the nuclear atom

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19.1 The atomic model

The nuclear atom

• More than 2000 years ago, the Greeks suggested that matter was made up of very tiny (small) particles which they called atoms.

What is inside the atom?What is the structure of

an atom?

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Models of an atom

Plum pudding model

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• The plum pudding model of the atom by J.J. Thomson, who discovered the electron in 1897, was proposed in 1904.

• The atom is composed of electrons, surrounded by a soup of positive charge to balance the electron's negative charge, like negatively-charged "plums" surrounded by positively-charged "pudding".

• The model was disproved by the 1909 gold foil experiment,

Plum pudding model

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Geiger-Marsden scattering experiment

• Experiment:Experiment:• particles are made to hit the thin (super particles are made to hit the thin (super

thin) gold foil.thin) gold foil.• Flashes of light will be observed when Flashes of light will be observed when

particles hit the zinc sulphide screen.particles hit the zinc sulphide screen.

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Geiger-Marsden scattering experiment

• We expectWe expect allall the the particles can pass particles can pass through the gold foil. through the gold foil.

Results:Results:

Nearly all Nearly all particles pass particles pass straight through the gold foil.straight through the gold foil.

Some α-particles (about Some α-particles (about 1/8000) were scattered 1/8000) were scattered by angles greater than by angles greater than 90∘and very few even 90∘and very few even rebounded back along rebounded back along

original paths.original paths.

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Rutherford’s Remark

• It was quite the most incredible event that has ever happened to me in my life.

• 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.

• On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus.

• It was then that I had the idea of an atom with a minute massive centre, carrying a charge.

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How can the heavy particles bounce back after hitting the thin gold foil? particles – He2+

Explanation:All positive charge of the atom and most of the mass were concentrated in a tiny core called nucleus.The rest of the atom was largely empty space.

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How can the heavy particle bounce back after hitting the thin gold foil?

• Explanation:• Most of particles

passed straight through the empty space of the gold atoms.

• Some come close to the nucleus were repelled by a strong electrostatic force, so they were deflected or bounced back.

Bounced back

deflected

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-particle lose their K.E. on approaching the +ve charged nucleus, being repelled by an electrostatic force.

• At P, distance of closest approach K.E. lost = P.E. due to -particle location in electric field of nucleus.

-particle is then ‘reflected’ away from nucleus and finally acquires the same K.E. as it had initially. Collision is elastic.

N u cleu s+ Z e

A lp h ap artic le s

+ e2 P

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Estimated Upper Limit of the Size of a Nucleus

• PE = 2Ze2/40r • At P, distance of nearest approach K.E. of ’s,

½mv2 = 2Ze2/40r (P.E.) • hence an estimate of r which is upper limit to

size of nucleus.

P ath fo r g lan c in gco llis io n

P ath fo rh ead -o n co llis io n

B

AP

pN

rC en tre

o f n u c leu s

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1 Nuclear fissionWhen a heavy nucleus splits up, huge amount of energy

is released.

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Two typical nuclear fission reactions are:

The total mass of the nuclear products is slightly less than the total mass of the original particles.

Energy is released and can be calculated by using Einstein’s mass-energy relation (E = mc2)

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3 Nuclear fusion

A huge amount of energy is also released when two light nuclei join together to form a heavy nucleus.

This process is called nuclear fusion.

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Example of Nuclear Fusion• Deuterium-Tritium Fusion Reaction.

energy10

42

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21 nHeHH

The total mass of the nuclear products is slightly less than the total mass of the original particles.

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3 Nuclear fusion

Since nuclei carry +ve charges, they repel each other.

For fusion to occur, the 2 hydrogen nuclei must approach each other with very high speed.

He42

+H21

n10+ +energyH3

1

hydrogen gas of high temperature (108 C)!

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3 Nuclear fusion

Nuclear fusion occurs in the core of the Sun, giving out heat and light. The reaction takes place continuously for billions of years.

(Photo credit: US NASA)

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Nuclear Power – controlledcontrolled fission

• The schematic diagram of a nuclear reactor is shown below:

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Fuel

• Enriched uranium is used as the fuel (uranium dioxide).

• Natural uranium contains only 0.7% uranium-235, which does not undergo fission in these circumstances.

• Treatment is required to increase the concentration to 3%.

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Moderator

• The probability that fission caused by a high energy neutron (fast moving neutron) is quite low.

• Use materials to slow down neutrons so that the probability of causing a fission is significantly higher.

• These neutron slowing down materials are the so called moderators.

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Moderator• These neutron slowing down materials are the so

called moderators.

umM

Mmv

umM

mV

2

u v VAt rest

Before collision

After collision

By momentum conservation: mu = mv + MV

By energy conservation: ½mu2 = ½mv2 + ½MV2

m M

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Moderator

• The choice of the moderator– The atoms of an ideal moderator should have the same

mass as a neutron (M = m). So a neutron colliding elastically with a moderator atom would lose almost all its KE to the moderator atom.

– The moderator atoms should not absorb neutrons but should scatter them instead.

– In practice, heavy water (D2O i.e. 2H2O) is chosen as the moderator. (Heavy water is different from hard water)

v VAfter collision

umM

Mmv

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Control rods

• The control rods are made of boron-steel, which absorbs neutrons.

• They are raised and lowered to vary the number of neutrons to control the rate of fission.

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2 Chain reactionWhen a uranium nucleus splits, 2 or 3 neutrons are emitted.

neutronU-235

nucleus splits

If these neutrons carry on splitting other uranium nucleus...

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2 Chain reaction

neutron

U-235 nucleus splits

self-sustaining chain reaction

escapes

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Chain reaction

• The fission neutrons enter the moderator and collide with moderator atoms, transferring KE to these atoms.

• So the neutrons slow down until the average KE of a neutron is about the same as that of a moderator atom.

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The fission neutrons could be absorbed by the U-238 nuclei without producing further fission.

The fission neutron could escape from the isolated block of uranium block without causing further fission.

The critical mass of fuel is the minimum mass capable of producing a self-sustaining chain reaction.

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• At the fission of U-235 on the average 2.5 neutrons are released but not all of these cause fission.

• multiplication factor (k)

Two typical nuclear fission reactions are:

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generation preceding in the neutons ofnumber

"generation"neutron ain neutrons ofnumber k

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neutron

U-235 nucleus splits

escapes



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• critical reactor• The number of neutrons in the system is

constant, i.e. they cause the same number of fissions in every second.



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• k > 1, the system is supercritical • k < 1, the system is subcritical

k can be varied by lowering or raising the control rods

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1 Nuclear powerb Potential hazards

The nuclear waste remains radioactive for thousands of years. serious handling and storage problems

Nuclear accidents lead to the leakage of radiation. widespread and long-lasting disasters

(Photo credits: BREDL; IAEA)

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1 Nuclear powerc Controlled nuclear fusion

Fuel = H-2, H-3 (plentiful in sea water)

Waste product = He-4(inert and non-radioactive)

Cheaper, abundant and safer, but not yet in practice

(Photo credit: Princeton Plasma Physics Lab)

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1 Nuclear powerd Benefits and disadvantages

Create serious social and environmental problems

Solve energy shortage crisis

contentious

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Solve energy shortage crisis

No fuel transportation problem

Cheaper than coal/oil for generating power in most cases

Little environment pollution

Benefits

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Accident serious consequence

Expensive in maintaining safety standards

Unnecessary as alternative energy sources exist

Lead to widespread of nuclear weapons

Disadvantages

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Pre-lesson assignment Einstein's mass-energy relation (E = mc2)

1. According to Einstein’s mass-energy equation E = mc2, what is the amount of energy produced if 1 kg of a certain element completely changes into energy?

Solution:Solution:Energy released = (1)(3 x 108)2

= 9 x 109 x 101616 J J

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Q2

• Mass of U-235: 3.9030 x 10-25 kgBa-144: 2.3899 x 10-25 kgKr-90: 1.4931 x 10-25 kg

Neutron: 1.6749 x 10-27 kg• Calculate the nuclear energy released in the

nuclear fission.Solution:Solution:• Mass difference

= 3.9030 x 10-25 – (2.3899 x 10-25 + 1.4931 x 10-25 + 1.6749 x 10-27)= 3.251 x 10-28 kg

• Energy released = (3.251 x 10-28 )(3 x 108)2

= 2.93 x 102.93 x 10-11-11 J J

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Nuclear Energy

1. unified atomic mass unit 2. unit of energy: eV3. Binding energy

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Atomic mass

Particle or element

Atomic mass / kg

Proton (p) 1.673 x 10-27

Neutron (n) 1.675 x 10-27

Electron (e) 9.14 x 10-31

Hydrogen (H)

1.674 x 10-27

Helium (He) 6.646 x 10-27

Lithium (Li) 1.165 x 10-26

1.00728u

1.00867u

0.00055u

1.00794u

4.00260u

7.01601u

1 u = unified atomic mass unit

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Question:Question:How to express the unified atomic mass

unit (1u) in kg?Solution:Solution:Given that:

mass of a C-12 atom = 1.9927 x 10-26 kg ∴1u = 1.9927 x 10-26 kg /12

= 1.66 x 10-27 kg

The unified atomic mass unit (u) is defined as one twelfth of the mass of the carbon-12 carbon-12 atomatom.

History

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Example 1

• Given that the atomic mass of a hydrogen atom is 1.00783u. Find the mass of a hydrogen atom.

• SolutionSolutionMass of a hydrogen atom = 1.00783 x 1.66 x 10-27

= 1.67 x 10-27 kg

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Units of Energy Units of Energy

Very large unit of energy: kWhkWh(electric bill)S.I. unit: JJVery small unit of energy: eVeV (Atomic physics)

Energy = charge Q x voltage V

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Energy in eV

1 V

+–

1.6 x 10-19 C–

1 eV is the energy (K.E.) gained by an electronelectron when it is accelerated through 1 V.

By W = QV

1 eV = (1.6 x 10-19)(1)

∴ 1 eV = 1.6 x 10-19 J

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Energy in eV

1 V

+

–

1.6 x 10-19 C

+

1 eV is the work done in moving a charge of 1.6 x 10-19 C through 1 V.

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Example 1 page 25• Show that 1 u of mass is equivalent to 931

MeV by the mass-energy relation.

• Solution:Solution: The energy equivalent to 1 u of mass= mc2

= (1.66 x 10-27)(2.998 x 108)2

= 1.492 x 10-10 J= (1.492 x 10-10 / 1.602 x 10-19) eV= 931 x 106 eV= 931 MeV

1 eV = 1.602 x 10-19 J1 u = 1.66 x 10-27 kg

1 u = 931 MeV

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Example 2Consider the following nuclear fission.

Given that: 1u = 931 MeVMass of neutron: 1.00866 u

U-235: 235.044 uBa-144: 143.923 uKr-90: 89.9195 u

(a) Find the amount of nuclear energy released in nuclear reaction.

Solution:Solution:Mass difference = (235.044 – 143.923 – 89.9195 – 1.00866)u= 0.19284uEnergy released = 0.19284 x 931 MeV= 180 MeV

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(b) Hence, show that the energy released from 1 kg of U-235 is about 7.4 x 1013 J which is the energy released by burning about 3 x 106 tonnes of coal.

Solution: Solution: • No of U-235 atom in 1 kg fuel

= 1/(235.044 x 1.66 x 10-27) = 2.563 x 1024

• Energy released = (2.563 x 1024) x (180 MeV) = (2.563 x 1024) x (180 x 106) x (1.6 x 10-19)= 7.4 x 1013 J

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(180 MeV)(180 MeV)

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Example 3Consider the following nuclear fusion.

Given that: 1u = 1.66 x 10-27 kg = 931 MeVMass of neutron: 1.00866 u

H-2: 2.01355 uH-3: 3.01605 uHe-4: 4.0026 u

(a) Find the amount of nuclear energy released in the reaction.

Solution:Difference in mass = (2.01355 + 3.01605 – 4.0026 – 1.00866)u= 0.01834uEnergy released = 0.01834 x 931 MeV= 17.1 MeV

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(b) Hence, show that the energy released from 1 kg of fuel is about 3.3 x 1014 J.

Solution:No of H-2 and H-3 in 1 kg fuel= 1/[(2.01355 + 3.01605 ) x 1.66 x 10-27] = 1.1977 x 1026

Total energy released= (1.1977 x 1026) x (2.74 x 10-12) = 3.3 x 1014 J

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Nuclear Energy – nucleus

• ProtonsProtons & neutronsneutrons are collectively called nucleonsnucleons.

• A = mass number / Nucleon number

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Nuclear Energy – nucleus• Consider a helium nucleus.

Mass of a helium atom (He – 4)

Mass of components

2 protons: 2 x 1.00728 u

2 neutrons: 2 x 1.00866 u

2 electrons: 2 x 0.00055 u

Total mass: 4.0330 uTotal mass: 4.0330 u4.0026 u4.0026 u

++

+

+

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• The difference between the mass of an atom and the total mass of the particles in the atom taken separately is known as mass defect mass defect mm.

Mass of a helium atom (He – 4)

Mass of components

4.0026 u4.0026 u 2 protons: 2 x 1.00728 u

2 neutrons: 2 x 1.00866 u

2 electrons: 2 x 0.00055 u

Total mass = 4.0330 Total mass = 4.0330 uu

Mass defect of He = 4.0330u – 4.0026 u Mass defect of He = 4.0330u – 4.0026 u = 0.0304 u = 0.0304 u

++

+

+ –m =

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Binding Energy

• mass of a nucleus < total mass of separated nucleons• The energy is required to separate the nucleons

(to overcome nuclear force binding the nucleons)

• This energy is called the binding energy Eb.

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Binding Energy

• binding energy = Δm c2

where Δm is the mass defect of the nucleus.

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Find the binding energy per nucleon in a He-4 atom.

m = 4.0330 u – 4.0026 u = 0.0304 u

(1u = 931 MeV) Eb = 0.0304 x 931 MeV = 28.3 MeV

Binding energy per nucleon = 7.08 MeV

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Binding Energy

• The greater the binding energy per nucleon, per nucleon, (Eb/A), the more / less stable the nuclei.

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Binding energy Curve

1. Fe – 56 has the largest binding energy per nucleon. Therefore, it is very stable.

2. Either side of maximum binding energy per nucleon are less stable.

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3. When light nuclei are joined together, the binding energy per nucleon is also increased and become more stable.

large binding energy per nucleon ⇔ nucleons are at low energy state. (energy is released.)

Fusion

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4. When a big nucleus disintegrates, the binding energy per nucleon increases and become more stable

(energy is released)

Fission

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Example 4• Find the binding energy per nucleon for a Pb-206

nucleus.

Solution:Solution:• Mass defect

= (82 x 1.00728 + 124 x 1.00866 u + 206 x 0.00055 u) – 205.969 u= 1.8151 u

• Binding energy = 1.8151 x 931 MeV= 1690 MeV

• Binding energy per nucleon Eb/A

= 1690 MeV / 206 = 8.20 MeV

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Conditions for a Fusion Reaction (2)

ConfinementThe hot plasma must be well isolated away from material surfaces in order to avoid cooling the plasma and releasing impurities that would contaminate and further cool the plasma.

In the Tokamak system, the plasma is isolated by magnetic fields.

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History of unified atomic mass unit

• The chemist John Dalton was the first to suggest the mass of one atom of hydrogen as the atomic mass unit.

• Francis Aston, inventor of the mass spectrometer, later used 1/16 of the mass of one atom of oxygen-16 as his unit.

• Oxygen was chosen because it forms chemical compounds with many other elements.

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The discovery of isotopes

• Oxygen-17 and Oxygen-18 was discovered.

• Chemists: one-sixteenth of the average mass of the oxygen atoms.

• Physicists: one-sixteenth of the mass of an atom of oxygen-16.

• Difference in atomic weights: about 275 parts per million

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• Since 1961, by definition the unified atomic mass unit is equal to 1/12 of the mass of a carbon-12 atom.

• Physicists: carbon-12 was already used as a standard in mass spectroscopy.

• Chemists: difference in atomic mass: 42 parts per million which was acceptable

the nuclear atom

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