The_Cambridge_Handbook_of_Physics_Formulas

4.4 Hydrogenic atoms
95
Harmonic oscillator
¯h (Planck constant)/(2π)
Schrödinger
equation
− ¯h2 ∂ 2 ψ n
2m ∂x 2 + 1 2 mω2 x 2 m mass
ψ n = E n ψ n (4.67)
ψ n nth eigenfunction
x displacement
(
Energy
levels a E n = n+ 1 )
n integer ≥ 0
¯hω (4.68) ω angular frequency
2
E n total energy in nth state
ψ n = H n(x/a)exp[−x 2 /(2a 2 )]
(4.69)
Eigenfunctions
( ) H 1/2 n Hermite polynomials
(n!2 n aπ 1/2 ) 1/2
¯h
where a =
mω
Hermite
polynomials
H 0 (y)=1, H 1 (y)=2y, H 2 (y)=4y 2 −2
H n+1 (y)=2yH n (y)−2nH n−1 (y) (4.70)
y
dummy variable
4
a E 0 is the zero-point energy of the oscillator.
4.4 Hydrogenic atoms
Bohr model a
Quantisation
condition
µr 2 nΩ=n¯h (4.71)
Bohr radius a 0 = ɛ 0h 2
πm e e 2 = α ≃ 52.9pm (4.72)
4πR ∞
Orbit radius
r n = n2
Z a m e
0
µ
Total energy E n = − µe4 Z 2
8ɛ 2 0 h2 n 2 = −R ∞hc µ m e
Z 2
Fine structure
constant
α = µ 0ce 2
2h = e2
4πɛ 0¯hc ≃ 1
137
(4.73)
n 2 (4.74)
(4.75)
r n nth orbit radius
Ω orbital angular speed
n principal quantum number
(> 0)
a 0 Bohr radius
µ reduced mass (≃ m e )
−e electronic charge
Z atomic number
h Planck constant
¯h h/(2π)
E n
ɛ 0
m e
α
total energy of nth orbit
permittivity of free space
electron mass
fine structure constant
µ 0 permeability of free space
Hartree energy E H = ¯h2
m e a 2 ≃ 4.36×10 −18 J (4.76) E H Hartree energy
0
Rydberg
constant
R ∞ = m ecα 2
2h = m ee 4
8h 3 ɛ 2 0 c = E H
2hc
(4.77)
R ∞
c
Rydberg constant
speed of light
(
Rydberg’s 1 µ 1
formula b = R ∞ Z 2
λ mn m e n 2 − 1 )
λ mn photon wavelength
m 2 (4.78)
m integer >n
a Because the Bohr model is strictly a two-body problem, the equations use reduced mass, µ = m e m nuc /(m e +m nuc ) ≃ m e ,
where m nuc is the nuclear mass, throughout. The orbit radius is therefore the electron–nucleus distance.
b Wavelength of the spectral line corresponding to electron transitions between orbits m and n.