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Mass-Energy Equivalence

Mass-Energy Equivalence - Linville Mass... · Mass-Energy Equivalence • Work is required to remove a nucleon from ... • Determine the mass defect and binding energy of an alpha

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Mass-Energy

Equivalence

• Work is required to remove a nucleon from

a stable nucleus because of the strong

nuclear force

• The binding energy of a nucleus is the

energy required to separate all of its

nucleons and move them infinitely far

apart

The maximum binding energy per nucleon is

between A =50 and A =74 (most stable)

Comparison of Atomic & Nuclear

binding energy

Mass Defect

• The mass of a nucleus is always less than

the mass of all the separate nucleons

(protons and neutrons)

• this difference in mass is called the mass

defect

• the binding energy is related to the mass

defect by the equation E = mc2

Example

• Determine the mass defect and binding

energy of an alpha particle.

Solution • alpha particle mass (2 protons, 2 neutrons) =

6.65 x 10-27 kg (data sheet)

• mass of 2 protons = 2 x 1.67 x 10-27 kg = 3.34 x 10-27 kg

• mass of 2 neutrons = 2 x 1.67 x 10-27 kg = 3.34 x 10-27 kg

• total mass of separate nucleons = 6.68 x 10-27

kg

• mass defect = separate nucleon mass -

mass = 0.03 x 10-27 kg

• E = mc2

• E = (0.03 x 10-27 kg)(3.00 x 108 m/s)2

• E = 2.70 x 10-12 J

• in nuclear reactions, mass is converted to

energy or energy is converted to mass

• Conservation of mass-energy

principle

Example

• Calculate the energy produced in the reaction

2H mass = 3.34341 x 10-27 kg 3H mass = 5.00661 x 10-27 kg

Solution

• Total mass of reactants = 8.35002 x 10-27 kg

• The total mass of the products = 8.3212x10-27 kg

• Mass defect

• =8.35002x10-27 kg – 8.3212x10-27 kg

• = 2.882 x 10-29 kg

• E = mc2

• E = (2.882 x 10-29 kg)(3.00 x 108 m/s)2

• E = 2.59 x 10-12 J

• in the form of kinetic energy of products

STS

• In a CANDU reactor, 1 kg of fuel (natural

uranium) produces 3.4 x 105 MJ of heat

that is converted to electricity

• in oil and coal power plants 1 kg of fuel

produces about 4 MJ of heat

Pair Production

• a very high energy photon may create

matter

• The process must produce 2 particles

whose total charge is zero, since charge &

momentum must be conserved.

• A particle and its antiparticle (antimatter) are

often produced (i.e. an electron and antielectron)

have the same mass and other properties, but

opposite signs

Bubble chamber track

Neutral particles do not leave a

track

Example

• A 8.50 x 1020 Hz photon produces an

electron and an anti-electron. Determine

the total kinetic energy of the particles.

• Conservation of Mass-Energy!!!

Solution

• photon energy = energy to make particles + Ek

• Ephoton = Eelectron + Eantielectron + Ek

• hf = mc2 + mc2 + Ek

• Ek =(6.63x10-34Js)(8.50x1020) - 2(9.11x10-31kg)c2

• = 4.00 x 10-13 J

Practice

• P 796: 1, 4, 7, 10

• See p 792 for the data table