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Mass and Energy

A LevelAQA

Mass and Energy

E=mc2E=mc^2 is an equation most people have heard of but many may not know it’s meaning or relevance. In this section we look at this equation and the link between energy and mass. 

Mass-Energy Equivalence

While experimenting on his famous theory of relativity, Einstein proposed mass and energy can be considered equivalent and are interchangeable. 

This idea is represented by the equation:

E=mc2E=mc^2

  • E=E= energy in joules (J)text{(J)}
  • m=m= mass in kilograms (kg)text{(kg)}
  • c=c= the speed of light (=3×108 ms−1)(=3times 10^8 text{ ms}^{-1})

The most useful practice from this is the ability to convert mass to energy. For a small amount of mass, a huge amount of energy can be produced due to the c2c^2 in the equation. 

Because of the production of huge amounts of energy, the mass-energy equivalence can be put to use in nuclear weapons and nuclear power, nuclear fusion in the sun and high energy collisions in particle accelerators. 

A LevelAQA
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Atomic Mass Unit

The atomic mass unit (u) is often used in nuclear physics instead of dealing with incredibly small masses in (kg)text{(kg)}. The a.m.u is equal to 112thdfrac{1}{12}th of the mass of a carbon−12-12 atom. The conversion you need to use is: (this is given on your data sheet)

1 u=1.661×10−27 kg1 text{ u}=1.661 times 10^{-27} text{ kg} 

Example: An electron has a mass of 5.5×10−4 u5.5 times 10^{-4} text{ u}. Convert this mass to kilograms.

[1 mark]

1 u=1.66×10−27 kg5.5×10−4×1.661×10−27=9.1355×10−31 kg1 text{ u} = 1.66 times 10^{-27} text{ kg} textcolor{10a6f3}{5.5 times 10^{-4}} times 1.661 times 10^{-27} = bold{9.1355 times 10^{-31}} textbf{ kg}

 

A LevelAQA

Mass Defect

When scientists measured the mass of the nucleus of an atom as a whole and then they measured the mass of the nucleus separated into its constituents and compared the two, they found that the mass of a nucleus as a whole was always less than the mass of its constituents. They named the difference in mass the mass defect. 

The diagram below shows a representation of carbon−12-12 as a whole nucleus on the left and separated into protons and neutrons on the right. If measured, the mass of the nucleus on the left will always be less than the mass of the protons and neutrons on the right. The difference is the mass defect of carbon−12-12.

The mass defect (Δm)(Delta m) can be calculated using the equation:

Δm=Zmp+(A−Z)mn−mtotalDelta m= Z m_p + (A-Z) m_n – m_{text{total}}

  • Δm=Delta m= the mass defect in kilograms kg)text{kg})
  • Z=Z= the number of protons
  • mp=m_p= the mass of a proton (=1.67×10−27 kg=1.00728 u)(=1.67 times 10^{-27} text{ kg}=1.00728 text{ u})
  • A=A= the nucleon number
  • mn=m_n= the mass of a neutron (=1.67×10−27 kg=1.00867 u)(= 1.67 times 10^{-27} text{ kg} = 1.00867 text{ u})
  • mtotal=m_{text{total}}= the mass of the nucleus as a whole

This equation can be simplified to:

mass defect=total mass of protons+total mass of neutrons−total mass of the nucleus as a wholetext{mass defect}=text{total mass of protons} + text{total mass of neutrons} – text{total mass of the nucleus as a whole}

Example: The mass of iron−56-56 is 55.845 u55.845 text{ u}. Calculate the mass defect of iron−56-56. Give your answer in kilograms. 

[3 marks]

Find A and Z for iron−56-56:

Z=26Z=26 (from datasheet)

A=56−26=30A=textcolor{7cb447}{56}-26=30

Substitute into the mass defect equation:

Δm=Zmp+(A−Z)mn−mtotal=(26×1.00728)+(30×1.00867)−55.845=0.6043 u=0.06043×1.661×10−27=1.0039×10−27 kgbegin{aligned} bold{Delta m} &= bold{Z m_p + (A-Z) m_n – m_{text{total}}} &= (26 times 1.00728)+(30times 1.00867) – textcolor{ffad05}{55.845} &= 0.6043 text{ u} &= bold{0.06043 times 1.661 times 10 {-27}} &= bold{1.0039 times 10^{-27}} textbf{ kg} end{aligned}

 

A LevelAQA
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Binding Energy

Binding energy is the energy needed to separate a nucleus into its components. The mass of the components is always greater than the mass of the nucleus as an energy input is needed to separate the nucleus into its components, and mass and energy are interchangeable. Therefore the input of energy needed to separate the components becomes the gain in mass. 

This also means that when a nucleus is formed, the equal but opposite amount of energy is released. The amount of energy can be calculated using the equation:

E=Δmc2E=Delta m c^2

  • E=E= energy in joules (J)text{(J)}
  • Δm=Delta m= change in mass in kilograms (kg)text{(kg)}
  • c=c= the speed of light (=3×108ms−1)(=3 times 10^8 text{ms}^{-1}) 

Example: Using the previous example, calculate the binding energy per nucleon (EA)(dfrac{E}{A})of iron−56-56. Δm=1.0039×10−27 kgDelta m = 1.0039 times 10^{-27} text{ kg}.

[2 marks]

E=Δmc2= 1.0039×10−27×(3×108)29.035×10−11JEA=9.035×10−1156=1.6×10−12 Jbegin{aligned} bold{E} &= bold{Delta mc^2} &=  textcolor{aa57ff}{1.0039 times 10^{-27}} times (3times 10^8)^2 9.035 times 10^{-11} text{J} dfrac{E}{A} &= dfrac{9.035 times 10^{-11}}{textcolor{f95d27}{56}} &= bold{1.6 times 10^{-12}} textbf{ J} end{aligned} 

 

A LevelAQA

Mass and Energy Example Questions

Question 1: What is the mass defect and how is it calculated?

[2 marks]

A Level AQA

The difference in mass between the nucleus of an atom and the mass of its constituents when separated.

Δm=Zmp+(A−Z)mn−mtotalbold{Delta m = Z m_p + (A-Z)m_n – m_{text{total}}}

where:

  •  ΔmDelta m is the mass defect
  • ZZ is the number of protons
  • mpm_p is the mass of a proton
  • AA is the nucleon number
  • mnm_n is the mass of a neutron
  • mtotalm_{text{total}} is the mass of the nucleus as a whole
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Question 2: The mass of carbon−12-12 is 12.011 u12.011 text{ u}. Calculate the mass defect of carbon−12-12. (Mass of proton =1.00728 u=1.00728 text{ u} and mass of a neutron =1.00867 u=1.00867 text{ u}

[3 marks]

A Level AQA

Δm=Zmp+(A−Z)mn−mtotal=(6×1.00728)+(6×1.00867)−12.011=0.0847 u= 0.0847×1.661×10−27=1.41×10−28 kgbegin{aligned} bold{Delta m} &= bold{Zm_p + (A-Z)m_n – m_{text{total}} } &= (6 times 1.00728) + (6 times 1.00867) – 12.011 &= bold{0.0847} textbf{ u} &=  0.0847 times 1.661 times 10^{-27} &= bold{1.41 times 10^{-28}} textbf{ kg} end{aligned}

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Question 3: Define the binding energy of a nucleus. 

[1 mark]

A Level AQA

Binding energy is the energy needed to separate a nucleus into its components.

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Specification Points Covered

AQA A-level: 

  • 3.8.1.6 Mass and energy (A-level only)
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