Transcript Today`s Objectives - RanelaghALevelPhysics

Stable and Unstable Nuclei
Particle Physics Lesson 2
Homework Reminder!

Ernest Rutherford

What important
experiment did he direct
in the early 20th Century?
What did Rutherford
conclude from his
analysis of the
observations?

Homework

Revise for the Skills Test next week.

Complete worksheets on Nuclear Physics
(during the weekend, after the test).

Recap today’s lesson using the slides on the
website.
Today’s Objectives

To recap specific charge.

To know what holds the nucleus together.

To know what happens when α, β and γ
radiation are emitted.
Problem Solving: Specific Charge
Particle
Charge Mass Specific Charge
Carbon-12 Nucleus
6e
12mP
6e/12mP
Carbon-14 Nucleus
Calcium-40 Atom
Calcium-40 Nucleus
Ca2+ Ion
e = charge on the electron (=1.60 × 10-19 C)
mP = mass of a proton/neutron (=1.67 × 10-27 kg)
Problem Solving: Specific Charge
Particle
Charge Mass Specific Charge
Carbon-12 Nucleus
6e
12mP
6e/12mP
Carbon-14 Nucleus
6e
14mP
6e/14mP
Calcium-40 Atom
0
40mP+ 0
20me
Calcium-40 Nucleus
20e
40mP
20e/40mP
Ca2+ Ion
2e
40mP
2e/40mP
+ 18em
e = charge on the electron (=1.60 × 10-19 C)
mP = mass of a proton/neutron (=1.67 × 10-27 kg)
The Nuclear Model of the Atom
Question
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Why don’t the electrons go flying off?
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Let’s now play with magnets…
What happens in each case?
S
N S
S
N N
N
S
S
N
S
N
Van der Graaf Generator

Are we still happy with our model?
Size of the nucleus...

If the atom were the size of the school canteen,
the nucleus would be the size of a pea dropped
in the middle.

Really small compared to the atom!!
For fun?...(don’t worry about the
equations)
1 Q1Q2
F
2
40 r
Gm1m2
F 
2
r
Repulsive force between two
protons.
Gravitational Attraction.
Q1,Q2=e
G is gravitational constant.
m are proton masses
R is distance between the
two protons (10-15m).
Force Comparison
19 2
(1.6 10 )
FC 
4 (8.85 1012 )(11015 ) 2
6.67 1011 (1.67 1027 ) 2
FG  
(11015 ) 2
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Coulomb force of repulsion is
about 200N

Gravitational Force of attraction is
only about 1.9×10-34 N.

So Coulomb force of repulsion is
much, much stronger!!
Video

In Search of Giants (9 of 15) The Weak and
Strong Nuclear Forces
Why doesn’t it fly apart?

The repulsive force between the two charges is
much larger than the gravitational force.

So why is it stuck together?

We need another force – the STRONG force.
The Strong Force Graph
Important to Note

It’s a weird force!

At very short ranges, below 0.5 femtometres
(0.5 × 10-15 m) the strong nuclear force is
repulsive.  what would happen otherwise?

It is attractive up to its maximum range of 3 fm
(3 × 10-15 m).
Strong Force



What would happen if the strong force wasn’t
repulsive at short distances?
(protons would get pushed together)
It is about 200 N between two protons – strong
enough to counter the repulsive Coulomb force.
For the exam, you need to know...
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The strong nuclear force:
-its role in keeping the nucleus stable;
- short-range attraction to about 3 fm, veryshort range repulsion below about 0.5 fm
Equations for alpha decay and beta - decay
including the neutrino.
Radioactive Decay

Some isotopes are stable but others are not.

Those which are not stable release radiation and
change into a more stable isotope (normally a
different element).
How much do you know?

Complete the table (in pencil) that describes the
properties of the three common radiations:Radiation
Particle
Range in
air
Stopped By
Answers
Radiation
Particle
Range in air Stopped By
Alpha
Helium
nucleus (P)
Few mm (P)
Paper (P)
Beta
High speed
electron (P)
Few cm (P)
Aluminium
sheet (P)
Gamma
Energetic
photon (P)
Infinite (P)
Several cm
lead (P)
What happens to the nucleus?
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
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When an α-particle is emitted?
When a β-particle is emitted?
When a γ-ray is emitted?
A
Z
X
Alpha radiation

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Mostly comes from heavy nuclides with proton
numbers greater than 82, but smaller nuclides
with too few neutrons can also be α-emitters.
The general decay equation is summarised
below.
A
A 4
4
Z

X  Z 2Y  2 
The α-particle is also sometimes written as
4
2
He
Notes on α-decay

The alpha particle is a helium nucleus (NOT
atom).

Energy is released in the decay. The energy is
kinetic, with the majority going to the alpha
particle and a little going to the decayed nucleus.

The velocity of the alpha particle is much greater
than that of the nucleus.
Example Decay


The nucleon number goes down by 4, the
proton number by 2.
A typical alpha decay is:
Th Ra  
228
90

224
88
4
2
Is this equation balanced? Explain your answer
Beta radiation
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
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Neutron rich nuclei tend to decay by beta
minus (β-) emission.
The beta particle is a high-speed electron
ejected from the nucleus, NOT the electron
clouds.
It is formed by the decay of neutrons, which are
slightly more energetic than a proton.
Isolated protons are stable; isolated neutrons last
about 10 minutes.
Equation for Beta decay.
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
The neutron, having emitted an electron, is
converted to a proton, and this results in the
proton number of the nuclide going up by 1.
A new element is formed. The general equation
is:
A
Z


X  Y   
A
Z 1
0
1
Note the β-particle can be written as
The  denotes a anti-neutrino.
0
1
e
Neutrinos

Neutrinos are neutral particles with a very small
mass.

This makes them very hard to detect.

An anti-neutrino is an anti-particle.

Don’t stress about this we will be revisiting all
this later.
Example Beta Decay
29
13
Al  Si    
29
14
0
1

Note that the β- particle is assigned a proton
number, Z = -1 to make the equation balance.

Remember that it is an electron.
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The anti-neutrino has A=0 and Z=0.
Notes
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The nucleon number remains the same ;
The proton number goes up by 1.
The beta particle is created at the instant of the
decay.
The antineutrino is very highly penetrating and
has a tiny mass. It is very hard to detect.
A precise amount of energy is released,
according to the nuclide.
That energy is shared among the nucleus, the
electron and the antineutrino.
Question
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
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What is the balanced nuclear equation for the
following decays?
(a) emission of a beta- particle from oxygen 19
(b) emission of an alpha particle from polonium
212
(c) emission of a beta + particle from cobalt 56
Proton numbers O – 8, F – 9, Fe – 26, Co –
27, Pb – 82, Po – 84
Answers

(a) 19
8

O F   
19
9
(b) 212
84

(c)
0
1
Po Pb 
208
82
4
2
Co Fe  
56
27
56
26
0
1

Summary
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A graph of neutron number against proton
number shows that there are more neutrons in
larger nuclei
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This is needed to ensure stability of the nuclei.
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Natural decay occurs with alpha decay
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Or beta minus decay.