +=Ψ

)()( tkxitkxiIII GeFe ωω −−− +=Ψ

to the left of the barrier

to the right of the barrier

Instructive to consider the probability of transmission and reflection…

R+T=1 of course…

2

2

*

*

incident*

reflected*

)(

)(

A

B

AA

BBR ==

ΨΨΨΨ

=

2

2

*

*

incident*

dtransmitte*

)(

)(

A

F

AA

FFT ==

ΨΨΨΨ

=

+ +

+ +

+ +

+ +

+ +

+

0

E

-U

U(x)=-exx

0

εeEx /2 −=

⎟⎠

⎞⎜⎝

⎛ −−≈ ∫ dxExUmET )(22

exp)(h

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−≈

ε1

3

24exp)(

2

3

he

EmET

Why does the half life vary more than the kinetic energy?

U(r)

r

E

R

rkZerU /2)( 2=

EkZeR /2 21 =

kinetic energy of escaping

alpha particle

⎭⎬⎫

⎩⎨⎧

+−=0

0 84exp)(r

ZR

E

EZET π

Separation of centers of alpha and nucleus at edge of barrier 9.1 fm

Height of barrier 26.4 MeV

Radius at which barrier drops to alpha energy 26.9 fm

Width of barrier seen by alpha 17.9 fm

Alpha's frequency of hitting the barrier 1.1 x 10^21/s

The photoelectric detector

Without smoke With smoke