Download PHY–309 L. Solutions for homework set # 10. Textbook question Q

PHY–309 L.
Solutions for homework set # 10.
Textbook question Q.21 ant the end of chapter 18:
The gold atoms — just like all the other atoms — are mostly empty space containing a few
electrons. But the electrons are so much lighter than the α particles — m(e) : m(α) ≈ 1 :
7300 — that when an α particle collides with an electron, it hardly notices the collision and
keeps moving in almost the same direction with almost the same speed. For comparison,
think of an M1 Abrams tank colliding with a grocery-store cart.
So when Rutherford, Geiger, and Marsden aimed a beam of α particles at a very thin
(less than a micrometer) gold foil, they expected all the α particles to go through the foil
with very little deflection. And indeed, most of the α particles did precisely that. The
surprise was that a small fraction of the α particles were somehow deflected through large
angles, and some even bounced back. In Rutherford’s words, “It was almost as incredible as
if you fired a 15–inch shell at a piece of tissue paper and it came back and hit you”.
For more details, please see Wikipedia page on Geiger–Marsden experiment.
Non-textbook problem #I:
(a) To find if the particles flying along some tube are electrically charged or neutral, we may
use a magnetic field in a direction ⊥ to the particle beam. We do not even need to open up
the tube, we may simply put a strong magnet outside the tube.
If the particles are neutral, the magnetic field would not exert any force on them, so the
particles would keep moving in the same direction as before. However, if the particles are
electrically charged, they would fill the Lorentz force
~
F~ = q~v × B
(1)
in a direction ⊥ to their velocity ~v , so their motion would be deflected sideways. Moreover,
the direction of the force (1) depends on the sign of the electric charge q, so by observing the
side towards which the particles are deflected we may find if they are positively or negatively
charged. For example, in a magnetic field directed vertically up, the positively charged
particles would turn right while the negatively charged particles would turn left.
1
(b) The acceleration due to Lorentz force (1) is proportional to the particles’ charge/mass
ratio,
F~
q
~
=
~v × B.
m
m
~a =
(2)
Consequently, the angle through which the particles are turned as they fly through the
magnetic field depends on the charge/mass ratio q/m. Specifically,
θturn =
a×t
q
=
× Bt
v
m
(3)
where t is the time it takes a particle to cross the magnetic field; it depends of the magnet’s
geometry and on the particles’ speed, but we do not need the exact formula here. Thus,
measuring the angle through which the particles change their direction in the magnetic field,
we may find their charge-to-mass ratio as
θturn
q
=
.
m
Bt
(4)
Non-textbook problem II:
Note: Modern flat-screen plasma or LCD TVs or monitors do not use tubes. Instead, the
brightness of each pixel on the screen is controlled by its own electronic circuit.
But from the invention of electronic TV by Zworykin in 1929 until a about 2005, all TV
sets and even computer monitors used Cathode Ray Tubes (CRT). At the back end of a
CRT tube, an electron gun produces a beam of electrons that fly to the screen (which acts
as the anode) and hit it at high speed (see homework set 2, problem II). Along the way,
there are magnetic coils that deflect the beam horizontally and vertically. The currents in
those coils change in repetitive pattern, which makes the beam to scan the phosphorescent
pixels on screen. When the beam hits a pixel, it lights up with a brightness proportional to
the beam’s intensity. The beam intensity varies with time according to the signal received
by the TV, so different pixels light up with different brightness, and that’s how the picture
appears on the screen.
2
In color TVs and monitors, there are three electron guns producing three electrons beams.
Just before the screen, there is a mask that allows each beam to strike only the phosphors
emitting a particular color — red for one beam, green for another, and blue for the third.
This wikipedia page shows a diagram of a color CRT tube.
In any CRT tube — a TV, a monitor, an oscilloscope, or an X-ray tube — a beam of
electrons hits the anode (a screen, or just a piece of metal) at high speed. When the atoms
in the anode are hit by fast electrons, sometimes the inner electrons of the atom change
their orbits. When the orbits change back to the lowest-energy state of the atom, the inner
electrons emit electromagnetic radiation of very high frequency — the X-rays.
Thus, all CRT tubes — including the old TVs and monitors — emit X-rays. But the
intensity of the X-rays depend on the speed of the electrons, on the anode material they hit,
and on the intensity (i.e., the electric current) of the beam itself. Modern X-ray tubes use
very short pulses of very intense beams of electrons accelerated to speeds up to 190,000 km/s
(63% of light speed) by very high voltages (up to 150 kilovolt). When such electrons hit
tungsten atoms in the anode, they produce a lot of X-rays.
In a TV or monitor, the electrons move slower, hit lighter atoms, and the beam intensity
is much lower, so they produce much fewer X-rays. But they do produce some X-rays, and
that could be hazardous to the health of a person — especially a child — who sits too close
to a TV or a monitor for too many hours. This was a particular concern with the earliest
models of color TV made in your grandparent’s time. The later models were engineered to
emit fewer X-rays, so the health damage came more from the programming than from the
sets themselves.
Non-textbook problem #III:
(a) Let’s start with the isotope naming conventions. A complete name of an isotope has
form
A
Eℓ,
Z
for example
16
1
H,
1
O,
35
Cl,
17
63
Cu, . . .
29
(5)
Here ‘El’ is the chemical symbol for the element — ‘H’ for hydrogen, ‘O’ for oxygen, ‘Cl’ for
chlorine, ‘Cu’ for copper, etc. In addition, to the left of the chemical symbol there are two
3
numbers describing the nucleus of the isotope. The upper left number A is the net number
of nucleons — i.e., protons and neutrons, — while the lower left number Z is the number of
protons. The number of neutrons in the nucleus is not written, but it can be easily obtained
as the difference N = A − Z.
63
Cu
29
In particular,
is a copper isotope whose nuclei have 63 protons and neutrons,
including 29 protons; the remaining 63 − 29 = 34 nucleons are neutrons. Likewise,
65
Cu
29
is
another copper isotope whose nuclei have 65 neutrons and protons, including 29 protons and
65 − 29 = 36 neutrons.
The number of protons is the nucleus controls its electric charge
Qnucleus = N × 0 + Z × (+e) = +Ze
(6)
and hence the number of electrons E in a neutral atom:
Qatom = Qnucleus + E × (−e) = 0
=⇒
E = Z.
(7)
This is the only number that matters for chemistry, so all isotopes with the same Z belong
to the same chemical element. Conversely, all isotopes of the same chemical element have
the same proton number Z. In fact, the serial number of an element in the Periodic Table
— usually called the atomic number — is precisely the proton number Z.
Thus, all isotopes of hydrogen have one proton in the nucleus and one electron in the
neutral atom, all isotopes of oxygen have 8 protons in the nucleus and 8 electrons in the
neutral atom, etc. In particular, all isotopes of copper have 29 protons in the nucleus and
29 electrons in the neutral atom.
Since the proton number Z is implicit in the chemical element symbol, it is usually
omitted from the isotope’s name. Thus, instead of
63
Cu,
29
1
people write simply
any copper isotope has proton number Z = 29. Likewise, H means
isotope has Z = 1,
16
O means
16
H
8
1
H
1
63
Cu since
since any hydrogen
since any oxygen isotope has Z = 8, etc., etc.
Please remember this rule for the textbook problem SP2. If an isotope’s name is written
down without a left subscript and you need to find its proton number, look up the chemical
4
element in the Periodic Table: The atomic number of the element — it’s sequential number
in the Table — is the proton number of any isotope of that element.
(b) Let’s start with atomic weights of pure isotopes. All atoms of a pure isotope have exactly
the same mass
Matom = Z × (Mproton + Melectron ) + (N = A − Z) × Mneutron −
binding energy
. (8)
c2
To a very good approximation — three or four significant figures — this mass is proportional
to the net number of nucleons A = Z + N, and the atomic mass unit is chosen such that
Matom ≈ A amu.
(9)
In other words, the atomic weight of a pure isotope is very close to the net number A of
protons and neutrons in the nucleus.
In particular, the copper isotope
isotope
65
Cu
29
63
Cu
29
has atomic weight 62.929, 597 ≈ 63.0, while the
has atomic weight 64.927, 789 ≈ 65.0. (The high-precision atomic weights are
taken from the NIST web page.)
But the natural sources (ores, etc.) for many chemical elements provide mixtures of
different isotopes, and since the isotopes of the same element have similar chemical properties,
they don’t get separated when the element is refined from its ore or participates in chemical
reactions. Although there are ways to separate isotopes in a lab, or even on the industrial
scale, this is very difficult and expensive, so it’s done only when the element is used for its
nuclear properties (for example, uranium). Thus, most chemical elements and compounds
you can find on Earth do not contain pure isotopes but rather natural mixtures of several
⋆
isotopes. In particular, all copper on this planet that hasn’t been through a nuclear lab is
a mixture of two isotopes: about 69% of
63
Cu and 31% of
65
†
Cu.
⋆ Some elements — beryllium, fluorine, sodium, aluminum, phosphorus, scandium, manganese, cobalt,
arsenic, yttrium, niobium, rhodium, iodine, cesium, gold, bismuth, and some lantanides — have only
one stable isotope. But all the other elements are present on Earth as isotope mixtures.
† Actually, there are small differences between isotope ratios of elements coming from different sources.
63
For example, copper from different mines can have from 68.98% to 69.38% of Cu.
5
As far as chemistry is concerned, the atomic weight of an element with multiple isotopes is
the weighted average of the isotopes’ atomic weights, weighted by their relative abundances,
µ(element) =
X
µ(isotope)×fraction(isotope) ≈
isotopes
X
A(isotope)×fraction(isotope).
isotopes
(10)
In particular,
µ(copper) = µ
63
63
65
65
Cu × fraction
Cu + µ
Cu × fraction
Cu
(11)
≈ 63 × 0.69 + 65 × 0.31
= 63.62 ≈ 63.6.
To see how this works, note that to a chemist, the atomic weight of an element — or a
molecular weight of a compound — is simply the mass (in grams) of 1 mol of the substance.
And 1 mol is the the Avogadro’s number 6.02 · 1023 of atoms (or molecules), regardless of
their isotopes. Indeed, one mol of copper — 6.02 · 1023 copper atoms — is what combines
with
1
2
mol of oxygen O2 to make copper oxide CuO, or with 1 mol of sulfuric acid to make
copper sulfate, etc., etc., and it does not make any difference what isotopes do those copper
atoms belong to, as long as they are all copper atoms and there are 6.02 · 1023 of them.
On the other hand, the net mass of those 6.02 · 1023 copper atoms depends on the isotope
mixture. One mol of natural copper comprises 0.69 mols of
63
Cu and 0.31 mols of
65
Cu, so
its net mass is
M(1 mol of natural Cu) = M 0.69 mol of Cu + M 0.31 mol of
63
65
= 0.69 × µ
Cu g + 0.31 × µ
Cu g
63
65
Cu
(12)
≈ 0.69 × 63 g + 0.31 × 65 g
≈ 63.6 g.
To a chemist, thus mass (in grams) of 1 mol of natural copper is its atomic weight, thus
µ(natural copper) ≈ 63.6.
6
(13)
Non-textbook problem #IV:
Note: the atom number Z is the sequential number of the chemical element in the periodic
table; it is equal to the number of electrons in a neutral atom, which in turn is equal to
the number of protons in the atom’s nucleus. The atom number is the same for all isotopes
of the same chemical element. OOH, the mass number A is different for different isotopes.
Specifically,
A = Z + N
(14)
is the net number of nucleons — protons and neutrons — in the nucleus. It’s called the
mass number because it governs the atomic mass of a pure isotope; to a 3-digit or 4-digit
accuracy, µ ≈ A.
(a) In an α decay
mother nucleus → daughter nucleus + α particle,
(15)
the net number of protons is conserved, i.e., Znet after the decay is the same as before the
decay. Since the α particle has 2 protons, this means
Zmother = Zdaughter + 2
(16)
Zdaughter = Zmother − 2,
(17)
and therefore
Similarly, the net number of neutrons is conserved in the α decay, and since the α particle
has 2 neutrons,
Ndaughter = Nmother − 2.
(18)
Finally, the mass number A = Z + N follows from the numbers of protons and neutrons,
Adaughter = Amother − 4.
(19)
Altogether, in an alpha decay, the daughter nucleus loses 2 units of atom number and 4 units
of mass number relative to the mother nucleus.
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(b) In a β decay, the weak forces turn one of the neutrons in the mother nucleus into a
proton, an electron, and an antineutrino,
n → p + e + ν̄.
(20)
The new proton remains in the nucleus while the electron and the antineutrino fly away; the
antineutrino is usually undetected while the electron shows up as the β particle.
In the nuclear context, the whole reaction has form
mother nucleus → daughter nucleus + e + ν̄,
(21)
where the daughter nucleus has one more proton but one less neutron than the mother
nucleus,
Zdaughter = Zmother + 1,
Ndaughter = Nmother − 1,
(22)
Adaughter = Amother .
Thus, in a beta decay, the daughter nucleus has the same mass number as the mother nucleus,
but its atom number increases by 1.
(c) In a γ decay, the protons and the neutrons do not leave the mother nucleus or change
into each other. They merely change the way they move inside the nucleus lower their net
kinetic + potential energy; the energy ‘saved’ by this reconfiguration is radiated away in the
form of γ-rays, which are very-high-frequency electromagnetic waves. Altogether, a γ decay
mother nucleus → daughter nucleus + γ
(23)
does not change the net numbers of protons and neutrons, thus
Zdaughter = Zmother ,
Ndaughter = Nmother ,
(24)
Adaughter = Amother ,
and the daughter nucleus has the same mass number and the same atom number as the
mother nucleus.
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Textbook question Q.5 at the end of chapter 19:
In terms of the atomic nucleus, the atom number Z is the number of protons. This number
determines the electric charge Q = +Ze of the nucleus and hence the number of electrons
#e = Z in the neutral atom. The nuclear charge also affects the electron’s orbits around
the nucleus, Which are very important for the way atoms combine with other atoms into
molecules. Thus, the atom number Z is very important for the chemical properties of an
element.
On the other hand, the mass number A = Z + N — the net number of protons and
neutrons in the nucleus — governs the atom’s mass, but it has almost no effect on the
electrons. Indeed, the Coulomb force between the electrons and the nucleus depend only
on the nucleus’s electric charge +Ze and don’t care about its mass as long as it’s much
heavier than the electrons (which is always true). Consequently, the chemical properties of
the element do not depend on the mass number A but only on the atom number Z. And
that’s why isotopes with the same atom number Z but different mass numbers A all belong
to the same chemical element. Indeed, the number of that element in the periodic table is
Z, regardless of A.
PS: While most kinds of chemical bonds depend only on the electrons of the atoms involved,
the so-called hydrogen bonds depend on the relatively low mass of the hydrogen atoms.
2
Consequently, the heavier isotopes of hydrogen — the deuterium D = 1 H and the tritium
1
3
T = 1 H — have slightly different chemical properties from the main hydrogen isotope 1 H.
This difference is important in biochemistry, so drinking too much heavy water (D2 O instead
of H2 O) would be bad for your health.
But this mass effect is limited to hydrogen isotopes. For all the other elements, isotopes
with similar atom numbers but different mass numbers are chemically indistinguishable.
Textbook problem SP.2 at the end of chapter 19:
(a) The chemical symbols in the thorium decay chain stand for the following elements: Th
is thorium, Z = 90; Ra is radium, Z = 88; Ac is actinium, Z = 89; Rn is radon, Z = 86; Po
is polonium, Z = 84; Pb is lead, Z = 82; Bi is bismuth, z = 83.
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(b) The α and β decays have different effects on the atomic number Z: In an α decay it
decreases by 2, while in a β decay, it increases by 1. Thus, comparing the atomic numbers
of a parent isotope and its decay product will immediately tells us if the decay in question
is α or β. In particular, in the thorium decay chain in question, in the decays
Th → Ra,
Ra → Rn,
Rn → Po,
Po → Pb
(25)
the atomic number decreases by 2, so these are α decays, while in the decays
Ra → Ac,
Ac → Th,
Pb → Bi,
Bi → Po
(26)
the atomic number increases by 1, so these are β decays.
(c–d) Now that we know which decay is α and which is β, we may determine the mass
numbers A of all the isotopes involved by following a simple rule: In an α decay A decreases
by 4, while in a β decay A does not change. Thus:
(α)
(β)
(β)
(α)
(α)
(α)
(α)
(β)
(β)
(α)
232
Th
90
228
Ra
88
228
Ac
89
228
Th
90
224
Ra
88
220
Rn
86
216
Po
84
212
Pb
82
212
Bi
83
212
Po
84
→
→
→
→
→
→
→
→
→
→
228
Ra
88
228
Ac
89
228
Th
90
224
Ra
88
220
Rn
86
216
Po
84
212
Pb
82
212
Bi
83
212
Po
84
208
Pb
82
10
4
+ 2 He,
+ e− + ν̄,
+ e− + ν̄,
4
+ 2 He,
4
+ 2 He,
4
+ 2 He,
4
+ 2 He,
+ e− + ν̄,
+ e− + ν̄,
4
+ 2 He.
(27)
And the whole decay chain can be summarized as
β 228
β 228
232
α
α 220
α 224
α 228
Th → 88 Ra → 89 Ac → 90 Th → 88 Ra → 86 Rn →
90
β 212
β 212
α 216
α 216
α 212
α 208
→ 84 Po → 84 Po → 82 Pb → 83 Bi → 84 Po → 82 Pb.
(28)
There is a Walter Fendt applet showing this decay chain step by step. It also shows several
other decay chains, in particular
238
U
92
→ ··· →
11
206
Pb
82
and
235
U
92
→ ··· →
207
Pb.
82