QUANTUM PHYSICS AND NUCLEAR PHYSICS
13.1
(SL Option B1) Quantum physics
13.2
(SL Option B2) Nuclear physics
13
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QUANTUM PHYSICS
AND NUCLEAR PHYSICS
13.1 (SL oPTIoN B.1) QUANTUM PHYSICS
The quantum nature of radiation
charge involved in the photo-electric efect consisted of
particles identical in every respect to those isolated by
J. J. homson two years previously, namely, electrons.
13.1.1 Describe the photoelectric effect.
13.1.2 Describe the concept of the photon, and
use it to explain the photoelectric effect.
13.1.3 Describe and explain an experiment to test
the Einstein model.
13.1.4 Solve problems involving the photoelectric
effect.
© IBO 2007
13.1.1 (B.1.1) THE PHoToELECTRIC
EffECT
I
n 1888, Hertz carried out an experiment to verify
Maxwell’s electromagnetic theory of radiation. Whilst
performing the experiment Hertz noted that a spark was
more easily produced if the electrodes of the spark gap
were illuminated with ultra-violet light. Hertz paid this fact
little heed, and it was let to one of his pupils, Hallswach, to
investigate the efect more thoroughly. Hallswach noticed
that metal surfaces became charged when illuminated with
ultra-violet light and that the surface was always positively
charged. He concluded therefore that the ultra-violet light
caused negative charge to be ejected from the surface in
some manner. In 1899 Lenard showed that the negative
Figure 1301 (a) shows schematically the sort of arrangement
that might be used to investigate the photo-electric efect
in more detail. he tube B is highly evacuated, and a
potential diference of about 10 V is applied between
anode and cathode. he cathode consists of a small zinc
plate, and a quartz window is arranged in the side of the
tube such that the cathode may be illuminated with ultraviolet light. he current measured by the micro-ammeter
gives a direct measure of the number of electrons emitted
at the cathode.
When the tube is dark no electrons are emitted at the
cathode and therefore no current is recorded. When ultraviolet light is allowed to fall on the cathode electrons are
ejected and traverse the tube to the anode, under the
inluence of the anode-cathode potential. A small current
is recorded by the micro-ammeter. Figure 1301 (b) shows
a plot of photoelectric current against light intensity
for a constant anode-cathode potential. As you would
expect, the graph is a a straight line and doubling the light
intensity doubles the number of electrons ejected at the
cathode. he graph of photoelectric current against light
frequency, Figure 1301 (c) is not quite so obvious. he
graph shows clearly that there is a frequency of light below
which no electrons are emitted. his frequency is called
the threshold frequency. Further experiment shows that
the value of the threshold frequency is independent of the
333
CHAPTER 13
intensity of the light and also that its value depends on the
nature of the material of the cathode.
U.V. source
μA
P. E. current
Cathode
B
13.1.2 (B.1.2) EINSTEIN AND THE
Anode
10 V
P.E. current
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Figure 1301 (a) Apparatus to investigate the
photo-electric effect
light intensity
P.E. current
Figure 1301 (b)
f0
frequency
Figure 1301 (c)
In terms of wave theory we would expect photo-emission
to occur for light of any frequency. For example: consider
a very small portion of the cathode, so small in fact that
it contains only one electron for photo-emission. If the
incident light is a wave motion, the energy absorbed by
this small portion of the cathode and consequently by the
334
electron will increase uniformly with time. he amount
of energy absorbed in a given time will depend on the
intensity of the incident light and not on the frequency.
If the light of a given frequency is made very very feeble
there should be an appreciable time lag during which the
electron absorbs suicient energy to escape from within
the metal. No time lag is ever observed.
PHoToN
he existence of a threshold frequency and spontaneous
emission even for light of a very low intensity cannot
be explained in terms of a wave theory of light. In 1905
Einstein proposed a daring solution to the problem. Planck
had shown that radiation is emitted in pulses, each pulse
having an energy hf where h is a constant known as the
Planck constant and f is the frequency of the radiation.
Why, argued Einstein, should these pulses spread out
as waves? Perhaps each pulse of radiation maintains a
separate identity throughout the time of propagation of
the radiation. Instead of light consisting of a train of waves
we should think of it as consisting of a hail of discrete
energy bundles. On this basis the signiicance of light
frequency is not so much the frequency of a pulsating
electromagnetic ield, but a measure of the energy of each
‘bundle’ or ‘particle of light’. he name given to these tiny
bundles of energy is photon or quantum of radiation.
Einstein’s interpretation of a threshold frequency is that
a photon below this frequency has insuicient energy to
remove an electron from the metal. he minimum energy
required to remove an electron from the surface of a metal
is called the work-function of the metal. he electrons in
the metal surface will have varying kinetic energy and so
at a frequency above the threshold frequency the ejected
electrons will also have widely varying energies. However,
according to Einstein’s theory there will be a deinite upper
limit to the energy that a photo-electron can have. Suppose
we have an electron in the metal surface that needs just φ
units of energy to be ejected, where φ is the work function
of the metal. A photon of energy hf strikes this electron
and so the electron absorbs hf units of energy. If hf ≥ φ the
electron will be ejected from the metal, and if energy is to
be conserved it will gain an amount of kinetic energy EK
given by
EK = hf - φ
Since φ is the minimum amount of energy required to
eject an electron from the surface, it follows that the above
QUANTUM PHYSICS AND NUCLEAR PHYSICS
electron will have the maximum possible kinetic energy.
We can therefore write that
EK
max
= hf - φ
Planck constant is measured. he modern accepted value
is 6.2660693 × 10-34 J s. he intercept on the frequency axis
is the threshold frequency and intercept of the Vs axis is
numerically equal to the work function measured in
electron-volt.
Or
EK
max
= hf – hf0
Where f0 is the threshold frequency. Either of the above
equations is referred to as the Einstein Photoelectric
Equation.
It is worth noting that Einstein received the Nobel prize
for Physics in 1921 for “his contributions to mathematical
physics and especially for his discovery of the law of the
photoelectric efect”.
Figure 1302 shows the typical results of Millikan’s
experiment which shows the variation with frequency f of
the maximum kinetic energy Ek .
It is let to you as an exercise, using data from this graph,
to determine the Planck constant, and the threshold
frequency and work function of the metal used for the
cathode.
(Ans 6.6 × 10-34 J s, 4.5 eV)
6
PHoToELECTRIC EQUATIoN
In 1916 Millikan veriied the Einstein photoelectric
equation using apparatus similar to that shown in Figure
1301 (a). Millikan reversed the potential diference
between the anode and cathode such that the anode was
now negatively charged. Electrons emiited by light shone
onto the cathode now face a ‘potential barrier’ and will only
reach the anode if they have a certain amount of energy.
he situation is analogous to a car freewheeling along the
lat and meeting a hill. he car will reach the top of the
hill only if its kinetic energy is greater than or equal to its
potential energy at the top of the hill. For the electron, if
the potential diference between cathode and anode is Vs
(‘stopping potential’) then it will reach the anode only if
its kinetic energy is equal to or greater than Vse where e is
the electron charge.
In this situation, the Einstein equation becomes
Vse = hf – hf0
Millikan recorded values of the stopping potential for
diferent frequencies of the light incident on the cathode.
For Einstein’s theory of the photoelectric efect to be
correct, a plot of stopping potential against frequency
should produce a straight line whose gradient equals eh .
A value for the Planck constant had been previously
determined using measurements from the spectra
associated with hot objects. he results of Millikan’s
experiment yielded the same value and the photoelectric
efect is regarded as the method by which the value of the
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13.1.3 (B.1.3) MILLIkAN’S
vERIfICATIoN of EINSTEIN’S
4
Ek /eV
3
2
1
0
0
0.5
1
1.5
2
2.5
15
f /10 Hz
Figure 1302 The results of Millikan’s experiment
In summary, Figure 1303 shows the observations associated
with the photoelectric efect and why the Classical theory,
that is the wave theory, of electromagnetic radiation is
unable to explain the observations i.e. makes predictions
inconsistent with the observations.
Observation
Emission of
electrons is
instantaneous
no matter
what the
intensity of
the incident
radiation.
he existence
of a threshold
frequency.
Classical theory predictions
Energy should be absorbed by the
electron continuously until it has
suicient energy to break free from the
metal surface. he less the intensity of
the incident radiation, the less energy
incident of the surface per unit time,
so the longer it takes the electron to be
ejected.
he intensity of the radiation is
independent of frequency. Emission
of electrons should occur for all
frequencies.
Figure 1303 Observations associated with the
photoelectric effect
335
CHAPTER 13
he fact that the photoelectric efect gives convincing
evidence for the particle nature of light, raises the question
as to whether light consists of waves or particles. If
particulate in nature, how do we explain such phenomena
as interference and difraction?. his is an interesting area
of discussion for TOK and it is worth bearing in mind that
Newton wrote in his introduction to his book Optics ‘It
seems to me that the nature of light be particulate’.
13.1.4 (B.1.4) SoLvE PRoBLEMS oN
THE PHoToELECTRIC EffECT
AHL
Example 1
Calculate the energy of a photon in light of wavelength
120 nm.
State and explain two observations associated with
the photoelectric efect that cannot be explained
by the Classical theory of electromagnetic
radiation.
3.
In an experiment to measure the Planck constant,
light of diferent frequencies f was shone on to the
surface of silver and the stopping potential Vs for
the emitted electrons was measured.
he results are shown below. Uncertainties in the
data are not shown.
Vs / V
1.2
1.6
2.0
2.5
3.0
3.2
f / 1015 Hz
0.25
1.7
3.3
5.6
7.7
8.4
Plot a graph to show the variation of Vs with f.
Draw a line of best-it for the data points.
Solution
f =
2.
c 3.0 ×10 8
=
= 2.5 × 1015 Hz
λ 1.2 ×10 −7
Use the graph to determine
E = hf = 6.6 × 10 × 2.5 × 10 = 1.7 × 10 J
-34
15
-18
Example 2
he photoelectric work function of potassium is 2.0 eV.
Calculate the threshold frequency of potassium.
(i)
(ii)
a value of the Planck constant
the work function of silver in electron-volt.
The wave nature of matter
13.1.5 Describe the de Broglie hypothesis and the
concept of matter waves.
13.1.6 Outline an experiment to verify the
de Broglie hypothesis.
Solution
13.1.7 Solve problems involving matter waves.
φ 2.0 ×1.6 ×10 −19
f0 = =
= 4.8 × 1014 Hz
h
6.6 ×10 −34
Exercises
1.
336
Use data from example 2 to calculate the
maximum kinetic energy in electron-volts of
electrons emitted from the surface of potassium
when illuminated with light of wavelength
120 nm.
© IBO 2007
13.1.5 (B.1.5) THE DE BRogLIE
HYPoTHESIS AND MATTER WAvES
In 1923 Louis de Broglie suggested that since Nature should
be symmetrical, that just as waves could exhibit particle
properties, then what are considered to be particles should
exhibit wave properties.
QUANTUM PHYSICS AND NUCLEAR PHYSICS
We have seen (Section 7.3.4) that as a consequence of
Special relativity the total mass-energy of a particle is
given by
nickel
crystal
accelerated
electron beam
E = mc2
We can use this expression to ind the momentum of a
photon by combining it with the Planck equation E = hf
such that
E = hf =
hc
= mc 2
λ
scattered
electrons
electron detector
Figure 1305 The scattering of electrons by a nickel crystal
h
mc =
λ
But mc is the momentum p of the photon, so that
p=
h
λ
Based on this result, the de Broglie hypothesis is that any
h
particle will have an associated wavelength given by p = λ
he waves to which the wavelength relates are called
matter waves.
For a person of 70 kg running with a speed of 5 m s-1, the
wavelength λ associated with the person is given by
λ=
h 6.6 ×10 −34
=
≈ 2 × 10-36 m
p
70 × 5
his wavelength is minute to say the least. However,
consider an electron moving with speed of 107 m s-1, then
its associated wavelength is
λ=
h
6.6 ×10 −34
=
≈ 7 × 10-11 m
p 9.1 ×10 −31 ×10 7
Although small this is measurable.
13.1.6 (B.1.6) ExPERIMENTAL
vERIfICATIoN of THE
DE BRogLIE HYPoTHESIS
In 1927 Clinton Davisson and Lester Germer who both
worked at the Bell Laboratory in New Jersey, USA, were
studying the scattering of electrons by a nickel crystal.
A schematic diagram of their apparatus is shown in
Figure 1305.
heir vacuum system broke down and the crystal oxidized.
To remove the oxidization, Davisson and Germer heated
the crystal to a high temperature. On continuing the
experiment they found that the intensity of the scattered
electrons went through a series of maxima and minimathe electrons were being difracted. he heating of the
nickel crystal had changed it into a single crystal and the
electrons were now behaving just as scattered X-rays do.
(See Chapter 18 Topic G.6). Efectively, that lattice ions of
the crystal act as a difraction grating whose slit width is
equal to the spacing of the lattice ions.
Davisson and Germer were able to calculate the de Broglie
wavelength λ of the electrons from the potential diference
V through which they had been accelerated.
Using the relationship between kinetic energy and
momentum, we have
E k = Ve =
p2
2m
herefore
p = 2mVe
Using the de Broglie hypothesis p =
λ=
h
2 mVe
h
, we have
λ
hey knew the spacing of the lattice ions from X-ray
measurements and so were able to calculate the predicted
difraction angles for a wavelength equal to the de Broglie
wavelength of the electrons. he predicted angles were
in close agreement with the measured angles and the
de Broglie hypothesis was veriied – particles behave as
waves.
Of course we now have a real dilemma; waves behave
like particles and particles behave like waves. How can
this be? his so-called wave-particle duality paradox was
not resolved until the advent of Quantum Mechanics in
337
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from which
CHAPTER 13
1926-27. here are plenty of physicists today who argue
that it has still not really been resolved. To paraphrase
what the late Richard Feynman once said, ‘If someone tells
you that they understand Quantum Mechanics, they are
fooling themselves’.
13.1.7 (B.1.7) SoLvE PRoBLEMS
INvoLvINg MATTER WAvES
Example
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Solution
λ=
energy states
13.1.8 Outline a laboratory procedure for
producing and observing atomic spectra.
13.1.9 Explain how atomic spectra provide
evidence for the quantization of energy in
atoms.
13.1.10 Calculate wavelengths of spectral lines
from energy level differences and vice
versa.
Calculate the de Broglie wavelength of an electron ater
acceleration through a potential diference of 75 V.
Use λ =
Atomic spectra and atomic
13.1.11 Explain the origin of atomic energy levels
in terms of the “electron in a box” model.
13.1.12 Outline the Schrödinger model of the
hydrogen atom.
h
2 mVe
6.6 ×10 −34
2 × 9.1 ×10 −31 ×75 ×1.6 ×10 −19
13.1.13 Outline the Heisenberg uncertainty
principle with regard to position–
momentum and time–energy.
© IBO 2007
= 1.4 nm.
Exercises
1.
Repeat the example above but for a proton.
2.
Determine the ratio of the de Broglie wavelength
of an electron to that of a proton accelerated
through the same magnitude of potential
diference.
13.1.8-10 (B.1.8-10) oBSERvINg
AToMIC SPECTRA
hese topics are discussed in Topic 7.1.4. he following
example will serve as a reminder and also reinforce the
concept of the photon.
Example
he diagram below shows some of the energy levels of the
hydrogen atom.
-0.85 eV
B
-1.50 eV
A
-3.4 eV
Calculate the frequency associated with the photon
emitted in each of the electron transitions A and B.
338
QUANTUM PHYSICS AND NUCLEAR PHYSICS
p2
Using E k = 2 m we have therefore that the energy of the
electron En with wavelength λn
Solution
En =
A
∆E = 3.4 – 0.85 = 2.6 eV = 2.6 × 1.6 × 10-19 J = 4.2 × 10-19 J
= hf to give f =
4.2 ×10 −19
= 6.4 × 1014 Hz
6.6 ×10 −34
13.3.12 (B.1.12) THE SCHRöDINgER
Transition A gives rise the blue line in the visible spectrum
of atomic hydrogen and B to a line in the infrared region of
the spectrum.
13.3.11 (B.1.11) THE oRIgIN of THE
ENERgY LEvELS
he electron is bound to the nucleus by the Coulomb
force and this force will essentially determine the energy
of the electron. If we were to regard the hydrogen atom for
instance as a miniature Earth-Moon system, the electron’s
energy would fall of with inverse of distance from the
nucleus and could take any value. However we can the
origin of dither existence of discrete energy levels within
the atom if we consider the wave nature of the electron.
To simplify matters we shall consider the electron to be
conined by a one dimensional box rather than a three
dimensional “box” whose ends follow a 1 shape.
r
In classical wave theory, a wave that is conined is a
standing wave. If our electron box is of length L then the
allowed wavelengths λn are give by (see Topic 11.1)
2L
where n = 1, 2 , 3 …)
n
However from the de Broglie hypothesis we have that
pn =
h nh
=
λn 2 L
In 1926 Erwin Schrödinger proposed a model of the
hydrogen atom based on the wave nature of the electron
and hence the de Broglie hypothesis. his was actually the
birth of Quantum Mechanics. Quantum mechanics and
General Relativity are now regarded as the two principal
theories of physics
he mathematics of Schrödinger’s so-called wavemechanics is somewhat complicated so at this level, the
best that can be done is to outline his theory. Essentially,
he proposed that the electron in the hydrogen atom
is described by a wave function. his wave function is
described by an equation known as the Schrödinger wave
equation, the solution of which give the values that the
wave function can have. If the equation is set up for the
electron in the hydrogen atom, it is found that the equation
will only have solutions for which the energy E of the
electron is given by E = (n + ½ )hf. Hence the concept of
quantization of energy is built into the equation. Of course
we do need to know what the wave function is actually
describing. In fact the square of the amplitude of the wave
function measures the probability of locating the electron
in a speciied region of space.
he solution of the equation predicts exactly the line
spectra of the hydrogen atom. If the relativistic motion
of the electron is taken into account, the solution even
predicts the ine structure of some of the spectral lines.
(For example, the red line on closer examination, is found
to consist of seven lines close together.)
he Schrödinger equation is not an easy equation to solve
and to get exact solutions for atoms other than hydrogen
or singly ionised helium, is well-nigh impossible.
Nonetheless, Schrödinger’s theory changed completely the
direction of physics and opened whole new vistas- and
posed a whole load of new philosophical problems.
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AHL
0.65 ×1.6 ×10 −19
= 1.6 × 1014 Hz
6.6 ×10 −34
he corresponding wavelengths are A = 470 nm and
B = 190 nm
λn =
Hence we see that the energy of the electron is quantized.
MoDEL of THE HYDRogEN AToM
B
f =
n2 h2
8me L
CHAPTER 13
13.3.13 (B.1.13) THE HEISENBERg
UNCERTAINTY PRINCIPLE
means is that if its momentum is deined precisely, then
its associated probability wave is ininite in extent and we
have no idea where it is. Efectively, the more precisely we
know the momentum of a particle, the less precisely we
know its position and vice versa.
In 1927 Werner Heisenberg proposed a principle that went
along way to understanding the interpretation of the
Schrödinger wave function. Suppose the uncertainty in
our knowledge of the position of a particle is ∆x and the
uncertainty in the momentum is ∆p, then the Uncertainty
Principle states that the product ∆x∆p is at least the order of
h, the Planck constant. A more rigorous analysis shows that
From the argument above, we see that in the real world
waves are always made up of a range of wavelengths and
form what is called a wave group. If this group is of length
∆x, then classical wave theory predicts that the wavelength
spread ∆λ in the group is given by
AHL
∆x∆p ≥
h
4π
To understand how this links in with the de Broglie
hypothesis and wave functions, consider a situation in
which the momentum of a particle is known precisely.
h
In this situation, the wavelength is given by λ = p and
is completely deined. But for a wave to have a single
wavelength it must be ininite in time and space. For
example if you switch on a sine-wave signal generator and
observe the waveform produced on an oscilloscope, you
will indeed see a single frequency/wavelength looking
wave. However, when you switch of the generator, the
wave amplitude decays to zero and in this decay there
will be lots of other wavelengths present. So if you want
a pure sine wave, don’t ever switch the generator of and,
conversely, never switch it on. For our particle what this
1
∆   ∆x ≈ 1
λ
If we consider the wave group to be associated with a
particle i.e. a measure of the particle momentum, then
 1  ∆p
∆  =
λ h
that is
∆p
∆x ≈ 1
h
or
∆p ∆x ≈ h
We have in fact used the classical idea of a wave but
we have interpreted the term ∆  1  as a measure of the
λ
Tok What is quantum mechanics?
he advent of quantum mechanics meant that the determinism of classical physics was a thing of the past. In classical
physics, it was believed that if the initial state of a system is known precisely, then the future behaviour of the system
could be predicted for all time. However, according to quantum mechanics, because of the inability to deine the initial
data with absolute precision, such predictions can no longer be made. In classical physics it was thought that the only
thing that limited knowing the initial state of the system with suicient precision was determined basically by precision
of the measuring tool. he Uncertainty principle put paid to this idea – uncertainty is an inherent part of Nature.
here have been many attempts to understand what quantum mechanics is really all about. On a pragmatic level, many
physicists accept that it works and get on with their job. Others worry about the many paradoxes to which it leads.
One of the true mysteries (apart from the ever famous Schrödinger cat) is the double slit experiment and its interpretation.
Fire electrons at a double slit and just like light waves, an observable interference pattern can be obtained. However, if
you observe through which slit each electron passes, the interference pattern disappears and the electrons behave like
particles. If you can interpret this then you can “understand” what quantum mechanics is all about. However, remember
what Richard Feynman had to say on this topic.
Finally, if quantum mechanics is the “correct” physics, why do we in this course spend so much time learning classical
physics? An example might suice to answer this question.
If you apply Newtonian mechanics to the motion of a projectile, you get the “right” answer, if you apply it to the motion
of electrons, you get the “wrong” answer. However, if you apply quantum mechanics to each of these motions, in each
case you will get the right answer. he only problem is, that using quantum mechanics to solve a projectile problem is like
using the proverbial sledge-hammer to crack a walnut. We must move on.
340
QUANTUM PHYSICS AND NUCLEAR PHYSICS
uncertainty in the momentum of a particle and this leads
to the idea of the Uncertainty Principle. You will not be
expected to produce this argument in an IB examination
and is presented here only as a matter of interest.
he Principle also applies to energy and time. If ∆E is the
uncertainty in a particle’s energy and ∆t is the uncertainty
in the time for which the particle is observed is ∆t, then
h
4π
path of deflected
α-particle
region
occupied by
gold nucleus
path of
Figure 1307 An α-particle colliding with a gold nucleus
his is the reason why spectral lines have inite width.
For a spectral line to have a single wavelength, there must
be no uncertainty in the diference of energy between
the associated energy levels. his would imply that the
electron must make the transition between the levels in
zero time.
13.2 (SL oPTIoN B.2)
NUCLEAR PHYSICS
he kinetic energy of the α-particle when it is a long way
from the nucleus is Ek. As it approaches the nucleus, due
to the Coulomb force, its kinetic energy is converted into
electrostatic potential energy. At the distance of closest
approach all the kinetic energy will have become potential
energy and the α-particle will be momentarily at rest.
Hence we have that
Ze × 2e
Ze 2
=
Ek =
4πε 0 d 2πε 0 d
AHL
∆E ∆t =
d = distance of
closest approach
Where Z is the proton number of gold such that the charge
of the nucleus is Ze . he charge of the α-particle is 2e.
13.2.1 Explain how the radii of nuclei may be
estimated from charged particle scattering
experiments.
For an α-particle with kinetic energy 4.0 MeV we have that
13.2.2 Describe how the masses of nuclei may
be determined using a Bainbridge mass
spectrometer.
=1.2 × 10-13 m
13.2.3 Describe one piece of evidence for the
existence of nuclear energy levels.
© IBO 2007
13.2.1 (B.2.1) NUCLEI RADII
2πε 0 E k 2 × 3.14 × 8.85 ×10 −12× 4.0 ×10
d=
=
Ze 2
79 ×1.6 ×10 −19
6
he distance of closest approach will of course depend on
the initial kinetic energy of the α-particle. However, as
the energy is increased a point is reached where Coulomb
scattering no longer take place. he above calculation is
therefore only an estimate. It is has been demonstrated at
separations of the order of 10-15 m, the Coulomb force is
overtaken by the strong nuclear force.
In 7.1.2 we outlined how the Geiger-Marsden experiment,
in which α-particles were scattered by gold atoms,
provided evidence for the nuclear model of the atom.
he experiment also enabled an estimate of the nuclear
diameter to be made.
13.2.2 (B.2.2) MEASURINg NUCLEAR
Figure 1307 shows an α-particle that is on a collision
course with a gold nucleus and its subsequent path. Since
the gold nucleus is much more massive than the α-particle
we can ignore any recoil of the gold nucleus.
he measurement of nuclear (isotope) masses is achieved
using a mass spectrometer. A form of mass spectrometer
is shown in Figure 1308.
MASSES
Positive ions of the element under study are produced in a
high voltage discharge tube (not shown) and pass through
a slit (S1) in the cathode of the discharge tube. he beam of
ions is further collimated by passing through slit S2 which
provides an entry to the spectrometer. In the region X, the
ions move in crossed electric and magnetic ields.
341
CHAPTER 13
S1
Plate
P1
X
S2
P2
S3
B’
Y
electron transitions in the hydrogen atom which are only of
the order of several eV.
he existence of nuclear energy levels receives complete
experimental veriication from the fact that γ-rays from
radioactive decay have discrete energies consistent with
the energies of the α-particles emitted by the parent
nucleus. Not all radioactive transformations give rise to
γ-emission and in this case the emitted α-particles all have
the same energies.
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Figure 1308 Measuring nuclear masses
he electric ield is produced by the plates P 1 and P 2
and the magnetic ield by a Helmholtz coil arrangement.
he region X acts as a velocity selector. If the magnitude
of the electric ield strength in this region is E and that
of the magnetic ield strength is B (and the magnitude of
the charge on an ion is e) only those ions which have a v
velocity given the expression Ee = Bev will pass through the
slit S3 and so enter the main body, Y, of the instrument.
A uniform magnetic ield, B´, exists in this region and in
such a direction as to make the ions describe circular orbits.
From Sections 2.4 and 6.3 we see that for a particular ion
the radius r of the orbit is given by,
mv2
mv
= B´ev that is r =
r
B´e
Since all the ions have very nearly the same velocity, ions
of diferent masses will describe orbits of diferent radii,
the variations in value depending only on the mass of the
ion. A number of lines will therefore be obtained on the
photographic plate P, each line corresponding to a diferent
isotopic mass of the element. he position of a line on the
plate will enable r to be determined and as B´, e and v are
known, m can be determined.
Radioactive decay
13.2.4 Describe β+ decay, including the existence
of the neutrino.
13.2.5 State the radioactive decay law as an
exponential function and define the decay
constant.
13.2.6 Derive the relationship between decay
constant and half-life.
13.2.7 Outline methods for measuring the half-life
of an isotope.
13.2.8 Solve problems involving radioactive halflife.
© IBO 2007
13.2.4 (B.2.4) β- DECAY
We say in 7.2.2 that β- decay results from the decay of a
neutron into a proton and that β+ decay results from the
decay of a proton in a nucleus into a neutron viz,
13.2.3 (B.2.3) NUCLEAR ENERgY LEvELS
he α-particles emitted in the radioactive decay of a
particular nuclide do not necessarily have the same energy.
For example, the energies of the α-particles emitted in the
decay of nuclei of the isotope thorium-C have six distinct
energies, 6.086 MeV being the greatest value and 5.481
MeV being the smallest value. To understand this we
introduce the idea of nuclear energy levels. For example, if
a nucleus of thorium emits an α-particle with energy 6.086
MeV, the resultant daughter nucleus will be in its ground
state. However, if the emitted α-particle has energy 5.481
MeV, the daughter will be in an excited energy state and will
reach its ground state by emitting a gamma photon of energy
0.605 MeV. Remember that the energy of photons emitted by
342
1
0
n → 11 p + −01 e + –
ν
1
1
p → 01 n + −01 e + ν
and
It is found that the energy spectrum of the β-particles is
continuous whereas that of any γ-rays involved is discrete.
his was one of the reasons that the existence of the
neutrino was postulated otherwise there is a problem with
the conservation of energy. α-decay clearly indicates the
existence of nuclear energy levels so something in β-decay
has to account for any energy diference between the
maximum β-particle energy and the sum of the γ-ray plus
intermediate β-particle energies. We can illustrate how
the neutrino accounts for this discrepancy by referring
QUANTUM PHYSICS AND NUCLEAR PHYSICS
to Figure 1309 that show the energy levels of a ictitious
daughter nucleus and possible decay routes of the parent
nucleus undergoing β+ decay.
therefore
ln N − ln N0 = −λ t
or
N = N0 e− λt
parent nucleus
β+
ν
γ
Figure 1309
excited level of
daughter nucleus
his is the radioactive decay law and veriies mathematically the exponential nature of radioactive decay that we
introduced in 7.2.6.
ground state of
daughter nucleus
13.2.6 (B.2.6) HALf-LIfE
β+
γ
ν
γ
Neutrinos and the conservation of energy
he igure shows how the neutrino accounts for the
continuous β spectrum without sacriicing the conservation of energy. An equivalent diagram can of course be
drawn for β- decay with the neutrino being replaced by an
anti-neutrino.
he radioactive decay law enables us to determine a
relation between the half-life of a radioactive element and
the decay constant.
If a sample of a radioactive element initially contains N0
atoms, ater an interval of one half-life the sample will
contain N2 atoms. If the half-life of the element is T½ from
the decay law we can write that
0
− λT1
N0
= N0 e 2
2
13.2.5 (B.2.5) THE RADIoACTIvE
DECAY LAW
We have seen in 7.2.6 that radioactive decay is a random
process. However, we are able to say that the activity of a
sample element at a particular instant is proportional to
the number of atoms of the element in the sample at that
instant. If this number is N we can write that
∆N
= −λ N
∆t
where λ is the constant of proportionality called the decay
constant and is deined as ‘the probability of decay of a
nucleus per unit time’.
he above equation should be written as a diferential
equation i.e.
dN
= −λ N
dt
his equation is solved by separating the variables and
integrating viz,
such that
dN
= −λ dt
N
ln N = − t + constant
It is then said that at time t = 0 the number of atoms is N0
such that
ln N0 = constant
or
e-λ T½ =
1
2
that is
T½ =
ln 2
λ
13.2.7 (B.2.7) MEASURINg
RADIoACTIvE HALf-LIfE
he method used to measure the half-life of an element
depends on whether the half-life is relatively long or
relatively short. If the activity of a sample stays constant
over a few hours it is safe to conclude that it has a
relatively long half-life. On the other had if its activity
drops rapidly to zero it is clear it has a very short half-life.
Elements with long half-lives
Essentially the method is to measure the activity of a
known mass of a sample of the element. he activity can
be measured by a Geiger counter and the decay equation
in its diferential form is used to ind the decay constant.
An example will help understand the method.
343
AHL
β+
CHAPTER 13
A sample of the isotope uranium-234 has a mass of 2.0
µg. Its activity is measured as 3.0 × 103 Bq. he number of
atoms in the sample is
13.2.8 (B.2.8) SoLvE PRoBLEMS
INvoLvINg RADIoACTIvE
2.0 ×10 −6 2.0 ×10 −6
=
= 3.3 × 1016
NA
6.0 ×10 23
∆N
Using
= −λ N we have
∆t
∆N
3.0 ×10 3
∆
t
= 9.0 × 10-14 s-1
λ=
=
3.3 ×10 16
N
HALf-LIfE
Example
1.
Using the relation between half-life and decay constant we
have
(i)
(ii)
0.69
ln 2
=
= 7.6 × 1012 s ≈ 2.4 × 105 years
T½ =
λ 9.0 ×10 −14
AHL
Elements with short half-lives
he above is just an outline of the methods available for
measuring half-lives and is suicient for the HL course.
Clearly in some cases the actual measurement can be very
tricky. For example, many radioactive isotopes decay into
isotopes that themselves are radioactive and these in turn
decay into other radioactive isotopes. So, although one
may start with a sample that contains only one radioactive
isotope, some time later the sample could contain several
radioactive isotopes.
344
the decay constant for radium-223
the fraction of a given sample that will have
decayed ater 3 days.
Solution
For elements that have half-lives of the order of hours,
the activity can be measured by measuring the number of
decays over a short period of time (minutes) at diferent
time intervals. A graph of activity against time is plotted
and the half-life read straight from the graph. Better is
to plot the logarithm of activity against time to yield a
straight line graph whose gradient is equal to the negative
value of the decay constant.
For elements with half-lives of the order of seconds, the
ionisation properties of the radiations can be used. If the
sample is placed in a tube across which an electric ield is
applied, the radiation from the source will ionise the air
in the tube and thereby give rise to an ionisation current.
With a suitable arrangement, the decay of the ionisation
current can be displayed on an oscilloscope.
he isotope radium-223 has a half-life of 11.2 days.
Determine
(i)
0.0616 day-1
(ii)
0.169
Exercises
1.
he isotope technetium-99 has a half-life of
6.02 hours. A freshly prepared sample of the
isotope has an activity of 640 Bq. Calculate the
activity of the sample ater 8.00 hours.
2.
A radioactive isotope has a half-life of 18 days.
2
Calculate the time it takes for of the atoms in a
5
sample of the isotope to decay.
3.
A nucleus of potassium-40 decays to a stable
nucleus of argon-40. he half-life of potassium-40
is 1.3 × 109 yr.
In a certain lump of rock, the amount of
potassium-40 is 2.1 µg and the amount of trapped
argon-40 is 1.7 µg. Estimate the age of the rocks.