Nuclear Reactions

Fission and Fusion

A brief history…• 1919: Ernest Rutherford experimented

with bombarding nitrogen gas molecules with alpha particles emitted from bismuth-214

• Discovery: faster moving particles were produced, and these could travel farther than the alpha particles!

• “New” particles also deflected in a magnetic field like a positive particle

A brief history…• Conclusion: The faster moving particles

were protons• Artificial Transmutation:

• The change of one element to another through the bombardment of a nucleus

• More experiments to determine exact nature of the particles and how they were “created” done with a cloud chamber…

Cloud Chambers• Invented ~1911 by a Scottish Atmospheric

Physicist (C.T.R. Wilson) to experiment with rain clouds.• Enclosed environment made to be

supersaturated (originally with water vapor, now commonly ethanol)

• Ions introduced to this environment would attract water molecules (which are polar), forming clouds…

• Earned a share in the 1927 Nobel Prize in Physics for the invention…

Cloud Chambers• Why would this be useful for Rutherford?

• Water vapor condenses around ions• An alpha particle is ionizing radiation, thus

leave a LOT of ions in its path• Water vapor would condense around these

ions, leaving a vapor trail showing where an alpha particle had been…

• Video

Rutherford’s Theories…• If proton was simply “chipped off” the

Nitrogen nucleus by the alpha particle, there should be 4 visible tracks in the cloud chamber:

• The original alpha particle BEFORE collision• The alpha particle AFTER the collision• The “chipped off” proton• The Nitrogen nucleus, now charged, as it

recoiled after the collision

Rutherford’s Theories…• If alpha particle was absorbed, and that

caused the proton to be pushed out, then there should be 3 visible tracks:

• The alpha particle before collision• The proton emitted after the collision• The path of the recoiling Nitrogen nucleus

• This theory was supported in 1925

4 14 17 12 7 8 1He N O H

Balancing Nuclear Equations:

• Note: Deuteron = Hydrogen-2 atom, a.k.a Deuterium

• Example problem:• A sample of Oxygen-16 is bombarded with

neutrons. If one of the resulting products is a deuteron, what is the resulting nucleus?

16 1 28 0 1

AZO n X H

16 1 15 28 0 7 1O n N H

Unified Mass Unit (u)

• A unit adopted by scientists that is more appropriate for masses along the order of magnitude of atomic masses

1 u = 1.66 x 10-27 kg

Mass of an electron (me) = 0.000549 u

Mass of a proton (mp) = 1.007277 u

Mass of a neutron (mn) = 1.008665 u

Mass of 1 H atom (mH) = 1.007825 u

Mass-energy equivalence• Einstein hypothesized a relationship

between mass and energy in 1905

• Many years later, data from nuclear reactions showed that his hypothesis was indeed true

c = 3.00 x 108 m·s-1

m = mass (kg)

E = Energy (J)

2E mc

Mass-energy equivalence

• Used to calculate the Rest Energy of a mass

• Used to calculate the amount of energy released in nuclear reactions

For Example:Calculate the amount of energy released when 1.00 kg of fuel is used up in a nuclear reactor…

2E mc

2 8 1 2(1.00 kg)(3.00 x 10 )E mc m s 169.00 x 10 J

The unified mass unit is defined as

A. the mass of one neutral atom of Carbon-12

B. 1/12 of the mass of one neutral atom of Carbon-12

C. 1/6 of the mass of one neutral atom of Carbon-12

D. The mass of the nucleus of Carbon-12

Binding Energy• All atomic nuclei have a total mass that is

lower than the sum of the masses of each individual particle• For example: The EXPECTED mass of an atom of

Helium would be the sum of the mass of 2 neutrons, 2 protons, and 2 electrons:2(0.000549 u) + 2(1.007277 u) + 2(1.008665 u) =

4.032982 u

The MEASURED mass of an atom of helium has been found to be 4.002602 u

a difference of 0.03038 u

This difference is known as the Mass Defect of the atom

Binding Energy• …a measure of the energy needed to

keep a nucleus together

• Binding Energy is the energy equivalent of the mass defectmass defect

E = mc2

E = (1.66 x 10-27 kg)(3.00 x 108 m·s-1)2

E = 1.49 x 10-10 J = 931 MeV

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

What is the energy equivalent of 1 u?

A. 319 MeVB. 931 eVC. 319 keVD. 931 MeV

A

B

C

D

E

F

Binding Energy Example:• Calculate the binding energy of Oxygen-16.

The measured mass of Oxygen-16 is 15.994915 u

8 electrons+8 protons+8 neutrons

8me + 8mp + 8mn = mexpected

= 8(0.000549 u) + 8(1.007277 u) + 8(1.008665 u)

= 0.004392 u + 8.058216 u + 8.069320 u

= 16.131928 u

Binding Energy Example:• Calculate the binding energy of

Oxygen-16. The measured mass of Oxygen-16 is 15.994915 u

mdefect = mexpected – mmeasured

= 16.131928 u – 15.994915 u= 0.137013 u

Eb = mdefect · (931 MeV·u-1)

Eb = (0.137013)(931)= 128 MeV

How many Joules of energy is 128 MeV?

A. 8.00 x 1020 JB. 8.00 x 1026 JC. 2.05 x 10-17 JD. 2.05 x 10-11 J

Nuclear Reactions• Fission: A reaction that involves the

splitting of a large, unstable nucleus into 2 or more smaller, more stable nuclei

Which nucleus is most likely to be part of a fission reaction?

A. Carbon-14B. DeuteriumC. PlutoniumD. Potassium-40

Nuclear Reactions• Fusion: A reaction that joins two very

light nuclei to form a heavier nucleus

• Picture source: www.atomicarchive.com

Nuclear Reactions and Binding Energy

• Nuclei with higher amounts of binding energy per nucleon are more stable than those with lower amounts of binding energy per nucleon.

• Fission and fusion processes each release large amounts of energy as the nuclei join or split to form more stable products.

• To predict how much energy can result from a nuclear reaction, we use a binding energy curve…

Binding Energy Curve

Binding Energy Curve• Example: Use the binding energy curve to

predict the amount of energy released when Uranium-235 undergoes fission to produce two Palladium-117 fragments.

• Eb for 235U = 7.6 MeV/nucleon

• Eb for 117Pd = 8.4 MeV/nucleon

The difference between these values, multiplied by the total number of nucleons, is equal to the amount of energy released in the reaction:

(0.8 MeV/nucleon) x (235 Nucleons) = 188 MeV

Nuclear Fission• Only takes place in certain very heavy

elements, such as Uranium-235

• Fissile Uranium-235 is used in nuclear reactions:

• Nucleus bombarded with a neutron to begin a chain reaction…

235 1 236 90 144 192 0 92 38 54 02U n U Sr Xe n

Fission Reactions• Self-sustaining (chain) reactions: when

enough neutrons are produced to naturally enable the reaction to continue until all fissile material is gone• Examples: Nuclear Reactors in Power

Plants; Bombs dropped on Hiroshima and Nagasaki in WWII

• Critical Mass: The amount of fissile material required to sustain a fission reaction

                                                                                                                           

                                                                                                                     Figure from Physics for Scientists and Engineers (6th ed.) by Serway and Jewett

(Thomson Brooks/Cole, 2004).

Nuclear Fusion Reactions• Conditions required for fusion

reactions:• Very high temperatures (because nuclei need

very high kinetic energies)• Very densely packed (to ensure that enough

collisions will occur), therefore:• Very high pressures

• Problems with creating fusion on Earth:

• Containment is a huge problem• At temps required, atoms would ionize and

technically would become a plasma

Nuclear Fusion Reactions• Proton-Proton Cycle = the fusion

reaction that is the source of energy in young/cool stars such as the sun:

• The first two reactions in the cycle must occur twice

• Total energy released = 24.7 MeV

1 1 2 01 1 1 1

1 2 31 1 2

3 3 4 12 2 2 1

0.4 MeV

5.5 MeV

2 12.9 MeV

H H H e

H H He

He He He H

Fusion Example• Calculate the energy released when a

proton and a deuteron undergo fusion to produce helium-3.