Fission and Fusion Book pg 286 - 287

©cg

rah

amp

hys

ics.

com

20

16

Review

• BE is the energy that holds a nucleus together. This is equal to the mass defect of the nucleus.

• Also called separation energy. The energy required to decompose an atom, or nucleus into its constituent particles, equal to the energy equivalent of the mass defect.

• Elements with a high binding energy per nucleon are very difficult to break up.

• Iron 56 is close to the peak of the curve and has one of the highest binding energies per nucleon of any isotope.

©cg

rah

amp

hys

ics.

com

20

16

Nuclear fission and nuclear fusion

Nuclear fission is the splitting of a large nucleus into

two smaller (daughter) nuclei.

An example of fission is

In the animation, 235U is hit

by a neutron, and capturing

it, becomes excited and

unstable:

It quickly splits into two

smaller daughter nuclei, and

two neutrons.

Nuclear fission

235U + 1n (236U*) 140Xe + 94Sr + 2(1n) 92

0 92 54 38 0

140Xe 94Sr ©cgrahamphysics.com 2016

Nuclear fission and nuclear fusion

Note that the splitting was triggered by a single neutron that had just the right energy to excite the nucleus.

Note also that during the split, two more neutrons were released.

If each of these neutrons splits subsequent nuclei, we have what is called a chain reaction.

Chain reaction

235U + 1n (236U*) 140Xe + 94Sr + 2(1n) 92

0 92 54 38 0

Primary Secondary Tertiary

1 2 4 8

Exponential Growth

©cg

rah

amp

hys

ics.

com

20

16

©cg

rah

amp

hys

ics.

com

20

16

Critical Mass • The smallest possible

amount of fissionable material that will sustain a chain reaction

• Too few neutrons released - reaction will slow down

• Too many neutrons released - reaction becomes uncontrolled - causes explosion - nuclear meltdown

©cg

rah

amp

hys

ics.

com

20

16

Atomic bomb – nuclear fision

©cg

rah

amp

hys

ics.

com

20

16

Fission

• Nucleon loses energy when split mass decreases high BE p.n.

• High nucleon number elements undergo fission

• Low nucleon number nuclides may undergo fusion they all try to reach the nuclide that is most stable: iron - 56

©cg

rah

amp

hys

ics.

com

20

16

Stars start to shine - fission

©cg

rah

amp

hys

ics.

com

20

16

Fission

• Fission brought on by absorption of neutrons is called induced fission

• Fission that happens if a nucleus splits is called spontaneous fission

• Induced fission is called neutron activation

©cg

rah

amp

hys

ics.

com

20

16

Neutron Activation

• Radioactive decay is spontaneously

• If atoms are bombarded with different particles, these change into new elements

• Neutrons are particularly effective in inducing artificial transmutation to produce radioactive isotopes

©cg

rah

amp

hys

ics.

com

20

16

Neutrons are produced

• Beryllium is bombarded with 𝛼 - particles

• 𝛼 + 𝐵𝑒 → 𝐶 + 𝑛01

612

49

24

Identify the bombarding particle:

• 𝑁 + 𝑛 → 𝐻 + 𝐶614

11

01

714

• 𝐿𝑖 +36 𝑛 + 𝑝1

1 → 𝛼24

01 + 𝛼2

4

• 𝑇𝑙 +81205 𝑝1

1 + 𝑛 → 𝑃𝑏 + 𝑛01

82206

01

©cg

rah

amp

hys

ics.

com

20

16

Liquid drop model The balance of electrostatic forces b/w protons and the short range nuclear force hold the nucleus together Absorbing a slow neutron the unstable atom U-236 is formed and the balance b/w forces is disturbed The long range Coulomb force of repulsion dominates now over short ranged nuclear force of attraction

©cg

rah

amp

hys

ics.

com

20

16

A possible fission reaction

• 𝑛 + 𝑈 → 𝐵𝑎 + 𝐾𝑟 + 2 𝑛01

3690

56144

92235

01

• n = 1.009u

• U - 235 = 235.044u

• Ba – 144 = 143.923u

• Kr – 90 = 89.920u

• Find the mass defect and energy equivalent

• ∆𝑚 = 235.044 + 1.009 − 143.923 + 89.920 + 2 𝑥 1.009 = 236.053 − 235.861 = 0.192𝑢

• E = 0.192 x 931.5 = 178.85 ~ 179 MeV

©cg

rah

amp

hys

ics.

com

20

16

Conservation laws govern fission • Momentum is conserved (relativistically)

• Total charge is conserved (charge of products is charge of reactants)

• The number of nucleons remains constant

• Mass – energy is conserved (∆𝐸 = 𝑚𝑐2)

The KE after a reaction is greater than the KE of

the initial neutron – the mass defect ∆m has been

changed into energy ∆𝐸.

• KE is due to energy of fired neutron

©cg

rah

amp

hys

ics.

com

20

16

Fusion • Two lighter nuclei fuse together

• Very high pressure and temperature are needed

• Nucleus must overcome Coulomb repulsion

• Nucleus must come under the influence of the strong nuclear force

If two nuclei with low atomic numbers could join together they would have - higher BE p.n - overall mass decrease - lose energy

©cg

rah

amp

hys

ics.

com

20

16

Example • For this reaction find the mass defect and the energy released

Mass of 𝐻2 = 2.014102u

Mass of 𝐻3 = 3.016049u

Mass of 𝐻𝑒4 = 4.002604u

Mass of n = 1.008665u ∆𝑚 = 2.014102 + 3.016049 − 4.002604 + 1.008665

= 5.030151 – 5.011269

= 0.01882u

BE = 0.01882 x 931.5 = 17.6MeV

• The energy released = KE of helium nucleus and neutron

• Advantage: no radioactive elements are produced

• Disadvantage: obtaining and maintaining high temperature and pressure to initiate fusion

©cg

rah

amp

hys

ics.

com

20

16

Revisit BE curve

- Nuclides with nucleon number ~ 60 are most stable

- High nucleon number nuclides undergo fission - Low nucleon number nuclides undergo fusion

©cg

rah

amp

hys

ics.

com

20

16

To reach most stable nuclide

• Fission: - loosing energy - increases BE of fission nuclei - becomes more stable

• Fusion - PE of new nucleus is less than that of individual atoms - increases BE of fused nucleus - becomes more stable

©cg

rah

amp

hys

ics.

com

20

16

Example 𝑈 + 𝑛 → 𝑆𝑟 + 𝑋𝑒 + 3 𝑛01

54146

3890

01

92238

Use the graph to compare the BE

238-U: 7.6 x 238 =1800MeV 90 – Sr: 8.7 x 90 = 780MeV 146 – Xe: 8.2 x 146 = 1200MeV 1200 + 780 = 1980MeV

The BE of fission nuclei is greater than BE of U – 238 losing energy increases BE nucleus more stable

©cg

rah

amp

hys

ics.

com

20

16

Fusion in the Sun • 𝑝:+ 𝑝:

𝐻12

• 𝐻12 + 𝑝:

𝐻23 𝑒

• 𝐻23 𝑒 + 𝐻2

3 𝑒 𝐻24 𝑒 +2 𝑝:

• For a complete cycle the first two

reactions must occur twice • Result:

- 1 helium nucleus - 2 positrones - 2 protons, which are available for future fusions - 2 neutrinos

©cg

rah

amp

hys

ics.

com

20

16

Example • 𝐻2

4 𝑒= 4.00260u

• 𝐻12 = 2.01410u

• 𝑛01 = 1.00867u

• 𝑝11 = 1.00728u

• 𝑒10 = 0.00055u

a) What is the mass of the 𝐻24 𝑒 nucleus?

b) What is the BE of the 𝐻24 𝑒 nucleus?

c) What is the BE p.n. for a 𝐻24 𝑒 nucleus?

d) What is the BE p.n. for a 𝐻12 nucleus?

e) When two deuterium nuclei fuse to form a helium nucleus how much energy is released?

a) 𝑚𝐻𝑒 = 4.00260 × 931.5 = 3728.4𝑀𝑒𝑉𝑐;2

b) 2p + 2n 𝐻𝑒24

∆𝑚 = 2 × 1.00728 + 2 × 1.00867 × 931.5 − 3728.4 = 3755.7 − 3728.4 = 27.3 𝑀𝑒𝑉𝑐;2 BE = 27.3 MeV c) BE =

𝐸

𝐴=

27.3

4= 6.8𝑀𝑒𝑉

d) 1p + 1n 𝐻12

∆𝑚 = 1.00728 + 1.00867 − 2.01410 = 2.01595 − 2.01410 = 0.00185𝑢

BE= 0.00185 x 931.5 = 1.723MeV and BE p.n. = 𝐸

𝐴=

1.273

2= 0.86𝑀𝑒𝑉

e) 𝐻12 + 𝐻1

2 𝐻24 𝑒

∆𝑚 = 2 𝑥 2.01410 − 4.00260 = 0.0256𝑢

BE = 0.0256 x 931.5 = 23.8 ~ 24MeV

©cg

rah

amp

hys

ics.

com

20

16

Example • Determine the number x of neutrons produced and calculate the energy

released in the following fission reaction:

𝑈 + 𝑛 → 𝐵𝑎 + 𝐾𝑟 + 𝑋 𝑛01

3690

56144

01

92235

• 235 – U = 235.043929u 144 – Ba = 143.922952u 90 – Kr = 89.919516u 1 – n = 1.008665u

• Solution

• Mass left side: 236

• Mass right side: 144 + 90 = 234

• Hence x = 2

• ∆𝑚 =235.043929 + 1.008665 −143.922952 + 89.919516 + 2 × 1.008665

= 236.052594 − 235.859798 = 0.192796u

• E = 0.192796 x 931.5 = 179.6MeV ~ 180 MeV

©cg

rah

amp

hys

ics.

com

20

16