Physics 1C Lecture 29B
"Nuclear powered vacuum cleaners will probably be a reality within 10 years. "
--Alex Lewyt, 1955
Problems with Bohr’s Model Bohr’s explanation of atomic spectra includes some
features of the currently accepted theory.
Bohr’s model includes both classical and non-classical ideas.
He applied Planck’s ideas of quantized energy levels to orbiting electrons and Einstein’s concept of the photon to determine frequencies of transitions.
Improved spectroscopic techniques, however, showed that many of the single spectral lines were actually groups of closely spaced lines.
Single spectral lines could be split into three closely spaced lines when the atom was placed in a magnetic field.
Quantum Numbers n is considered the principal quantum number.
n can range from 1 to infinity in integer steps.
But other quantum numbers were added later on to explain other subatomic effects (such as elliptical orbits; Arnold Sommerfeld (1868-1951).
The orbital quantum number, ℓ, was then introduced. ℓ can range from 0 to n-1 in integer steps.
Historically, all states with the same principal quantum number n are said to form a shell. Shells are identified by letters: K, L, M, …
All states with given values of n and ℓ are said to form a subshell. Subshells: s, p, d, f, g, h, …
Quantum
Numbers Largest angular
momentum will be
associated with:
ℓ = n – 1
This corresponds to
a circular orbit.
ℓ = 0 corresponds
to spherical
symmetry with no
axis of rotation.
Quantum Numbers
Suppose a weak magnetic field is applied to an atom and its direction coincides with z axis.
Then direction of the angular momentum vector relative to the z axis is quantized!
Lz = mℓ ħ Note that L cannot be
aligned with the z axis.
Because angular momentum is a vector, its direction
also must be specified.
Quantum Numbers States with quantum numbers
that violate the rules below cannot
exist.
They would not satisfy the
boundary conditions on the wave
function of the system.
Quantum Numbers In addition, it becomes
convenient to think of the electron as spinning as it orbits the nucleus.
There are two directions for this spin (up and down).
Another quantum number accounts for this, it is called the spin magnetic quantum number, ms.
Spin up, ms = 1/2
Spin down, ms = –1/2.
Wave Functions for Hydrogen
13
1( ) or as
o
r ea
The simplest wave function for hydrogen is the one
that describes the 1s state:
All s states are spherically symmetric.
The probability density for the 1s state is:
2 2
1 3
1or a
s
o
ea
Wave Functions for Hydrogen The radial probability density for the 1s state:
The peak at the Bohr radius indicates the most
probable location.
22
1 3
4( ) or as
o
rP r e
a
The atom has
no sharply
defined boundary.
The electron
charge is
extended through
an electron cloud.
Wave Functions for Hydrogen The electron cloud model is quite different from the
Bohr model.
The electron cloud structure does not change with time and remains the same on average.
The atom does not radiate when it is in one particular quantum state.
This removes
the problem of the
Rutherford atom.
Radiative
transition causes
the structure to
change in time.
Wave Functions for Hydrogen The next simplest wave function for hydrogen is for
the 2s state n = 2; ℓ = 0.
This radial probability
density for the 2s state
has two peaks.
In this case, the most
probable value is r 5a0.
The figure compares 1s
and 2s states. 3
22
2
1 1( ) 2
4 2
or a
s
o o
rr e
a a
Clicker Question 29B-1
Which of the following hydrogen atom series can
emit a photon with a wavelength in the infrared part
of the spectrum?
A) Lyman
B) Balmer
C) Paschen
D) All of the above series can emit a photon
with a wavelength in the infrared region
Problem
If an electron has a measured deBroglie wavelength
of 0.850x10–10m, what is its kinetic energy?
a. 55.0 eV.
b. 104 eV.
c. 147 eV.
d. 207 eV.
e. 18.8 eV.
Solution If an electron has a measured deBroglie wavelength of
0.850x10–10m, what is its kinetic energy?
What do we know?
deBroglie wavelength ->
Momentum – l = h/p
What do we want?
KE. Which is related to momentum
K.E. = (1/2)p2/m = (1/2)(h/l)2/m
= (1/2)(6.626x10-34Js/0.85x10-10m)2/(9.11x10-31kg)
x(eV/1.602x10-19J)
= 207 eV (choice d)
Electron Paramagnetic
Resonance (EPR) Apply magnetic field to
remove energy
degeneracy of spin up
and spin down electrons
When “splitting” matches
photon energy, get
absorption
Seen as a signal
Relatively rare in nature
The Nucleus All nuclei are composed of protons and neutrons
(they can also be called nucleons).
The atomic number, Z, is the number of protons in
the nucleus.
The neutron number, N, is the number of neutrons
in the nucleus.
The mass number, A, is the number of nucleons in
the nucleus (A = N + Z).
In symbol form, it is: XAZ where X is the chemical symbol of the element.
The Nucleus Isotopes of an element have the same Z but
differing N (and thus A) values.
For example,
both have 92 protons, but U-235 has 143 neutrons
while U-238 has 146 neutrons.
Nucleons and electrons are very, very small. For
this reason, it is convenient to define a new unit of
mass known as the unified mass unit, u.
Where: 1 u = 1.660559x10-27 kg (exactly 1/12 of
the mass of one atom of the isotope C-12)
92
235U
92
238U
The Nucleus Einstein’s mass-energy equivalence tells us that all
mass has an energy equivalence. This is known as
the rest energy.
This means mass can be converted to some
form(s) of useable energy.
The equation is given by: From here we get that the energy equivalent of 1u
of mass is:
ER mc2
ER 1 u c2 1.4924311010 J 931.494 MeV
This gives us the relationship between mass and
energy to also be:
1 u 931.494 MeVc2
Nuclear Stability Example
What is the binding energy (in MeV) of the
Helium-4 nucleus? The atomic mass of
Helium is 4.002602u.
Answer
We need to calculate the total rest energy of the
Helium nucleus and the total rest energy of the
separated nucleons.
We need to turn to: ER = mc2
Nuclear Stability Answer
For the Helium nucleus we find:
For the separated protons and neutrons:
ER mHec2 4.002602 u c2
ER 4.002602 u c2 931.494
MeVc2
1 u 3,728.4 MeV
ER 2mpc2 2mnc
2
ER 2 938.28 MeV
c2 c2 2 939.57 MeV
c2 c2 3,755.7 MeV
Calculate the difference:
Eb EHe Esep 3,728.4 MeV 3,755.7 MeV27.3 MeV
Ebnucleon
27.3 MeV4 6.825 MeVA
The Nucleus This table gives the following masses for the
selected particles:
Note that the neutron is slightly heavier than the
proton.
Nuclear Magnetic
Resonance Imaging Nuclei also have quantum
numbers
Detect nuclear magnetic
moments similar to NMR
Nuclear Magnetic
Resonance Imaging
magnetic moments of some of these protons change
and align with the direction of the field
radio frequency transmitter is briefly turned on,
producing a further varying electromagnetic field
photons of this field have just the right energy, known
as the resonance frequency, to be absorbed and flip
the spin of the aligned protons
Nuclear Magnetic
Resonance Angiography
generates pictures of
the arteries
administration of a
paramagnetic contrast
agent (gadolinium) or
using a technique
known as "flow-related
enhancement"
For Next Time (FNT)
Finish the homework for Chapter 29
Read Chapter 30 through page
1022
Wednesday – Overview of quarter