Transcript The birth of quantum mechanics

The birth of quantum mechanics
Until nearly the close of the 19th century, classical
mechanics and classical electrodynamics had been largely
successful in describing phenomena in the world. Material
particles were determinate objects that obeyed the laws of
classical mechanics. Electromagnetic waves were traveling
waves of electric and magnetic fields, in which the waves
were continuous and exhibited phenomena of interference
and refraction that could be explained from their
wavelength and frequency.
In 1890-1910, there were problems!
There were situations where electromagnetic waves exhibited
properties that should be associated with particles!
•Black body radiation
•Photoelectric effect
•Frank-Hertz experiment
•Spectra of emission and absorption by atoms
There were situations where material particles exhibited
properties that should be associated with waves!
•Electron diffraction
We will study several of these paradoxes,
and arrive at the wave-particle duality that
spawned quantum mechanics:
Light behaves like waves much of the time,
but like particles some of the time;
Material particles behave like particles much
of the time, but like waves some of the time.
A successful description of both light and
matter must somehow weave together both
kinds of properties!
First let’s review the statistical
mechanics of material particles.
(section 1.12 in the text)
Suppose we want to calculate the dependence
of the density of air as a function of altitude
on Earth.
We can get it by using the ideal gas law,
pV=NRT
and the principle of detailed balance
The ideal gas law relates pressure p (force per unit area)
to the number of moles N, the volume V, and the temperature
T. R is the universal gas constant.
For describing a gas, it is more convenient to use the number
density of particles n=NaN/V of particles:
p  nkT
where k = 1.4 x 10-23 J/oK = 10-4 eV/oK
is Boltzmann’s constant.
How does the atmosphere vary as a function of altitude?
1) A molecule must have work done upon it to elevate by a
distance dz:
F  mg
dE  mg dz
2) The density of air decreases with altitude: n(z)
Detailed Balance
We require that the force on each particle of gas be in
balance (otherwise it would rise or fall).
Consider a horizontal slice of the atmosphere at altitude z:
face area A, thickness dz, mass of each particle m
Downward force due to gravity pulling on each particle:
Fdown  Mg  nz  A dz mg
Upward force due to the pressure difference between the
top and bottom of the slice:
Fup  ( pbottom  ptop ) A  dp( z )  A
Now require Fup = Fdown in each slice of the sky:
n( z ) mg dz  dp ( z )
Now connect n(z) and p(z) through the ideal gas law:
p ( z )  n( z )kT
So the ideal gas law becomes
Solution:
mg
dn( z )  
n( z )dz
kT
n( z )  n0 e  mgz / kT
The air density decreases exponentially, with scale length
(1.4 1023 J / K )(300K )
L  kT / mg 
 8,400m
2
 27
(9.8m / s )(30 1.7 10 kg)
This result is easily generalized to any situation
where material particles are distributed in a

volume of space where the potential energy U (r )
that varies over the region:


U ( r ) / kT
n(r )  n0 e
This is the Boltzmann distribution function. It
governs the distribution of particles in the
presence of any interaction potential.
Black Body Radiation
When a material body is heated, it emits
electromagnetic radiation with a broad
spectrum.
Mystery #1. It is observed experimentally that
the total intensity (power per unit area) radiated
by a black body is determined solely by its
absolute temperature. There is no way to
explain this result by treating light as a wave!
Stephan-Boltzmann law
I  T
4
 = 5.7 x 10-8 W/m2/oK4
Example: A steel rod is heated red hot (T ~ 700 oC ~1,000 oK).
The rod is 1 cm diameter and 1 m long. How much power does
it radiate as blackbody radiation?
P =  T4 A = (5.7 x 10-8 W/m2/K4)(1,000 K)4 ( x 10-2m)(1 m)
= 1,800 W.
Mystery #2. The spectrum of light from
blackbody radiation cannot be explained by
assuming that the light is composed of waves.
The spectrum of light is the pattern of intensity as
a function of wavelength:
The spectrum of black-body radiation can be explained
(up to a point!) if we consider the radiation to be produced
by oscillations of the atoms in the material.
• The light emitted should be proportional to the
number of modes in which the oscillations of a
given wavelength can be excited:
Consider the modes that can be
excited within a cubic cavity of
dimension L.
#modes in x: Nx = L/l
#modes in y: Ny = L/l
#modes in z: Nz = L/l
The power radiated in wavelength interval dl is proportional
to the fraction of a wavelength: dP ~ dl/l
So the energy density  (J/m3) is
 (l )dl  N x N y N z
 (l)dλ  8
dl
l
4
dl
l

dl
l
4
in a rectangula r geometry;
in a spherical geometry
This energy spectrum was derived by Rayleigh and Jeans by
assuming that blackbody radiation is emitted from atomic
oscillators as a wave process, and that there must be detailed
balance between the standing waves that can be supported
inside the solid and the emitted radiation that comes out.
This Rayleigh-Jeans theory is not too bad in its description of
the spectrum of long-wavelength light.
Unfortunately, it leads to an ultraviolet catastrophe:
The power radiated at short wavelength (high energy)
increases without bound!
High-frequency (low-l) cutoff
a / l
requires e
factor
1 / l4
Plank realized that he could (empirically)
obtain the observed spectrum IF he assumed
that blackbody radiation behaved as if it were
emitted by oscillators that could only change
energy by integer multiples of some minimum
energy step u: E = mu
Then he would have from the Boltzmann
distribution:
n(u )  n0 e
 mu / kT
The total energy of all the oscillators emitting
this particular energy E=mu is then
mu n(u )  mu n0 e  mu / kT
Following the derivation in the book, we calculate the
average energy w of an oscillator:

w
 mu / kT
mu
n
e
 0
m 0

n e
m 0
0
 mu / kT

u
eu / kT  1
Put this together with the Rayleigh-Jeans result
for the number of oscillators of wavelength l:
 ( l ) dl 
8
l e
4
u
u / kT
1
dl
This spectrum matched experimental observation only if
the energy u were inversely proportional to wavelength l:
u
hc
l
 h
h = 6.6 x 10-34 J s = 2,000(2) eV Å/c
This only makes sense if light is emitted in quantized packets:
it behaves like a particle when it is emitted!
Example: We can estimate the temperature at the
surface of a star by determining the wavelength
corresponding to the maximum intensity in its
spectrum, and assume that the emission is a
blackbody spectrum. This wavelength is 550 nm
(red) for the Sun, 430 nm (blue) for the North Star,
and 290 nm (ultraviolet) for Sirius. Calculate the
surface temperatures.
8hcl5
 (l )  hc / lkT
e
1
d
8hc
 6 hc / lkT
dl l (e
 1) 2
hc hc / lkT 

hc / lkT
 1) 
e
0
 5(e

kTl


hc

lkT
 5(1  e  )    0

5
 1  e 
This is a transcendental equation. We must obtain an
approximate solution. The solution will be near  = 5.
hc = 2000 eV Å = 200(2) eV nm
hc
200  2 eV nm 2.5x106 K
T


4
5kl 5  (10 eV / K )l
lnm 
Sun
lpk = 550 nm
T = 4,600 oK
North Star
lpk = 430 nm
T = 5,800 oK
Sirius
lpk = 290 nm
T = 8,700 oK