Formula Sheet for final Exam

REVIEW MATERIAL FOR PHYSICS 250
1. Lorentz transformation:
S' moving along
+ x axis
x = γ ( x′ + cβt′ )
ct = γ ( ct′ + βx′ )
x 2 – c 2 t 2 = x′ 2 – c 2 t′ 2
2. addition of velocities
u'
u x + v
x
= --------------------------------
1 + ( u′ x v ⁄ c 2 )
1 u′
u y
-- y
= -------------------------------------
γ 1 + (( u′ x v) ⁄ c 2 )
3. Momentum and energy: p˜ = ( E ⁄ c,
p)
, E = γ m o c 2 , p = βγm o c , β
E 2 p 2 c 2 2 4
= + m o c
E 2 – p 2 c 2 = E′ 2 – p′ 2 c 2
E = γ ( E′ + βp′ x
c)
p x
c = γ ( p′ x
c + βE′ )
=
pc
-----
E
mv
=
qBR
pc( GeV) = 0.3q B( Tesla) Rm ( )
1 + β
4. Doppler Shift: f = ⎛-----------
⎞ 1 ⁄ 2
f′ , λ
⎝1 – β⎠
=
⎛1 – β
----------- ⎞ 1 ⁄ 2
λ′
⎝1 + β⎠
5. Statistical Physics
Maxwell-Boltzmann distribution
2πN
n( E)dE = g( E)f( E)dE
= ----------------------
( πkT) 3 ⁄ 2 E e – E ⁄ kT dE
n( v)dv = 4πNv 2 ⎛----------------
m ⎞ 3 2
⎝2πk B T⎠
⁄ 1
–--mv 2
⁄ k
2 B T
e
1
〈 K〉
--mv 2 3
3k
= 〈 〉 = --k , in 3D
2 2 B T v B T
rms = ------------
m
1 2 1
〈--mv 2 x
〉 = --k
2 B
T
1

6. Early Quantum Physics
7. Stefan-Boltzmann Law:
R σT 4 =
8. The Photelectric effect:
Particle Properties of Waves
hν
= --------------------------
e hν ⁄ k BT
– 1
8πhν 3 1
= ---------------
c 3 --------------------------
⁄
– 1
〈 εν ( )〉
u( ν)
hc
---------------- = 4.97
λ m
k B
T
σ = 5.67∗10 – 8 W ⁄ m 2 K 4
max
hν = φ+
T e
e hν k BT
(a) Compton scattering:
(b) Absorption of Photons:
h
λ – λ 0
= ------ ( 1 – cosθ)
mc
N( x) = N 0
e – µx
(c) Gravitational Red Shift:
gL
ν 2
= ν ⎛
1
1 + ----- ⎞
⎝ ⎠
C 2
GM
ν' = ν⎛1
– --------- ⎞
⎝
Rc 2 ⎠
Rutherford Scattering;
N( θ)
=
k 2 Z 2 e 4 Nnt
----------------------------------------
4r 2 2 4
T α sin ( θ ⁄ 2)
N = # alphas/m 2 , n = # atoms/m 3 in foil , t = thickness of foil
b
kZe 2
= ----------- cot θ ⁄ 2 r = distance of detector from foil
T α
2

Atoms and Line Spectra:
1
--
λ
f ⎛ 1 1 ⎞
= -- = R
c H ⎜----
– ----⎟
R H 109,500 cm – 1
–2
= = 1.095×10 nm – 1
⎝ ⎠
n f
2
n i
2
The Bohr model:
a.
L = mvr = mωr 2 = nh
n 2
h 2
nh
r n
= ----a ;
Z 0
a 0
= ------------
mke 2 = 0.529A v n
= --------
mr n
Z 2 mk 2 Z 2 e 4 ke 2
E n
= -----E ;
n 2 B
= –---------------------
2n 2 h 2 E B
= -------- = 13.6eV
2a 0
Wave Nature of particles:
a. p = h ⁄ λ = hk, E = hω = hf ; k = 2π ⁄ λ
b. Bragg Scattering:
2d sinθ
=
mλ
Wave Packets, Interference, and Uncertainty:
v p
a. = ω ⁄ k = phase velocity ; v g = dω ⁄ dk = group velocity
b. uncertainty principle
h
∆x∆p x
≥ --
2
h
∆E∆t
> --
2
Particle in a box: ψ( x) = 0 for x < 0orx > L
2
h 2
ψ n
( x)
= -- sin( nπx ⁄ L)
; E
L
n
------ ⎛-----
nπ ⎞ 2 n 2 h 2
= = -------------
2m⎝
L ⎠
8mL 2
Harmonic Oscillator:
E n
⎛ 1
n + -- ⎞ 1 –
= hω ; ω = C ⁄ m ; ψ0 ( x)
= ----------------------e ( 1 ⁄ 2)
( x ⁄ a)2
;
⎝ 2⎠
( πa) 1 ⁄ 2
a
⎛ h 2
-------- ⎞ 1⁄
2
= for V( x)
⎝mC⎠
=
1
--Cx 2
2
Reflectivity and Transmission probability:
R
J r
= ---- = B
--- 2
, T
A
J i
J t
= --- = ---- C Ā -- 2 = 1 – R
J i
k 2
k 1
3

Tunneling:
Schrodinger Equation in higher dimensions:
1. Operators and Expectation Values
T( E) 16 ----- E V 1
– 2k
⎛ E
– ----- ⎞
2 L
2mV ( – E)
≈ e k
0
⎝ ⎠
2
= --------------------------
V 0
h 2
p
h ∂
= -- ⎛----- ,----- ∂ ,----
∂ ⎞ p 2 h 2 ⎛
⎞
= – -------
i ⎝∂x
∂y ∂z⎠
⎜
∂x 2 + -------
∂y 2 + -------
⎝
∂z 2 ⎟
⎠
∂ 2
∂ 2
∂ 2
= –h 2 ∇ 2
h
L z
= xp y
– yp z
= -- -----
∂
i ∂φ
Two dimensional box:
E
=
h 2
------
2m
2πn
----------- x
⎝
⎛ ⎠
⎞ 2 + 2πn ----------- y
⎝
⎛ ⎠
⎞2
L x
L y
=
h 2
------
8m
n x
----
⎝
⎛ ⎠
⎞ 2 + ----
n y
⎝
⎛ ⎠
⎞2
L x
L y
ψ( xy , )
=
2
----
L x
2 πn x
x πn
---- ----------- y
y
sin sin-----------
L y
L x
L y
Central Forces:
Px ( 1
< x < x 2
, y 1
< y<
y 2
) = dx dy ψ( xy , ) 2
Spherical harmonics and total angular momentum
Y lm
( θφ , ) = Θ lm
( θ)e imφ , m = 0,±
1,±
2 ,…,±
l
∫
x 2
x 1
∫
y 2
y 1
L 2 Y lm
( θφ , ) = h 2 ll ( + 1)Y lm
( θφ , )
L z
Y lm
( θφ , ) = hm l
Y lm
( θφ , )
Probabilities:
P( θ 1 < θ< θ 2 , φ 1 < φ < φ 2 ) = dθsinθ dφ Y lm ( θφ , ) 2
∫
θ 2
θ 1
∫
φ 2
φ 1
Expectation values:
∫
〈 f ( θφ , )〉 = sinθ( dθ) dφf ( θφ , ) Y lm ( θφ , ) 2
Spherically symmetric potential:
U( r) = U( r)
ψ nlm ( r) = R nl ( r)Y lm ( θφ , )
4

∇ 2 [ Rr ( )Y lm
( θφ , )]
=
2
------- -- ∂ ll ( + 1)
----
∂r 2 + – ----------------- Rr ( )Y
r ∂r
lm
( θφ , )
∂ 2
r 2
h 2
–------
2m
d 2 R
--------
dr 2
+
2
-- dR ------
r dr
ll ( + 1)h 2
+ ----------------------
2mr 2 Rr ( ) + U( r)Rr
( ) = ER( r)
Radial probabilities and averages:
Pr ( 1 < r < r 2 ) = r 2 drRr
( ) 2
∫
r 2
r 1
Hydrogenic atoms:
∫
∞
〈 f( r)
〉 = drr 2 f( r) Rr ( ) 2
0
U( r) = – k--------
Ze2 , E
r n
= – k-------- e2 -----
Z2
2a 0
n 2
Magnetic moments:
–5
µ B
= 5.788×10 eV/T
Orbital:
E
= – µ ⋅ B µ = – µ B
( L/h)
E = µ B
m l
B
Electron Spin S:
S 2 = ss ( + 1)h , S z = m s h , s = 1 ⁄ 2,
m s = ± 1⁄
2
µ = – gµ B
( S ⁄ h) = – 2µ B
( S ⁄ h)
, E = 2µ B
m s
B
Protons and Neutrons:
µ n = eh ⁄ ( 2m p ) = 3.152×10 eV/T and s = 1 ⁄ 2
–8
, = 2g , p µ ( S ⁄ h ) , g p = 2.79 , g n = – 1.91 , ∆E = 2g pn , µ n B
µ pn
n
n
Electron and orbital spin:
E = ( m l + 2m s )µ B B
Fermi Sea: k ⎛
F 3π 2N --- ⎞ 1 ⁄ 3
π
= , 3D; k , ,
⎝ V⎠
F
-- N 2
---
h 2 k 2
= 1D E
L f = ----------
2m
n( E)dE
=
3N – 3 ⁄ 2
------- E
2 f
EdE
--------------------------------------
( – E f ) ⁄ ( kT)
+ 1
e E
5

R = R 0
A 1 ⁄ 3
T( E) = e
{– 4πZ E 0 ⁄ E + 8 ZR ⁄ r 0 }
6