7.3 Nuclear Reactions

A nucleus is more than just mass

Objectives

0 Define the unified mass unit;0 State the meaning of the terms mass defect and binding energy

and solve related problems;0 Write nuclear reaction equations and balance the atomic and

mass numbers;0 Understand the meaning if the graph of binding energy per

nucleon versus mass number;0 State the meaning of and difference between fission and fusion;0 Understand that nuclear fusion takes place in the core of stars; 0 Solve problems of fission and fusion reactions.

The unified mass unit

0Abbreviation: u or amu (atomic mass unit)0Definition: 1/12 the mass of an atom of carbon-120A mole of is 12 g and the number of molecules is the

Avogadro constant0An atom of carbon-12 has a mass of 6.02 x1023 x M =

12 g0M = 1.99 x 10-26 kg0Hence M/12 = 1.66 x 10-27 kg

Example

0Find in units of u the masses of the proton, neutron, or electron.

Unified Mass Unit 1.6605402 x 10-27 kg 1 u

electron 9.1093897 x 10-31 kg

proton 1.6726231 x 10-27 kg

neutron 1.6749286 x 10-27 kg

The mass defect and binding energy

0To find the mass of a nucleus: Mnucleus = Matom – Zme

0Matom given in periodic table

0For helium: Mnucleus = 4.0026 – 2 x .0005486 = 4.00156 u

0Lets check it by recalling that a helium nucleus is made up of 2 protons and 2 neutrons and adding their masses: 2 mp + 2 mn = 4.0320 u

0But that’s larger than the mass of the nucleus by .0304 u! This leads to the concept mass defect.

The mass defect and binding energy

0The mass of the protons plus neutrons is larger than the mass of the nucleus. This difference is defined as mass defect: δ = total mass of nucleons – mass of nucleus = Zmp + (A-Z)mn - Mnucleus

Example

0Find the mass defect of the nucleus of gold, .

Einstein’s mass-energy formula

0 Where is the missing mass?0 Einstein’s theory of special relativity states that mass and

energy are equivalent and can be converted into each other: E = mc2

0 The mass defect of a nucleus has been converted into energy and is stored in the nucleus. This energy is called the binding energy of the nucleus and is denoted by Eb

0 Thus Eb = δc2

0 Binding energy = the work (energy) required to completely separate the nucleons of a nucleus 0 higher binding energy, more stability

Einstein’s mass-energy formula

0How much energy corresponds to 1 u? 1 u x c2 = 1.6605402 x 10-27 x (2.9979 x 108)2 J

= 1.4923946316 x 10-10 J= 931.5 MeV

Example

0Find the energy equivalent to the mass of the proton, neutron, and electron.

Example

0Find the binding energy of the nucleus of carbon-12.

The binding energy curve

0We saw that the mass defect of helium is 0.0304 which corresponds to a binding energy of 0.0304 x 931.5 MeV = 28.32 MeV

0The alpha particle has an unusually high stability which is why unstable particles decay into them

0There are 4 nucleons in the helium nucleus so the binding energy per nucleon is 28.32 MeV/4 = 7.1 MeV

0Most nuclei have a binding energy per nucleon of approx. 8 MeV

The Binding Energy Curve

The binding energy curve

0Maximum: 62 (Nickel) 0 If a nucleus heavier than this splits into two lighter

ones, or if two lighter nuclei fuse together, then energy is released as a result.

Energy released in a decay

0Consider the alpha decay of radium: 0For any decay, the total energy to the left of the arrow

must equal the total decay on the right. Here, that energy corresponds to each mass according to Einstein’s formula + the kinetic energies of each mass

0 If the decaying radius is at rest, then the total energy available is Mc2 where M is the mass of the radium nucleus. To the right, we have the energies of the masses of the radon and the helium nuclei. Both have KE too.

Energy released in a decay

0Thus, to be possible, the decay must be such that at the very minimum the energy corresponding to the radium mass is larger than the energies corresponding to the radon plus alpha particle masses.

0Remember that your periodic table gives atomic, not nuclear, masses. You need to subtract the mass of the electrons…unless you’re only interested in mass differences (which we are), then atomic masses are ok.

Energy released in a decay

Mass of radium = 226.0254 u

Mass of radon = 222.0176 u+ Mass of helium = 4.0026 u

sum = 226.0202 u

We see that the mass of radium exceeds that of

radon + helium by 0.0052 u. Thus, there is kinetic energy released equal to 0.0052 x 931.5

MeV = 4.84 MeV

If 50 g of radium were to decay in this way, the total energy released in this way would be N x 4.84 MeV where N is the total number of nuclei in the 50 g of radium. (1.3 x 1023).The momenta of Rn and He are in opposite directions with equal magnitudes (momentum conservation). Vhelium ≈ 55 Vradon

In Summary0For the decay to take place the mass of the decaying

nucleus has to be greater than the combined masses of the products (binding energy of the decaying nucleus must be less than the binding energies of the product nuclei.

0This is why radioactive decay is possible for heavy elements lying to the right of nickel in the binding energy curve.

Assignment

0Questions 1,2,3,5,6

Nuclear reactions

0 If a nucleus cannot decay by itself, it can still do so if energy is supplied to it by a fast moving particle that collides with it

0Transmutation of nitrogen: an alpha particle colliding with nitrogen produces oxygen and hydrogen (a proton)

0Atomic and mass numbers on both sides match0Left mass = 18.0057 u, right mass = 18.0070 u (larger) so

the alpha must have enough kinetic energy to make up for the mass imbalance. KE>δ because products will also have KE

Nuclear reactions

0Generally, in a reaction , energy will be released if is positive.

0The amount of energy released is equal to 02 kinds of energy producing nuclear reactions: Fission &

Fusion

Nuclear Fission

0A heavy nucleus splits up into lighter nuclei0Uranium-235 absorbs a neutron, momentarily becoming

uranium-236, which then splits into lighter nuclei.0Many possibilities for daughter nuclei

0Production of neutrons is characteristic of fission; they can cause other fission reactions – chain reaction

0Minimum mass of uranium- 235 is needed, otherwise, neutrons escape without causing further reactions – critical mass

Nuclear fission

0The energy released can be calculated as follows:

0This energy appears as the kinetic energy of the products

0This is LOTS of energy (1 kg = 7 x 1013 J)

Mass of uranium plus neutron

= 236.0526 u

Mass of products = 143.92292 u + 88.91781 u + 3 x 1.008665 u= 235.8667250 u

Mass difference = 236.0526 u – 235.8667250 u = 0.0185875 u

Energy released = 0.185875 x 931.5 MeV = 173.14 MeV

Nuclear fission

0Nuclear reactors – controlled rate of reaction0Nuclear bomb – too much energy produced at 1 time –

uncontrolled rate of reaction0Alpha particles cannot be used to start fission because

protons would be repelled by the uranium nucleus0An electron would not sufficiently perturb the heavy

nucleus.0MUST be a neutron

Nuclear Fusion

0The joining of 2 light nuclei into a heavier one + energy

01 kg deuterium ≈ 1013 J

2 x mass of deuterium = 4.0282 u

Mass of helium + neutron

= 4.0247 u

Mass difference = 4.0282 u – 4.0247 u = 0.0035 u

Energy released = 0.0035 x 931.5 MeV = 3.26 MeV

Example

0Another fusion reaction is , where 4 h nuclei fuse into a He plus 2 positrons (electron antiparticles – same mass, opposite charge), 2 electron neutrinos and a photon. Calculate the energy released in this reaction.

4 x mass of hydrogen = 4.029104 u

Mass of helium + positrons + neutrinos

= (4.0026 – 2 x 0.0005486) + (2 x 0.0005486)

= 4.002600 u

Mass difference = 0.026504 u

Energy released= 0.026504 x 931.5 MeV

= 24.7 MeV

Nuclear Fusion

0For light nuclei to fuse, very high temperatures are required to overcome the electrostatic repulsion between nuclei.

0The enormous temperature causes the nuclei to move fast enough to approach each other sufficiently for fusion to occur.

0Temperatures > 107 K = plasma state (ionized atoms)0Plasma cannot contact other things – must be

contained by magnetic fields (tokamaks)

Tokamaks

Serious unsolved problems with prolonged confinement of

plasmas – not a viable source of commercial energy any time

soon

Fusion in Stars0The high temperatures and pressures in the interior of stars

make stars ideal places for fusion.0High temperature needed to overcome electric repulsion0High pressure ensures that sufficient numbers of nuclei are

found close to each other, increasing probability of fusion0 - common reaction in stellar cores 0Fusion = source of star’s energy, prevents it from collapsing

under its own weight and provides energy that star sends out as light and heat.

0Stars = element factories

Summary

Assignment

0Questions: 7,8,9,10,12