{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions_Quiz8

# Solutions_Quiz8 - since G has to be zero no wave coming...

This preview shows pages 1–3. Sign up to view the full content.

Region I ( x < 0 ) TISE can be written ! h 2 2 m d 2 " dx 2 + U " = E " where U= 6 eV and E = 10 eV so d 2 ! dx 2 = " k 2 ! where k 2 = 2 m h 2 ( E ! U ) so solution in Region I can be written as ! I = Ae ikx + Be " ikx Region II ( x > 0 ) TISE can be written ! h 2 2 m d 2 " dx 2 + U " = E " where U= 0 so d 2 ! dx 2 = " k 1 2 ! where k 1 2 = 2 m h 2 E so solution in Region II can be written as ! II = Ce ik 1 x + De " ik 1 x = Ce ik 1 x since D must be zero (no wave coming in from large x region) At x= 0, ! I = ! II and d ! I dx = d ! II dx Thus A + B = C ikA ! ikB = ik 1 C We need only B/A and C/A Solutions are B A = k ! k 1 k + k 1 and C A = 2 k k + k 1 so T = C A 2 = 4 k 2 ( k + k 1 ) 2 = 4 (1 + k 1 / k ) 2 and R = B A 2 = ( k ! k 1 ) 2 ( k + k 1 ) 2 = (1 ! k 1 / k ) 2 (1 + k 1 / k ) 2

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

View Full Document
Now k 1 / k = E E ! U = 10 4 = 1.58 So, T = 0.6 and R = 0.05 Obviously R + T is not = 1 But (k 1 /k) T + R = 1 This expresses the fact that current conservation at the interface requires that k A 2 = k B 2 + k 1 C 2 so ( k 1 / k ) C A 2 + B A 2 = 1 2. (a) in Region I (x < 0 ) , solution of TISE for U = 0 gives ! I = Ae ikx + Be " ikx where k 2 = 2 m h 2 E similarly in region III ( x > L ) where also U = 0 ! III = Fe ikx
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: since G has to be zero ( no wave coming back from large x) In region II ( 0 < x < L ) where U = U (non-zero and assumed to be greater than E) II = Ce " x + De # x where 2 = 2 m h 2 ( U " E ) (b) Wavefunction and its derivative continuous at x=o gives A + B = C + D (1) ikA ! ikB = " C ! D (2) Same conditions at x=L gives Ce ! L + De " L = Fe ikL (3) Ce L " De " L = ikFe ikL (4) (c) We are given that α L >> 1, so we can set e ! L to zero in above equations Then the only way (3) and (4) can be simultaneously satisfied is if C = F = 0. Then Eqs. (1) and (2) become: A + B = D A – B = (i α /k)D Solution gives B A = k ! i k + i so R = B A 2 = 1...
View Full Document

{[ snackBarMessage ]}