Download Ch. 30 The Nature of the Atom

Ch. 30 The Nature of the Atom
30.1 Rutherford Scattering
We know there are two types of charge, + and -, and that matter is composed
of both.
English physicist J.J. Thompson discovered the electron and measured its
mass in
i 1897.
1897
He then suggested the Plum-Pudding Model of the atom:
Negative
egat e e
electrons
ect o s were
e e scatte
scattered
ed like
ep
plums
u s in a pudd
pudding
go
of pos
positive
t ec
charge
a ge
that was equally distributed throughout the atom
electrons
To this point, no one had detected the positively
positivelycharged particles in an atom, but since they knew
what the mass of the electron was, they knew
most of the atom’s mass was due to the positive
charges.
h
Smearedout positive
charge
Then Rutherford comes along (a New Zealander), working in London in 1907.
He had
H
h d been
b
studying
t d i with
ith Marie
M i and
d Pierre
Pi
Curie
C i in
i Montreal
M t
l who
h discovered
di
d
radioactivity.
They observed two types of radiation. Rutherford called them α and β.
Rutherford soon realized that β-radiation or β-rays were really just electrons.
But α-particles were much heavier and positively charged.
charged
(α-particles are really the nucleus of a helium atom. They consist of 2 protons and 2 neutrons.
No one knew this at the time.)
Rutherford
R
th f d decided
d id d to
t use the
th α-particles
ti l to
t investigate
i
ti t the
th internal
i t
l structure
t
t
off
atoms.
He shot the α-particles at a thin film of gold:
If Thompson’s model of the atom was
correct, very little should happen to the αparticles. They
p
y should basically
y travel
straight thru the film.
Most of them did just that, but some of the aparticles were deflected at large angles
angles. In
some cases almost 180o!
p
y shocking!
g Rutherford would later say
y that, “It was as if
This was completely
we were firing shells at tissue paper and having them bounce straight back at
you!”
So how could these large deflections occur?
Really only one solution presented itself, and that was to have all the positive
charges in the atom concentrated in a very small region – called the atomic
nucleus.
The nucleus was much smaller than the entire atom, but it accounted for
almost all of the atom’s mass.
Rutherford named the positively
positively-charged
charged particles in the nucleus Protons.
Protons
This led to the model of the atom that is still widely accepted today – the
nuclear model (planetary model).
(The model is qualitatively accurate, although the electrons
do not orbit the nucleus in pretty, constant, circular orbits.
Th actually
They
t ll exist
i t in
i a cloud
l d off probability.)
b bilit )
An atom is mostly empty space!!!
Nuclear radii ~1 × 10-15 m
Electron orbit ~1
1 × 10-10 m
But, there are problems with the Planetary Model of the Atom:
Consider an electron orbiting some nucleus at a radius r:
The electron has both KE and EPE.
r
Nucleus
Electron
v
me v 2
And…..it is accelerating: a =
r
But we know from previous chapters that accelerating
charges produce electromagnetic waves, and those
waves carry energy.
This carries some of the electron’s energy away, it slows down, its orbit
decreases, since opposite charges attract, and the electron spirals into the
nucleus……..The atom collapses!
The planetary model is not stable based on classical electrodynamics!
We’ll come back to this soon, but first let’s talk about line spectra.
p
30.2 Line Spectra
Remember, all objects emit EM radiation. Hot filaments in incandescent light
bulbs emit a continuous range of wavelengths, some of which are in the visible
spectrum.
This range of different wavelengths results from the electrons being “excited” in
all the different atoms that make up the solid.
In contrast to this, single atoms only emit radiation at certain wavelengths.
Low-pressure gases behave this way, for example, and only emit certain
wavelengths of light when exited.
And, we can use a diffraction grating to separate these wavelengths and produce
a characteristic line spectra for those atoms.
Take mercury (Hg) for example:
This is mercury’s line spectra.
Hg source
Hydrogen gas, being the simplest
element, also has the simplest line
spectra.
spectra
Diffraction grating
It forms 3 groups of lines, called the Lyman,
Balmer, and Paschen Series.
Only the Balmer series lies in the visible spectrum.
Balmer series for
hydrogen gas.
Lyman, Balmer
Lyman
Balmer, and Paschen
discovered relationships
between the different
wavelengths of light in the
line spectra.
Lyman
y
Series:
⎛1 1 ⎞
= R⎜ 2 − 2 ⎟, n = 2,3,4,...
, , ,
λ
⎝1 n ⎠
Balmer Series:
⎛ 1 1 ⎞
= R⎜ 2 − 2 ⎟, n = 3,4,5,...
λ
⎝2 n ⎠
Paschen Series:
⎛1 1 ⎞
= R⎜ 2 − 2 ⎟, n = 4,5,6,...
λ
⎝3 n ⎠
1
1
1
R = Rydberg Constant = 1.097 ×107 m -1
Since only the Balmer series is in the visible spectrum, what would be the
longest and shortest wavelengths we can see from hydrogen emission?
The longest would occur for n = 3, and the shortest will occur for n = ∞.
For n = 3, the Balmer series is:
1
⎛ 1 1⎞
⎛1 1⎞
= R⎜ 2 − 2 ⎟ ⇒ = R⎜ − ⎟
λ
λ
⎝2 3 ⎠
⎝4 9⎠
1
7.2
⇒λ =
= 656 nm
R
For n = ∞, the Balmer series is:
Red part of visible spectrum
1 ⎞
⎛ 1
= R⎜ 2 − 2 ⎟
λ
⎝2 ∞ ⎠
1
4
⇒ λ = = 365 nm
R
⎛1⎞
⇒ = R⎜ ⎟
λ
⎝4⎠
1
UViolet p
part of visible spectrum
p
D
30.3 Bohr Model of the Atom
Niels
Ni
l B
Bohr
h di
died
d in
i 1962.
1962 He
H was a native
ti off Denmark
D
k who
h wentt to
t England
E l d
on scholarship to work with J.J. Thompson. He and Thompson didn’t get
along, so Bohr went to work for Rutherford.
He worked
H
k d on th
the planetary
l
t
model
d l and
d tried
t i d to
t understand
d
t d how
h
matter
tt
could be stable if electrons are accelerating around the nucleus.
Bohr started with hydrogen, since it is the simplest of all elements. We
have just 1 electron orbiting 1 proton.
proton
Hydrogen atom
r
Proton
Electron
v
Like Einstein did with the photoelectric effect, Bohr adopted Planck’s idea
of quantized energy levels, and assumed that the electron in the hydrogen
atom could only have certain values of energy, i.e. they are quantized.
Each of these energy levels corresponds to a unique orbit that the electron
moves in around the proton. The larger the orbit, the larger the energy.
So here are Bohr’s assumptions:
1. Electrons travel in fixed orbits around the proton, each orbit being
defined by a unique radius and energy. These orbits are called stationary
orbits or states, and while in these orbits, the electrons do not emit
radiation.
How can we have radiationless orbits??? This goes against classical
physics!!!!!
Furthermore, we know that all matter absorbs and emits radiation, so
how do Bohr’s atoms do this??
To explain this,
this Bohr incorporated Einstein’s
Einstein s photon model:
2. An electron in an atom emits radiation (emits photons) when and only
when it moves from a higher energy level to a lower one
one, ii.e.
e it moves from
a larger radius orbit to a smaller one.
As the electron makes a transition from a higher
energy level to a lower one, it emits a photon.
By conservation of energy, the energy of the photon
must be equal to the difference in the energies
between the two stationary states Ei and Ef:
Photon Energy = Ei − E f ⇒
hf = Ei − E f =
hc
λ
Remember,
R
b this
hi is
i all
ll theoretical,
h
i l and
d Bohr
B h could
ld use this
hi equation
i to
calculate the wavelengths (colors) of light emitted by a hydrogen atom.
But first, he needed a way to calculate the energies, Ei and Ef.
To do this, he ends up using both classical and quantum physics.
Question 30 - 1
An electron sitting in the n = 2 energy level of a hydrogen
atom has an energy of -3.4
3.4 eV. This electron makes a
transition to the n=1 state by emitting a photon. The n=1
state has an energy of -13.6 eV. What is the energy of the
emitted photon?
1. 10.2 eV
2. 17.0 eV
3. -10.2 eV
4. -17.0 eV
Photon Energy = Ei − E f
= −3.4 eV − (−13.6 eV) = 10.2 eV
30.3 (continued) The Bohr Model of the Atom
hf = Ei − E f =
So Bohr came up
p with
ith the follo
following
ing expression:
e pression
hc
λ
In order to predict the wavelengths of light emitted by hydrogen, he first
had to come up with an expression for the energy levels.
c
Here is his simple model, where Z represents the # of
protons in the nucleus.
nucleus
There is an electrostatic force – attractive Coulomb
force – between the proton and electron.
This is directed in toward the center of motion and thus
provides the centripetal force.
Q1Q2
ETot = KE + EPE = mv + k
r
2
Ze
2
Here, Q1 = + Ze, and Q 2 = −e , so:
ETot = 12 mv − k
r
mv 2
Ze 2
But from forces, we know that: Fc = FE ⇒
=k 2
r
r
2
1
2
Solve for mv and plug into the energy equation:
2
Ze 2
E = −k
2r
This is a classical result for the total energy in a hydrogen
atom. It is negative, because the magnitude of the negative
EPE is
i larger
l
than
th the
th KE.
KE
But what about the radius? Here’s where the quantum physics comes in.
r not only defines a unique orbit,
orbit but also a unique angular momentum.
momentum
Remember from Physics 1, L = Iω, where I = mr2 for the electron, since it is
a point particle, and ω = v/r.
Thus,
L = (mr )( vr ) = mvr
2
Angular Momentum
This is the classical result for angular momentum.
But since Bohr assumed the energy of the electron in the hydrogen atom
was quantized, that meant r was quantized, which means the angular
momentum must be quantized.
Thus, L can only assume discreet values of angular momentum.
But what would those discreet values be???? A clue is g
given if we look
at the units of angular momentum: ⎡ kg ⋅ m 2 ⎤
⎢
⎣
s
⎥
⎦
What else has these units??? Well, oddly enough, look at h:
⎡ kg ⋅ m 2 ⋅ s ⎤ ⎡ kg ⋅ m 2 ⎤
h [J ⋅ s] = [N ⋅ m ⋅ s] = ⎢
⎥=⎢
⎥
2
s
s
⎣
⎦ ⎣
⎦
Thus, h has the same units as angular momentum!!!
In other words, h is the fundamental quantum value of angular momentum.
This led to Bohr’s third assumption:
Ln = mvn rn = nh, n = 1,2,3,...
In the quantum world, angular momentum comes in little packets of h.
Solve the above for v and plug into the force equation:
2
⎛
⎞ n2
h2
n
−11
⎟
=
5
.
29
×
10
m
, n = 1,2,3,...
rn = ⎜⎜ 2
2 ⎟
Z
⎝ 4π mke ⎠ Z
(
)
So here is the equation that gives us the different quantized orbits.
For the hydrogen atom, Z = 1, and for the lowest quantized orbit (n = 1) we find that:
r1 = 5.29 ×10 −11 m
This is the first Bohr radius for hydrogen.
Remember, each n-level will correspond to a
different Bohr radius.
Now that we have our expression for r, we can
plug it into our equation for the total energy:
⎛ 2π 2 mkk 2 e 4 ⎞ Z 2
⎟⎟ 2 , n = 1,2,3,...
En = −⎜⎜
2
h
⎠n
⎝
This represents the quantized Bohr energy levels of the hydrogen atom.
The stuff in parentheses is just a constant:
2
Z
En = −(2.18 × 10 −18 J) 2
n
Z2
En = −(13.6 eV) 2
n
Bohr energy levels in Joules.
Bohr energy levels in electron volts.
Again for hydrogen, Z = 1, and n = 1:
E1 = −13.6 eV
This is the energy
gy of the electron in the first Bohr level. It
is also called the ground-state energy.
Notice, because of the 1/n2 dependence, the spacing
between higher and higher energy levels decreases.
So, if I add 13.6 eV to an electron in the ground state of a
hydrogen atom, then it will have zero total energy and no
longer be bound to the proton → The electron is removed.
This process is called ionization.
So the Bohr theory predicts that the first ionization energy
of hydrogen is 13.6
13 6 eV.
eV This was in excellent agreement
with experiment.
So clearly
S
l
l he
h was on the
th right
i ht track,
t
k but
b t did his
hi theory
th
predict the correct emission wavelengths for hydrogen???
From his second assumption:
⎞
⎛ 2π 2 mk 2 e 4 ⎞ 2 ⎛ 1
1
⎟⎟( Z )⎜ 2 − 2 ⎟, n f , ni = 1,2,3,...
= Ei − E f ⇒ = ⎜⎜
3
⎜n
⎟
λ ⎝ hc ⎠
n
λ
f
i
ni > n f
⎝
⎠
1
hc
⎛ 1
⎞
1
⇒ (Constant)⎜ 2 − 2 ⎟
⎟
⎜n
n
i ⎠
⎝ f
So what is this constant:
⎛ 2π 2 mk 2 e 4 ⎞
7
−1
⎜⎜
⎟
=
1
.
097
×
10
m
=R
3
⎟
hc
⎝
⎠
⎛ 1
⎞
1
= R⎜ 2 − 2 ⎟, ni = 2,3,4,...
⎜n
⎟
n
λ
f
i
⎝
⎠
1
⎛ 2π 2 mk 2 e 4 ⎞
⎟⎟
⎜⎜
3
hc
⎝
⎠
Holy crap!!! This is just:
Lyman Series!!!
Bohr’s theory reproduces the experimental result!!!
So the Lyman series consists of transitions to the ground state (n = 1) starting
with the first excited state (n = 2):
The Balmer series consists of transitions to the first
excited state (n = 2) starting with the 2nd excited
state (n = 3).
The wavelengths we observe from hydrogen are
called emission lines.
Atoms can also absorb photons (absorption lines).
If a photon is incident on an atom and has an
gy equal
q
to the energy
gy difference between two
energy
quantum energy levels of the atom, then the
electron can get “bumped up” to that higher energy
level, called an excited state.
Typically though, the lifetime of the electron in the
excited state is finite and rather short. It quickly
drops back down to a lower energy state and emits
a photon whose energy is equal to the difference in
the energies of the two levels.
30.4 deBroglie’s Explanation of Bohr’s Angular Momentum Assumption
Based on Bohr’s work, deBroglie pictured the electron orbiting the proton as
a particle-wave.
For certain wavelengths, the electron will form standing waves around the
nucleus:
A standing
t di wave occurs when
h the
th distance
di t
the
th wave
travels is an integral number of its wavelength.
For this condition,
But,
h
h
λ= =
p mv
Plug in above:
⇒ L = nh
2πr = nλ , n = 1,2,3,,...
nh
nh
2πr =
⇒ mvr =
mv
2π
Which is just the Bohr assumption on angular momentum!!!
So Bohr’s assumption of quantized angular momentum is exactly equivalent to a
standing-wave condition for electron waves around the nucleus.
I believe de Broglie’s hypothesis is the first feeble ray of light on this worst of our
physical enigmas. It may look crazy, but it is really sound!
-Albert Einstein
30.5 The Quantum-Mechanical Description of the Atom
Considering the scope of the problem, Bohr’s achievements with such a simple
model were remarkable.
However, while the model worked well for simple hydrogen – one electron
However
orbiting one proton – it failed for more complicated atoms.
The electron states in the hydrogen atom were denoted by one unique number, n.
But, iit was llater determined
B
d
i d that
h for
f more complicated
li
d atoms (and
( d hydrogen
h d
too))
quantum mechanics would rely on 4 quantum numbers to describe the electron
states in an atom.
1. Principle Quantum #, n: Similar to the Bohr model. It gives the total energy of
the atom.
2. Orbital Quantum #, l: This # determines the angular momentum of the orbit.
The values
al es of l depend on n: l = 0,1,2,…,(n-1).
012
( 1) Notice
Notice, there are n values
al es of l.
l
Example: What would be the values of l for n = 3?
l = 0,1, and 2
The magnitude of L (the angular momentum) of the electron is then given by:
L = l (l + 1) h
3. Magnetic Quantum #, ml: This becomes important when we place the atom in a
magnetic field
field. The field alters the electron energy levels
levels. This is known as the
Zeeman Effect. If B = 0, then ml plays no role. The value of ml depends on l:
ml = -l, -l+1,…,-1,0,1,…+l, in integer steps.
Notice, there are (2l+1) total values of ml.
If an atom is placed in a magnetic field pointing in the z-direction, then the
p
of the angular
g
momentum which would arise along
g the z-direction
component
due to the Zeeman effect is:
L =mh
z
l
Example:
p
What would be the values of ml if ll=3?
3?
ml =-3, -2, -1, 0, 1, 2, and 3
y of spin. The orbital
3. Spin Quantum #, ms: An electron has the intrinsic property
and spin motions of the electron combine to produce magnetism in materials.
The spin quantum # can only take two possible values:
1
s
2
m =±
The ms = +1/2 is often called the “spin-up” state, and ms = -1/2 is referred to
as the “spin-down” state.
Spin, like mass and charge, is an intrinsic property of all electrons. Each
electron has the same spin.
Example:
Solution:
Determine completely the number of electronic states in a hydrogen atom
for n = 2.
l = 0 → ml = 0, ms = ± 12
n = 2 → l = 0,1
l = 1 → ml = −1,0,1
ml = −1, ms = ± 12
ml = 0, ms = ± 12
ml = 1, ms = ± 12
Th
Thus,
there
th
are 8 total
t t l states
t t (i
(in general,
l the
th total
t t l # off states
t t is
i (2n
(2 2).
)
n
l
1.
ml
ms
2
0
0
+1/2
2
2.
2
0
0
-1/2
1/2
3.
2
1
-1
+1/2
4.
2
1
-1
-1/2
5.
2
1
0
+1/2
6.
2
1
0
-1/2
7.
2
1
1
+1/2
8.
2
1
1
-1/2
According to the Bohr model, the nth energy level is defined by the orbit whose
radius is rn. Each time we make a measurement of the position of the electron, it is
always in a circular orbit at a distance rn from the nucleus
nucleus.
This picture is not correct. The quantum mechanical picture dictates that due to
the HUP, the position of the electron is uncertain. If we were to take many
measurements of the electron’s
electron s position
position, sometimes it would be very close to the
nucleus, and sometimes it would be far away.
So the electron exists as a particle-wave in a probability cloud around the
nucleus.
nucleus
If the electron is in the n=1 state, then after many
measurements, the most probable distance for the
electron would be at r1 = 5.29×10-11 m.
The higher-density cloud regions indicate areas of
higher probability.
There are different probability clouds for different
quantum
t
states.
t t
Warm-up question
Who is the cat in the cartoon?
1.
2.
3.
4.
Felix
Garfield
Socola
Heathcliff
Clicker Question 30 - 3
For the hydrogen atom, how many total electron states
e ist for n = 3?
exist
1.
2.
3.
4.
3
6
9
18
Clicker Question 30 -2
1.
2.
3.
4
4.
Shown below is the probability cloud for one of the
q ant m states with
quantum
ith n = 2.
2 At what
hat location would
o ld you
o
most likely find the electron in this state?
A
B
C
D
B
C
A
D
This l =1 state is a p-orbital which has a
dumbbell shape.
Don’t be a p-orbital!
30.6 The Pauli Exclusion Principle (PEP)
The Bohr model was derived explicitly for one electron and one proton. Will the
energy levels apply to atoms with more than one electron?
The answer is NO!
Once you add more than one electron to an atom, the electrons interact with
each other due to Coulombic repulsion
repulsion. This results in additional energy terms
in En.
Quantum mechanics does describe the energy levels correctly in multi-electron
atoms and it uses the 4 quantum #’s:
atoms,
# s: n,
n l,
l ml, and ms.
The energy of the electrons depends on n and l, and in general, the energy
increases as n and l increase, but there are exceptions.
So how do we categorize the energy levels of electrons
in multi-electron atoms?
Each value of n corresponds to a single “shell”
shell , called
the K-shell for n = 1, L-shell for n = 2, and the M-shell for
n =3.
Each shell has subshells defined by the orbital quantum
#, l.
For example, the K-shell is the n =1, and it has one subshell, l = 0.
The L-shell is the n = 2, and it has two subshells defined by l = 0,1
When atoms are not subjected to violent collisions, high temperature, or high
electric fields, characteristics of say room temperature, then the electrons in
the atoms tend to spend most of their time in the low energy
gy levels of the
atom. The lowest energy state is called the ground state.
However, when a multi-electron atom is in its ground state, not every electron
can be in the n = 1 level. They
y must obey
y a rule known as the Pauli Exclusion
Principle (PEP).
No two electrons in an atom can have the same set of
values for their 4 quantum numbers: n, l, ml, and ms. In
other
th words,
d no two
t
electrons
l t
can be
b in
i the
th same
quantum state.
Because of the PEP, there is a maximum # of electrons that can fit into each
energy level and subshell:
The Periodic Table
30.6 (cont.) The PEP
Because of the PEP, there is a maximum # of electrons that can fit into each
energy level and subshell:
Example:
Take n = 1: then l = 0, ml = 0, and ms = ±1/2.
Thus, each l = 0 subshell can hold a maximum of 2 electrons, one with spin-up
and one with spin-down.
Again I get two
states for l = 0.
Now take n = 2:
n = 2 → l = 0,1
l = 0 → ml = 0, ms = ± 12
l = 1 → ml = −1,0,1
ml = −1, ms = ± 12
ml = 0, ms = ± 12
ml = 1, ms = ± 12
Thus we get 8 total states for n = 2
Thus,
2.
In general the # of possible states for any given subshell l, is:
We get 6 states for
the l = 1 subshell.
2(2l + 1)
Representation of the l-subshell is customarily done by letter:
l
Letter Designation
Max. # of e-s: 2(2l+1)
0
s
2
1
p
6
2
d
10
3
f
14
4
g
18
5
h
22
In general, the maximum # of electrons in any shell (n) is 2n2:
K-shell, n = 1
→
2 electrons
L-shell, n = 2
→
8 electrons
M-shell, n =3
→
18 electrons
etc….
So here is an example of the spectroscopic notation used to specify the
electronic states in multi-electron atoms:
n =3
3p
4
# of e-s in
the subshell
l =1
1
Specifying the n and l value for each electron in an atom in this way is called the
electronic configuration.
Let’s Play!!!
Example:
Write down the full electronic configuration for carbon.
Carbon’s atomic # is 6, so it has 6 electrons to put into shells:
What order do the shells fill up???
It is given by the following construction:
The order is shown by diagonal arrows
starting in the upper right and pointing
toward the lower left:
1s
2s
2p
3s
3p 3d
4s
4p 4d 4f
5s
5p 5d 5f
5g
6s
6p 6d 6f
6g
etc.
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s etc. …
So we start filling in these shells with carbon’s 6 electrons, keeping in mind
th t s’’s hold
that
h ld 2 electrons,
l t
p’’s hold
h ld 6,
6 d’’s hold
h ld 10,
10 etc.
t
So for carbon, we find the electronic configuration to be:
1s2 2s2 2p2
Example:
Write down the full electronic configuration for calcium.
Calcium’s atomic number is 20, so we have 20 electrons to put into shells:
11s2 22s2 22p6 33s2 33p6 44s2
Elements in the Periodic Table are arranged by their electronic configurations:
It’s the outermost electrons in each
atom (valence electrons) that
determines the element’s
element s physical
properties.
The original periodic table was
arranged by a Russian chemist
chemist, Dmitri
Mendeleev, during the late 1800’s.
He arranged elements with similar
properties into col
columns
mns → they
the ha
have
e
the same # of valence electrons.
So Quantum
S
Q
t
Mechanics
M h i and
d the
th Pauli
P li
Exclusion Principle provide a natural
explanation for why the elements are
arranged as they are in the periodic
table.
30.7 X-Rays
Wilhelm Roentgen was a Dutch physicist working with cathode ray tubes
back in the late 1800’s.
The cathode ray tube looked essentially like this:
Electrons
El
t
gett accelerated
l
t d through
th
h a large
l
voltage and then collide with a metal target,
such as Cu, Mo, or Pd.
When the
Wh
th electrons
l t
collided
llid d with
ith the
th target,
t
t
Roentgen noticed certain “rays” were
produced.
He found that they could travel thru certain
materials that were opaque to light.
He thought they might be charged particles of
some kind
kind, but they didn’t
didn t deflect in a magnetic
field.
Since he couldn’t figure out what they were, he
called them “X-Rays”
X-Rays .
We now know that X-Rays are EM radiation –
light in the non-visible part of the spectrum.
Since they are EM radiation, they must be produced when charges are being
accelerated. This occurs when the electrons slow down upon collision with
the metal target.
If we measure the intensity of the X-Rays as a function of wavelength, we get
the following characteristic plot:
As the electrons slow down (decelerate)
(
)
upon striking the metal target, the broad
background part of the curve, called
Bremsstrahlung radiation, is produced.
Bremsstrahlung comes from the German
word for “braking”.
But where do those large peaks in the
intensity spectrum come from???
They depend on the material of the target,
and are called Kα and Kβ.
They are labeled “K” because they involve electrons in the n = 1 (K-shell).
As the electrons strike the atoms in the metal target, they slow down. But, if they
ha e enough
have
eno gh energy,
energ the
they can knock an electron in the atom o
outt of the K-shell.
K shell
Then another electron in the atom sitting in a higher energy state falls down to
replace it. When it makes this transition, a photon is emitted with X-Ray energies.
Clicker Question 30 - 3
Which characteristic X-Ray line, the Kα or Kβ, has the larger
energy?
?
Th
ey
ha
ve
th
e
sa
m
e
...
K
b
33% 33% 33%
K
a
1. Kα
1
2. Kβ
3 They have the
3.
same energy.
The energy of the incident electrons must be large if they are to access the Kshell of the atoms in the metal.
The energies of those levels is ~45
45,000
000 eV!
The Kα line occurs when an n = 2 electron falls back into the K-shell.
The Kβ line occurs when an n =3
3 electron falls back into the K
K-shell.
shell.
Ch t 31 – Nuclear
Chapter
N l
Ph
Physics
i and
d Radioactivity
R di
ti it
31.1 Nuclear Structure
We’ve been studying atomic structure – the Rutherford model of the atom,
where electrons orbit around a nucleus.
And now we know that it takes Quantum Mechanics to correctly describe the
nature of these orbits.
But what about the nucleus?
Well, so far we just know about protons.
But, in 1932 scattering experiments by English physicist James Chadwick
discovered another particle in the nucleus which had no charge → neutron.
So the nucleus is composed of positive protons and neutral neutrons.
These are called nucleons.
What distinguishes different elements in the periodic table is the # of protons
they have in their nucleus. This is called the Atomic Number (Z).
Each element also has an Atomic Weight or Atomic Mass Number (A).
Let N be the # of neutrons in the nucleus, then: A = Z + N
Take oxygen, for example:
Atomic number
Atomic mass number
8
15.9994
-2
2
MP
90.18
BP
50.15
1.429
D
Density
it
O
Symbol
1s22s22p4
O
Oxygen
Name
Electronic configuration
So oxygen has 8 protons
and N = A – Z = 16 – 8 = 8
8 Neutrons!
Shorthand notation for element representation:
So oxygen would be:
16
8
A
Z
X
O
The mass of the elements is usually given in atomic mass units (u).
1u = 1.6605 ×10-27 kg
Nuclei that contain the same number of protons but a different number of
neutrons are called isotopes.
For example, boron exists in nature as two stable isotopes:
11
5
B
10
5
B
Most boron atoms have 6 neutrons (81.1%), but some (18.9%) have only 5
neutrons.
The atomic mass number (A) listed on the periodic table is the average atomic
mass of the isotopes, for boron this is 10.811.
30.2 Strong Nuclear Force and Stability
So what does the nucleus look like???
The protons and neutrons are clustered together in a blob that is approximately
spherical. The radius of the nucleus is ~ 1 fm = 1 × 10-15 m.
Experiments show that:
r ≈ (1.2 ×10 −15 m) A1/ 3
The Nucleon Density, which is the # of protons
and neutrons per unit volume, is the same for
all atoms!
So we have this spherical blob of positive
protons and neutral neutrons. How does it
stay together???
For hydrogen, there’s no problem. It’s just one
proton.
B t what
But
h t about
b t the
th nextt simplest
i l t element,
l
t helium
h li
(He)?
(H )?
Now I have 2 protons and 2 neutrons:
+
+
He nucleus, 2 protons and 2 neutrons.
N
Now
Ih
have 2 protons
t
very close
l
together.
t
th
We know that like charges repel, and we can calculate the repulsive force
between the two protons using Coulomb’s Law.
Assuming the charges are separated by 1 fm, we get a force of 231 N. This
results in an acceleration of the protons = 1.4 × 1029 m/s2. HUGE!!!
So why don’t the protons, and thus the nucleus, just fly apart???
There must be an attractive force holding them together, and it’s not gravity,
because the gravitational attraction between to subatomic particles is very
small: ~1.9 × 10-34 N!
So, we need a new force. It’s the Strong Nuclear Force.
It is one of the fundamental forces, along with gravity and the electroweak force.
Characteristics of the strong nuclear force:
1. It is independent of charge. Thus, the attractive force between 2 protons is
the same as that between 2 neutrons or between a neutron and a proton.
2. The range of the force is extremely short. It is very strong for distances ~1 fm
and essentially zero at farther distances.
Most nuclei listed in the periodic table are stable, but some are not.
The stability of a nucleus depends on the balancing between the electrostatic
repulsion between protons and the strong nuclear attractive force between all
nucleons.
Every proton in the nucleus feels Coulombic
repulsion from every other proton, since the electrostatic force is long-ranged.
But each p
proton and neutron
only feels the strong nuclear
force from its closest
neighbors.
To compensate for this, the
more protons I add to the
nucleus, an even greater
number of neutrons must be
added to try and balance the
electrostatic force.
g and larger
g nuclei, the neutron
As I create larger
number keeps getting bigger and bigger.
Notice how the neutron # deviates from the N = Z line
for large nuclei.
Eventually, as more and more protons are added, no # of extra neutrons can
compensate for the large electrostatic repulsion, and the nucleus breaks apart.
This occurs at Z = 83
83, which is bismuth (Bi).
(Bi)
Any element with an atomic number Z > 83 will be unstable and break apart over
time. It will rearrange itself into a stable nuclei. This process is called
Radioactivity.
Clicker Question 31 - 1
Two different nuclei have different numbers of protons and
a different
diff
t number
b off neutrons.
t
Which
Whi h off the
th following
f ll i
could be true?
33% 33% 33%
..
de
ns
i
ty
e.
..
nu
cl
eo
n
Th
ei
r
ha
ve
th
e
sa
m
e
so
...
Th
ey
di
ffe
re
nt
i
3.
ar
e
2.
They are different isotopes
of the same element.
Theyy have the same electric
charge.
Their nucleon density is the
same.
Th
ey
1
1.