Download Bohr Half Quantum Physics

Quantum Physics
Schrödinger
Quantum Physics
Particle in 1D box - Infinite Potential
Classical
p
Classical:
x0
p  mv
1 2
1
p2
2
E  mv 
(mv) 
2
2m
2m
p
xL
x
p  2mE
x
p   2mE
Bohr
Half Quantum Physics:
 pdx  nh
Bohr
n 2 2 h 2
2  L  2mE  n2h  E 
2mL2
Quantum Physics
Particle in 1D box - Infinite Potential
Scrödinger Equation

 h2 d 2
)
x
(
V


 ( x)  E ( x)

2
dx
m
2


 d2
2mEn
2
2
kn 
k

n  n ( x )  0
 2
h2

 dx
 n ( x)  Aeik x  Be ik x  C cos k n x  D sin k n x
 n ( 0)   n ( L )  0

n
n
C  0 sin k n L  0 k n  n
L

 n ( x) 
nx
2
sin
L
L
n 2 2 h 2
En 
2mL2
Quantum Physics
Particle in 1D box - Infinite Potential
Scrödinger Equation
 h2 d 2



V
(
x
)

 ( x)  E ( x)
2
2
m
dx


 d2
2mEn
2
2

k

(
x
)

0
k

n  n
n
 2
h2
 dx

 n ( x)  Aeik x  Be ik x  C cos k n x  D sin k n x
 n (0)   n ( L)  0
n
n
C  0 sin k n L  0 k n  n

L
2
nx
sin
L
L
 n ( x) 
 n ( x) 
n 2 2 h 2
En 
2mL2
2
nx
sin
L
L
 ( x, t )  e
i
En t
h
 n ( x)  e
i
En t
h
2
nx
sin
L
L
U0
Quantum Physics
Particle in 1D box - Finite Potential
Scrödinger Equation
 h2 d 2



V
(
x
)

 ( x)  E ( x)
2
2
m
dx


Inside: V(x) = 0
 d2
2mEn
2
2

k
kn 
n  n ( x )  0
 2
h2
 dx

 n ( x)  Aeikn x  Be ikn x  C cos k n x  D sin k n x
 h2 d 2



V
(
x
)

 ( x)  E ( x)
2
2
m
dx


 d2

2m(U 0  E )
2



(
x
)

0


 2

h2
 dx

 ( x)  Re x  Se x
Outside: V(x) = U0
x  0 : S  0x  L : R  0
The solutions and it’s derivative must match at the boundary
Quantum Physics
Tunneling - Def
 h2 d 2



V
(
x
)

 ( x)  E ( x)
2
2
m
dx


Outside: V(x) = 0
 d2
2mEn
2
2

k
kn 
n  n ( x )  0
 2
h2
 dx

 n ( x)  Aeikn x  Be ikn x  C cos k n x  D sin k n x
 h2 d 2



V
(
x
)

 ( x)  E ( x)
2
2
m
dx


 d2

2m(U 0  E )
2



(
x
)

0


 2

h2
 dx

Inside: V(x) = U0
 ( x)  Re x  Se x
Tunneling probability T that the particle gets through the barrier is
proportional to the square of the ratio of the amplitudes of the
sinusoidal wave function on the two sides of the barrier.
T  Ge2L
G  16

E 
E 
1 

U0  U0 
2mU 0  E 
h
Quantum Physics
Tunneling - Example - Electron
A 2.0 eV electron encounters a barrier 5.0 eV heigh.
What is the probability that it will tunnel through the barrier
if the barrier width is a) 1.00 nm b) 0.50 nm ?
T  Ge
 2L
E 
E  2
1 
e
 16
U0  U0 
2 m U 0  E 
h
L
U 0  5.0eV  5.0 1.602 10 19 J
E  2.0eV  2.0 1.602 10 19 J
m  9.1110 31 kg
h  6.626 10 34 Js
h
h
2
a)
G  16
b)
L  1.00m  1.00 10 9 m
T  7.1 10
T  Ge2L
8
L  0.50nm  0.50 10 9 m
T  5.2 10
4

E 
E 
1 

U0  U0 
2mU 0  E 
h