Mass defect and nuclear binding energy
Why is a helium-4 atom lighter than two hydrogen atoms and two neutrons put together, even though it consists of exactly those particles?
Prior knowledge
- A nucleus consists of protons and neutrons, and in a decay equation the mass number and the charge number are conservedchapter The atomic nucleus and radioactive decay
- Energy can be given in electronvolts: $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$chapter The photoelectric effect and the photon
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 4.
Think about it
A helium-4 atom contains two protons, two neutrons and two electrons. The same particles are in two hydrogen atoms and two neutrons, and their total mass is $4.032980\,\text{u}$. The mass of the helium-4 atom, however, is only $4.002603\,\text{u}$. The difference is less than one per cent, but the masses are known to a millionth, so the difference is real.
Where has the missing mass of helium-4 gone when protons and neutrons join to form a nucleus?
Optional Review: alpha decay
Predict
Make your predictions before investigating 2 questions
Prediction 1: from light to heavy
Prediction 2: helium or lithium
Investigate
Choose a nucleus. The balance compares the mass of the separate nucleons with the mass of the atom, and the graph shows the binding energy per nucleon of every nucleus you have studied.
Simulation
How to use
Choose a nuclide from the Nucleus menu. In the left pan are the $Z$ hydrogen atoms and $N$ neutrons of the nucleus apart (red proton, grey neutron), in the right pan the same atom whose nucleons are bound into a compact nucleus. Under the pans are their masses and at the top the mass defect $\Delta m$. The balance tips towards the heavier side, that is, the separate nucleons; the tilt is exaggerated so that it can be seen. The graph keeps a point for every nucleus you have studied, and the current nucleus is circled. Show all nuclei also draws the other nuclei of the table as pale rings, and Clear the graph removes the studied points.
Assumptions of the model
The masses of the atoms are values from the atomic mass table AME2020 (IAEA Atomic Mass Data Center) to six decimals. The mass defect is calculated from the masses of hydrogen atoms ${}^{1}\text{H}$, not protons, so that the masses of the electrons cancel. The very small binding energy of the electrons themselves in the atom is not taken into account. The table has $32$ nuclides, one or two of each element. The readings are model values calculated from these masses, not your own measurements.
How the calculation works
The mass of the separate nucleons is $Z\,m_\text{H} + N\,m_\text{n}$, where $m_\text{H} = 1.007825\,\text{u}$ and $m_\text{n} = 1.008665\,\text{u}$. The mass defect is $\Delta m = Z\,m_\text{H} + N\,m_\text{n} - m_\text{atom}$, and the binding energy is obtained by putting $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$ into the equation $E_\text{B} = \Delta m\,c^2$. On the vertical axis of the graph is $\tfrac{E_\text{B}}{A}$ and on the horizontal axis the mass number $A$.
Experiment 1: light nuclei
Test prediction 2. Choose the nuclides in the order hydrogen-2 (deuterium), helium-4, lithium-7, carbon-12 and oxygen-16. Record for each the mass defect $\Delta m$, the binding energy $E_\text{B}$ and the binding energy per nucleon $\tfrac{E_\text{B}}{A}$, at least for the first three.
Initial state: nuclide: hydrogen-2. Kept constant: nothing. Varied: nuclide.
| nuclide | $\Delta m$ $(\text{u})$ | $E_\text{B}$ $(\text{MeV})$ | $E_\text{B}/A$ $(\text{MeV})$ |
|---|
Experiment 2: medium-heavy and heavy nuclei
Test prediction 1. Record the same quantities for the nuclides iron-56, nickel-62, silver-107, lead-208 and uranium-238, at least for three. Finally press Show all nuclei and look at the whole graph.
Initial state: nuclide: iron-56. Kept constant: nothing. Varied: nuclide.
| nuclide | $\Delta m$ $(\text{u})$ | $E_\text{B}$ $(\text{MeV})$ | $E_\text{B}/A$ $(\text{MeV})$ |
|---|
Conclusion: the binding energy curve
Write on the basis of your tables: How does the binding energy per nucleon change as the mass number increases, and for which nuclei is it highest? Which light nucleus differs from its neighbours? Which is more tightly bound, iron-56 or uranium-238, although the whole binding energy of uranium is much greater?
Model answer
As values of the model, the binding energy per nucleon rises steeply for light nuclei: hydrogen-2 $1.112\,\text{MeV}$, lithium-7 $5.607\,\text{MeV}$, carbon-12 $7.680\,\text{MeV}$ and oxygen-16 $7.976\,\text{MeV}$. Helium-4 is an exception: its value $7.074\,\text{MeV}$ is greater than that of the heavier lithium-7. The highest value is for medium-heavy nuclei, $8.795\,\text{MeV}$ for nickel-62 and $8.790\,\text{MeV}$ for iron-56. From there the value falls gently: silver-107 $8.554\,\text{MeV}$, lead-208 $7.868\,\text{MeV}$ and uranium-238 $7.570\,\text{MeV}$. The whole binding energy of uranium-238, $1801.7\,\text{MeV}$, is greater than that of iron, $492.3\,\text{MeV}$, because there are more nucleons, but per nucleon iron is more tightly bound. In light nuclei a nucleon has few neighbours to which the strong interaction binds it; in heavy nuclei the electric repulsion of the protons, which reaches across the whole nucleus, weakens the binding.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
Atomic mass unit
The masses of atoms and nuclei are given in atomic mass units. The atomic mass unit $\text{u}$ is defined so that the mass of a carbon-12 atom is exactly $12\,\text{u}$, which gives $1\,\text{u} = 1.66054 \cdot 10^{-27}\,\text{kg}$. The masses of the proton and the neutron are each about $1\,\text{u}$, so the mass of an atom in atomic mass units is close to its mass number $A$ but not exactly equal to it. The values in the table are masses of atoms: they include the electrons too. The mass of a hydrogen atom ${}^{1}\text{H}$ is the mass of a proton plus the mass of an electron.
| Particle | Symbol | Mass ($\text{u}$) |
|---|---|---|
| hydrogen-1 | ${}^{1}\text{H}$ | $1.007825$ |
| neutron | $\text{n}$ | $1.008665$ |
| hydrogen-2 | ${}^{2}\text{H}$ | $2.014102$ |
| helium-4 | ${}^{4}\text{He}$ | $4.002603$ |
| carbon-12 | ${}^{12}\text{C}$ | $12.000000$ |
| oxygen-16 | ${}^{16}\text{O}$ | $15.994915$ |
| iron-56 | ${}^{56}\text{Fe}$ | $55.934936$ |
| uranium-238 | ${}^{238}\text{U}$ | $238.050787$ |
Source: the atomic mass table AME2020 (Atomic Mass Evaluation 2020, IAEA Atomic Mass Data Center), rounded to six decimals. This many decimals are needed for nuclear masses because the important differences are small.
Mass defect
A helium-4 atom contains two protons, two neutrons and two electrons. The same particles are in two hydrogen atoms and two neutrons, whose total mass is $2 \cdot 1.007825\,\text{u} + 2 \cdot 1.008665\,\text{u} = 4.032980\,\text{u}$. The mass of the helium-4 atom, however, is only $4.002603\,\text{u}$. When the nucleons are bound into a nucleus, their mass is smaller than when they are apart. The difference $4.032980\,\text{u} - 4.002603\,\text{u} = 0.030377\,\text{u}$ is the mass defect $\Delta m$ of the nucleus, about $0.76\,\%$ of the mass of the atom. Hydrogen atoms are used instead of separate protons so that the masses of the electrons cancel: there are two electrons on both sides.
Mass and energy
The mass defect is not matter that has vanished. According to Einstein's theory of relativity, mass and energy are equivalent: a change $\Delta E$ in the energy of an object changes its mass by $\tfrac{\Delta E}{c^2}$, where $c$ is the speed of light. When protons and neutrons bind into a nucleus, the strong interaction pulls them together and energy is released, for example as gamma photons. The energy, and therefore also the mass, of the resulting nucleus is smaller than that of the separate nucleons. The same energy is needed if the nucleus is broken back into separate nucleons. This energy is the binding energy $E_\text{B}$ of the nucleus.
Why a change of mass is not noticed in everyday life
The equivalence of mass and energy applies to every change of energy, but $c^2$ is so large that in ordinary phenomena the change of mass is too small to measure. A mass of one gram corresponds to the energy
$$\begin{aligned} E &= 1.00 \cdot 10^{-3}\,\text{kg} \\ &\quad \cdot (2.998 \cdot 10^{8}\,\text{m/s})^2 \\ &\approx 9.0 \cdot 10^{13}\,\text{J}. \end{aligned}$$
When carbon burns, the mass decreases by only about one ten-billionth. In nuclei the binding energy is so large that the mass defect is about $0.8\,\%$ of the mass of the atom for helium-4 and uranium-238 alike. This is why nuclear physics calculates energy directly from the difference of masses.
Binding energy
Mass defect and binding energy
A nucleus ${}^{A}_{Z}\text{X}$ has $Z$ protons and $N = A - Z$ neutrons. The mass defect is calculated from atomic masses:
$$\Delta m = Z\,m_\text{H} + N\,m_\text{n} - m_\text{atom}$$
where $m_\text{H}$ is the mass of a hydrogen atom ${}^{1}\text{H}$, $m_\text{n}$ the mass of a neutron and $m_\text{atom}$ the mass of the atom ${}^{A}_{Z}\text{X}$. The binding energy of a nucleus is the energy corresponding to the mass defect
$$E_\text{B} = \Delta m\,c^2$$
where $E_\text{B}$ is the binding energy and $c = 2.998 \cdot 10^{8}\,\text{m/s}$ the speed of light. The binding energy of the electrons themselves in the atom is not taken into account, because it is at most hundreds of kiloelectronvolts, and even for the heaviest atoms less than a thousandth of the binding energy of the nucleus.
The atomic mass unit as energy
The energies of nuclei are given in megaelectronvolts: $1\,\text{MeV} = 10^{6}\,\text{eV}$ and $1\,\text{eV} = 1.602177 \cdot 10^{-19}\,\text{J}$. Let us calculate the energy that corresponds to a mass of $1\,\text{u}$. Because the mass defect is a small difference, six significant figures are used here for the constants:
$$\begin{aligned} E &= 1\,\text{u} \cdot c^2 \\ &= 1.66054 \cdot 10^{-27}\,\text{kg} \\ &\quad \cdot (2.99792 \cdot 10^{8}\,\text{m/s})^2 \\ &= 1.49241 \cdot 10^{-10}\,\text{J} \\ &= \frac{1.49241 \cdot 10^{-10}}{1.602177 \cdot 10^{-19}}\,\text{eV} \\ &= 9.3149 \cdot 10^{8}\,\text{eV} \\ &\approx 931.5\,\text{MeV} \end{aligned}$$
So $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$. When the mass defect is in atomic mass units, this is put in place of the unit, and $c^2$ cancels: for example $0.001000\,\text{u} \cdot c^2 = 0.001000 \cdot 931.5\,\tfrac{\text{MeV}}{c^2} \cdot c^2 = 0.9315\,\text{MeV}$. With four-figure constants $1.661 \cdot 10^{-27}\,\text{kg}$, $2.998 \cdot 10^{8}\,\text{m/s}$ and $1.602 \cdot 10^{-19}\,\text{J}$ the result would be $931.9\,\text{MeV}$, so four-figure constants already change the fourth figure.
Binding energy per nucleon
The strength of the binding of nuclei is compared with the binding energy per nucleon
$$\frac{E_\text{B}}{A}$$
where $A$ is the mass number, that is, the number of nucleons. The graph shows it calculated from atomic masses for $32$ nuclei. For light nuclei the value rises steeply, because each nucleon has more and more neighbours to which the strong interaction binds it. Helium-4 is exceptionally tightly bound. The value is highest, about $8.8\,\text{MeV}$, for iron-56 and nickel-62. For heavier nuclei it falls gently, because the strong interaction acts only between nearest neighbours, but the electric repulsion of the protons reaches across the whole nucleus.
The graph shows when a nuclear reaction releases energy: when the nuclei formed are more tightly bound than the original ones, the binding energy increases and the difference is released. The joining of light nuclei, fusion, and the splitting of a heavy nucleus, fission, move nucleons towards the peak of the graph. They are discussed in the chapter Fission, fusion and nuclear power.
Reaction energy
In a nuclear reaction nuclei change: a nucleus decays, or another nucleus or particle hits a nucleus. In a reaction the mass number and the charge number are conserved (chapter The atomic nucleus and radioactive decay), but mass is not. The reaction energy is
$$Q = (m_\text{initial} - m_\text{final})\,c^2$$
where $m_\text{initial}$ is the sum of the masses of the starting materials and $m_\text{final}$ the sum of the masses of the particles formed. When $Q > 0$, energy is released as the kinetic energy of the particles formed and as gamma radiation; when $Q < 0$, the reaction needs energy. Atomic masses can be used when the number of electrons is the same on both sides, as in alpha decay, beta minus decay and most nuclear reactions. The mass of the electron formed in beta minus decay is not added separately, because the atomic mass of the daughter atom already includes one more electron. For example, in the alpha decay of uranium-238, ${}^{238}_{92}\text{U} \to {}^{234}_{90}\text{Th} + {}^{4}_{2}\text{He}$, the mass decreases by $0.004583\,\text{u}$, so $Q = 0.004583 \cdot 931.5\,\text{MeV} \approx 4.27\,\text{MeV}$.
Common misconception
Many think that binding energy is energy stored in the nucleus, which is released when the nucleus is broken into pieces. The opposite is true. The binding energy was released when the nucleus formed, and the same energy is needed to break the nucleus into separate nucleons. This is why the nucleus is lighter than its parts. A large binding energy per nucleon means a tightly bound, stable nucleus, not a store of energy. In a nuclear reaction energy is released when the total binding energy of the nuclei formed is greater than that of the original ones.
Another mistake is to think that mass is conserved in a nuclear reaction as in chemistry. In a nuclear reaction the mass number, that is, the number of nucleons, and the charge number are conserved, but the mass changes by the amount of the energy released, $\Delta m = \tfrac{Q}{c^2}$.
Summary
- Atomic masses are given in atomic mass units: $1\,\text{u}$ is $\tfrac{1}{12}$ of the mass of a carbon-12 atom, that is, $1\,\text{u} = 1.66054 \cdot 10^{-27}\,\text{kg}$.
- A nucleus is lighter than its separate nucleons. The mass defect is $\Delta m = Z\,m_\text{H} + N\,m_\text{n} - m_\text{atom}$.
- The binding energy $E_\text{B} = \Delta m\,c^2$ is needed to break a nucleus into nucleons, and it is released when a nucleus forms. Putting in $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$ gives the energy in megaelectronvolts.
- The binding energy per nucleon is highest, about $8.8\,\text{MeV}$, for iron and nickel; fusion and fission release energy because they move nucleons into more tightly bound nuclei.
- The reaction energy is $Q = (m_\text{initial} - m_\text{final})\,c^2$: mass is not conserved in a nuclear reaction, although the mass number and the charge number are.
Optional Return to your predictions
Compare your predictions with the model values of the experiments. Then return to the opening question: where has the mass $0.030377\,\text{u}$ missing from the helium-4 atom gone, and why is helium-4 lighter than its parts?
Would you change your prediction?
Apply
Worked example
Problem
Calculate the binding energy and the binding energy per nucleon of the helium-4 nucleus ${}^{4}_{2}\text{He}$. Use the atomic masses $m({}^{4}\text{He}) = 4.002603\,\text{u}$, $m_\text{H} = 1.007825\,\text{u}$ and $m_\text{n} = 1.008665\,\text{u}$ (AME2020) and $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$.
Step 1/7
Known and asked. The atomic number is $Z = 2$ and the mass number $A = 4$, so the number of neutrons is $N = A - Z = 2$. The atomic masses are $m_\text{atom} = 4.002603\,\text{u}$, $m_\text{H} = 1.007825\,\text{u}$ and $m_\text{n} = 1.008665\,\text{u}$. The binding energy $E_\text{B}$ and the binding energy per nucleon $\tfrac{E_\text{B}}{A}$ are asked in megaelectronvolts.
Step 2/7
Principle. The nucleus is lighter than its separate nucleons, and the mass defect $\Delta m$ corresponds to the binding energy $E_\text{B} = \Delta m\,c^2$. We calculate with atomic masses: two hydrogen atoms contain two electrons, just like the helium-4 atom, so the masses of the electrons cancel.
Step 3/7
Formula solved. The mass defect, the binding energy and its share per nucleon:
$$\begin{aligned} \Delta m &= Z\,m_\text{H} + N\,m_\text{n} - m_\text{atom} \\ E_\text{B} &= \Delta m\,c^2 \\ \frac{E_\text{B}}{A} &= \frac{\Delta m\,c^2}{A} \end{aligned}$$
Step 4/7
Unit conversions. The mass is not converted to kilograms: $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$ is put in place of the unit of the mass defect, so $c^2$ cancels and the energy is obtained directly in megaelectronvolts.
Step 5/7
Substitution. First the mass defect, then the energy:
$$\begin{aligned} \Delta m &= Z\,m_\text{H} + N\,m_\text{n} - m_\text{atom} \\ &= 2 \cdot 1.007825\,\text{u} \\ &\quad + 2 \cdot 1.008665\,\text{u} \\ &\quad - 4.002603\,\text{u} \\ &= 0.030377\,\text{u} \\[6pt] E_\text{B} &= \Delta m\,c^2 = 0.030377\,\text{u} \cdot c^2 \\ &= 0.030377 \cdot 931.5\,\tfrac{\text{MeV}}{c^2} \cdot c^2 \\ &= 28.296\,\text{MeV} \\[6pt] \frac{E_\text{B}}{A} &= \frac{28.296\,\text{MeV}}{4} \\ &= 7.074\,\text{MeV} \end{aligned}$$
Step 6/7
Result. The binding energy of the helium-4 nucleus is $28.30\,\text{MeV}$ and the binding energy per nucleon $7.074\,\text{MeV}$, four significant figures like the conversion factor $931.5\,\tfrac{\text{MeV}}{c^2}$.
Step 7/7
Reasonableness. With rounded numbers $\Delta m \approx 0.03\,\text{u}$ and $0.03 \cdot 900\,\text{MeV} = 27\,\text{MeV}$, close to the result. The mass defect is about $0.8\,\%$ of the mass of the atom, as it usually is for nuclei, and the value $7.07\,\text{MeV}$ per nucleon is the helium-4 point of the graph, clearly higher than its neighbours.
Optional Find the error
Transfer: the energy of an alpha decay
Older models of smoke detectors and the heat sources of space probes have used alpha emitters. Polonium-210 ${}^{210}_{84}\text{Po}$ decays by alpha decay into lead-206. The atomic masses are $m({}^{210}\text{Po}) = 209.982874\,\text{u}$, $m({}^{206}\text{Pb}) = 205.974465\,\text{u}$ and $m({}^{4}\text{He}) = 4.002603\,\text{u}$ (AME2020).
How much energy is released in one decay? Give the answer in megaelectronvolts.
Hint 1: a guiding question
Write the decay equation. Which particles are at the start and which at the end?
Hint 2: what is the connection
The reaction energy is $Q = (m_\text{initial} - m_\text{final})\,c^2$, and at the end there are both lead-206 and the alpha particle, that is, helium-4.
Hint 3: an intermediate step
The difference of the masses is $0.005806\,\text{u}$. Put in $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$.
Show the solution
Given
$m({}^{210}\text{Po}) = 209.982874\,\text{u}$, $m({}^{206}\text{Pb}) = 205.974465\,\text{u}$, $m({}^{4}\text{He}) = 4.002603\,\text{u}$ (AME2020), $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$
Asked
the energy released in the decay, that is, the reaction energy $Q$ in megaelectronvolts
Principle
In the alpha decay ${}^{210}_{84}\text{Po} \to {}^{206}_{82}\text{Pb} + {}^{4}_{2}\text{He}$ the mass number and the charge number are conserved, but the mass of the atoms formed is smaller than that of the polonium atom. The energy corresponding to the decrease of mass is released, mainly as the kinetic energy of the alpha particle. Atomic masses can be used, because there are $84$ electrons on both sides: $82$ in the lead atom and $2$ in the helium atom.
Formula
$Q = (m_\text{initial} - m_\text{final})\,c^2$, where $m_\text{initial} = m({}^{210}\text{Po})$ and $m_\text{final} = m({}^{206}\text{Pb}) + m({}^{4}\text{He})$
Unit conversions
The mass is not converted to kilograms: $931.5\,\tfrac{\text{MeV}}{c^2}$ is put in place of the unit $\text{u}$, so $c^2$ cancels.
Substitution
so
Answer
$5.408\,\text{MeV}$
Four significant figures like the mass difference $0.005806\,\text{u}$.
Reasonableness
$Q > 0$, as it must be in a spontaneous decay. Because of the conservation of momentum, the light alpha particle gets about the share $\tfrac{206}{210}$ of the energy, that is, about $5.30\,\text{MeV}$; the measured energy of the alpha particles, $5.30\,\text{MeV}$, agrees. The result is of the same order of magnitude as the $4.27\,\text{MeV}$ of the alpha decay of uranium-238.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- The binding energy of the bismuth-209 nucleus is $1640\,\text{MeV}$ and that of the calcium-40 nucleus $342\,\text{MeV}$. Which nucleus is more tightly bound?
Answer
Calcium-40. The strength of binding is described by the binding energy per nucleon: for bismuth $1640\,\text{MeV} : 209 \approx 7.85\,\text{MeV}$ and for calcium $342\,\text{MeV} : 40 \approx 8.55\,\text{MeV}$. The whole binding energy of bismuth is greater only because it has more nucleons.
- Why is the mass of a hydrogen atom used in calculating the mass defect, and not the mass of a proton, when the mass of the nucleus is taken to be the mass of the atom?
Answer
The mass of an atom includes its $Z$ electrons. When $Z$ hydrogen atoms are used as the separate particles, they also contain $Z$ electrons, so the masses of the electrons cancel in the difference and what is left is the mass defect of the nucleus.
1. Concept
2. Read the binding energy curve
Optional 3. Explain and compare
When carbon burns, $\text{C} + \text{O}_2 \to \text{CO}_2$, energy of $393.5\,\text{kJ}$ is released for every $44.0\,\text{g}$ of starting materials.
Does the mass change in burning? Why is it said in chemistry that mass is conserved, whereas in nuclear reactions the energy is calculated from the change of mass?
Model answer
The mass changes, because the equivalence of mass and energy applies to every change of energy. The mass of the combustion products is smaller by $\Delta m = \tfrac{E}{c^2} = \tfrac{393.5 \cdot 10^{3}\,\text{J}}{(2.998 \cdot 10^{8}\,\text{m/s})^2} \approx 4.4 \cdot 10^{-12}\,\text{kg}$, which is about $10^{-10}$, a ten-billionth of the mass of the starting materials. Such a small change cannot be weighed, so in chemistry mass is conserved in practice. In nuclear reactions the relative change of mass is hundreds of thousands or millions of times greater: about $2 \cdot 10^{-5}$ in the alpha decay of uranium-238 and about $10^{-3}$ of the mass of the starting materials in the fission of uranium-235. This is why the energy of a nuclear reaction is calculated from the difference of masses.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 10-1Core
The mass of a carbon-12 atom is exactly $12.000000\,\text{u}$. The mass of a hydrogen atom is $m_\text{H} = 1.007825\,\text{u}$ and that of a neutron $m_\text{n} = 1.008665\,\text{u}$.
What is the mass defect of the carbon-12 nucleus ${}^{12}_{6}\text{C}$? Give the answer in atomic mass units.
Answer
$0.098940\,\text{u}$
Show the solution
Given
mass of the carbon-12 atom $m_\text{atom} = 12.000000\,\text{u}$, $m_\text{H} = 1.007825\,\text{u}$, $m_\text{n} = 1.008665\,\text{u}$; the nucleus ${}^{12}_{6}\text{C}$ has $Z = 6$ protons and $N = 12 - 6 = 6$ neutrons
Asked
the mass defect $\Delta m$ in atomic mass units
Principle
A nucleus is lighter than its separate nucleons. When the masses of hydrogen atoms are used, they contain six electrons just like the carbon atom, so the masses of the electrons cancel and the difference is the mass defect of the nucleus.
Formula
$\Delta m = Z\,m_\text{H} + N\,m_\text{n} - m_\text{atom}$
Unit conversions
None needed: all masses are in atomic mass units.
Substitution
$\Delta m$$= 6 \cdot 1.007825\,\text{u} + 6 \cdot 1.008665\,\text{u} - 12.000000\,\text{u}$$= 6.046950\,\text{u} + 6.051990\,\text{u} - 12.000000\,\text{u}$$= 0.098940\,\text{u}$Answer
$0.098940\,\text{u}$
Six decimals like the given masses.
Reasonableness
The mass of the separate nucleons is about $12.10\,\text{u}$, so the mass defect is about $0.8\,\%$ of it, as it usually is for nuclei. The result is positive, because a bound nucleus is lighter than its parts.
- Exercise 10-2Core
The mass defect of the oxygen-16 nucleus is $0.137005\,\text{u}$. What is the binding energy of the nucleus? Use $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$. Give the answer in megaelectronvolts.
Answer
$127.6\,\text{MeV}$
Show the solution
Given
the mass defect of the oxygen-16 nucleus $\Delta m = 0.137005\,\text{u}$, $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$
Asked
the binding energy $E_\text{B}$ in megaelectronvolts
Principle
By the equivalence of mass and energy, the mass defect corresponds to the energy $\Delta m\,c^2$. This is exactly the energy needed to break the nucleus into separate nucleons.
Formula
$E_\text{B} = \Delta m\,c^2$
Unit conversions
The mass is not converted to kilograms: $931.5\,\tfrac{\text{MeV}}{c^2}$ is put in place of the unit $\text{u}$, so $c^2$ cancels.
Substitution
$E_\text{B}$$= 0.137005\,\text{u} \cdot c^2$$= 0.137005 \cdot 931.5\,\tfrac{\text{MeV}}{c^2} \cdot c^2$$= 127.620\,\text{MeV}$Answer
$127.6\,\text{MeV}$
Four significant figures like the conversion factor $931.5\,\tfrac{\text{MeV}}{c^2}$.
Reasonableness
By estimation $0.14 \cdot 900\,\text{MeV} \approx 126\,\text{MeV}$. Per nucleon this is $127.6\,\text{MeV} : 16 \approx 8.0\,\text{MeV}$, which fits the binding energy graph at oxygen-16.
- Exercise 10-3
The binding energy of the uranium-238 nucleus ${}^{238}_{92}\text{U}$ is $1801.7\,\text{MeV}$. What is its binding energy per nucleon? Give the answer in megaelectronvolts.
Answer
$7.570\,\text{MeV}$
Show the solution
Given
the binding energy of the uranium-238 nucleus $E_\text{B} = 1801.7\,\text{MeV}$, mass number $A = 238$
Asked
the binding energy per nucleon $\tfrac{E_\text{B}}{A}$ in megaelectronvolts
Principle
The binding energy per nucleon tells how much energy is needed on average to remove one nucleon. It is obtained by dividing the whole binding energy by the number of nucleons, that is, the mass number.
Formula
$E_\text{B}$$= A \cdot \dfrac{E_\text{B}}{A}$so the asked quantity is $\dfrac{E_\text{B}}{A}$, the binding energy divided by the mass number
Unit conversions
None needed: the energy is already in megaelectronvolts, and the mass number is a pure number.
Substitution
$\dfrac{E_\text{B}}{A}$$= \dfrac{1801.7\,\text{MeV}}{238}$$= 7.5702\,\text{MeV}$Answer
$7.570\,\text{MeV}$
Four significant figures like the conversion factor $931.5\,\tfrac{\text{MeV}}{c^2}$ with which the binding energy $1801.7\,\text{MeV}$ was calculated.
Reasonableness
In the binding energy graph the value for the heaviest nuclei is about $7.6\,\text{MeV}$, less than the about $8.8\,\text{MeV}$ of iron. The estimate $1800 : 240 = 7.5$ agrees.
- Exercise 10-4
What energy corresponds to a mass of $1.00\,\text{g}$? The speed of light is $c = 2.998 \cdot 10^{8}\,\text{m/s}$. Give the answer in the unit $10^{13}\,\text{J}$.
Answer
$8.99 \cdot 10^{13}\,\text{J}$
Show the solution
Given
mass $m = 1.00\,\text{g}$, speed of light $c = 2.998 \cdot 10^{8}\,\text{m/s}$
Asked
the energy $E$ corresponding to the mass, in the unit $10^{13}\,\text{J}$
Principle
Mass and energy are equivalent: a mass $m$ corresponds to the energy $mc^2$. We calculate in SI units so that the energy is obtained in joules.
Formula
$E = mc^2$
Unit conversions
$1.00\,\text{g} = 1.00 \cdot 10^{-3}\,\text{kg}$, and $1\,\text{kg} \cdot \tfrac{\text{m}^2}{\text{s}^2} = 1\,\text{J}$.
Substitution
$E$$= 1.00 \cdot 10^{-3}\,\text{kg} \cdot (2.998 \cdot 10^{8}\,\text{m/s})^2$$= 1.00 \cdot 10^{-3}\,\text{kg} \cdot 8.988 \cdot 10^{16}\,\tfrac{\text{m}^2}{\text{s}^2}$$= 8.988 \cdot 10^{13}\,\text{J}$Answer
$8.99 \cdot 10^{13}\,\text{J}$
Three significant figures like the mass $1.00\,\text{g}$.
Reasonableness
By estimation $10^{-3} \cdot (3 \cdot 10^{8})^2 = 9 \cdot 10^{13}$. The energy is enormous: since
$1\,\text{GWh}$$= 10^{9}\,\text{W} \cdot 3600\,\text{s}$$= 3.6 \cdot 10^{12}\,\text{J}$it is about $25\,\text{GWh}$, so a nuclear power plant unit with an electric power of $1600\,\text{MW}$ produces it as electricity in about $16\,\text{h}$.
- Exercise 10-5
The binding energy of the deuterium nucleus ${}^{2}_{1}\text{H}$ is $2.224\,\text{MeV}$. What is its mass defect? Use $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$. Give the answer in the unit $10^{-3}\,\text{u}$.
Answer
$2.388 \cdot 10^{-3}\,\text{u}$
Show the solution
Given
the binding energy of the deuterium nucleus $E_\text{B} = 2.224\,\text{MeV}$, $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$
Asked
the mass defect $\Delta m$ in the unit $10^{-3}\,\text{u}$
Principle
The binding energy is the energy corresponding to the mass defect. When the energy is known, the mass defect is obtained by dividing it by $c^2$.
Formula
$E_\text{B} = \Delta m\,c^2$, from which $\Delta m = \dfrac{E_\text{B}}{c^2}$
Unit conversions
$1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$, that is, $1\,\tfrac{\text{MeV}}{c^2} = \tfrac{1}{931.5}\,\text{u}$; finally $1\,\text{u} = 1000 \cdot 10^{-3}\,\text{u}$.
Substitution
$\Delta m$$= \dfrac{2.224\,\text{MeV}}{c^2}$$= 2.224 \cdot \tfrac{1}{931.5}\,\text{u}$$= 0.0023876\,\text{u}$$= 2.3876 \cdot 10^{-3}\,\text{u}$Answer
$2.388 \cdot 10^{-3}\,\text{u}$
Four significant figures like the binding energy and the conversion factor.
Reasonableness
Deuterium has only two nucleons, so the mass defect is small: about $0.12\,\%$ of the atomic mass $2.014\,\text{u}$, much less than for heavier nuclei. Check: $0.002388 \cdot 931.5\,\text{MeV} = 2.224\,\text{MeV}$.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 10-6Core
The mass of an iron-56 atom ${}^{56}_{26}\text{Fe}$ is $55.934936\,\text{u}$. The mass of a hydrogen atom is $m_\text{H} = 1.007825\,\text{u}$ and that of a neutron $m_\text{n} = 1.008665\,\text{u}$.
What is the binding energy per nucleon of the iron-56 nucleus? Use $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$. Give the answer in megaelectronvolts.
Answer
$8.790\,\text{MeV}$
Show the solution
Given
mass of the iron-56 atom $m_\text{atom} = 55.934936\,\text{u}$, $Z = 26$, $A = 56$, so $N = 30$; $m_\text{H} = 1.007825\,\text{u}$, $m_\text{n} = 1.008665\,\text{u}$, $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$
Asked
the binding energy per nucleon $\tfrac{E_\text{B}}{A}$ in megaelectronvolts
Principle
The mass defect of the nucleus corresponds to the binding energy $E_\text{B} = \Delta m\,c^2$. The masses of hydrogen atoms are used so that the masses of $26$ electrons cancel with the electrons of the iron atom. Finally we divide by the number of nucleons.
Formula
$\Delta m = Z\,m_\text{H} + N\,m_\text{n} - m_\text{atom}$ and $E_\text{B} = \Delta m\,c^2$, so
$\dfrac{E_\text{B}}{A}$$= \dfrac{(Z\,m_\text{H} + N\,m_\text{n} - m_\text{atom})\,c^2}{A}$Unit conversions
$931.5\,\tfrac{\text{MeV}}{c^2}$ is put in place of the unit $\text{u}$, so $c^2$ cancels.
Substitution
$\Delta m$$= 26 \cdot 1.007825\,\text{u} + 30 \cdot 1.008665\,\text{u} - 55.934936\,\text{u}$$= 26.203450\,\text{u} + 30.259950\,\text{u} - 55.934936\,\text{u}$$= 0.528464\,\text{u}$so
$\dfrac{E_\text{B}}{A}$$= \dfrac{0.528464 \cdot 931.5\,\text{MeV}}{56}$$= \dfrac{492.264\,\text{MeV}}{56}$$= 8.7904\,\text{MeV}$Answer
$8.790\,\text{MeV}$
Four significant figures like the conversion factor $931.5\,\tfrac{\text{MeV}}{c^2}$.
Reasonableness
Iron-56 is at the peak of the binding energy graph, about $8.8\,\text{MeV}$, which agrees. Estimate: $\Delta m \approx 0.53\,\text{u}$ and $0.53 \cdot 930 : 56 \approx 8.8$.
- Exercise 10-7Core
Radium-226 decays by alpha decay: ${}^{226}_{88}\text{Ra} \to {}^{222}_{86}\text{Rn} + {}^{4}_{2}\text{He}$. The atomic masses are $m({}^{226}\text{Ra}) = 226.025410\,\text{u}$, $m({}^{222}\text{Rn}) = 222.017578\,\text{u}$ and $m({}^{4}\text{He}) = 4.002603\,\text{u}$.
How much energy is released in the decay? Use $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$. Give the answer in megaelectronvolts.
Answer
$4.871\,\text{MeV}$
Show the solution
Given
$m({}^{226}\text{Ra}) = 226.025410\,\text{u}$, $m({}^{222}\text{Rn}) = 222.017578\,\text{u}$, $m({}^{4}\text{He}) = 4.002603\,\text{u}$, $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$
Asked
the energy released in the decay, that is, the reaction energy $Q$ in megaelectronvolts
Principle
In the reaction the mass number and the charge number are conserved, but the mass decreases, and the energy corresponding to the decrease of mass is released as the kinetic energy of the particles. Atomic masses can be used, because in alpha decay there are $88$ electrons on both sides: $86$ in the radon atom and $2$ in the helium atom.
Formula
$Q = (m_\text{initial} - m_\text{final})\,c^2$, where $m_\text{initial} = m({}^{226}\text{Ra})$ and $m_\text{final} = m({}^{222}\text{Rn}) + m({}^{4}\text{He})$
Unit conversions
$931.5\,\tfrac{\text{MeV}}{c^2}$ is put in place of the unit $\text{u}$, so $c^2$ cancels.
Substitution
$m_\text{initial} - m_\text{final}$$= 226.025410\,\text{u} - 222.017578\,\text{u} - 4.002603\,\text{u}$$= 0.005229\,\text{u}$so
$Q$$= 0.005229 \cdot 931.5\,\text{MeV}$$= 4.8708\,\text{MeV}$Answer
$4.871\,\text{MeV}$
Four significant figures like the mass difference $0.005229\,\text{u}$.
Reasonableness
$Q$ is positive, as it must be in a spontaneous decay. The energies of alpha particles are typically $4$–$9\,\text{MeV}$, and the result is of the same order of magnitude as the $4.27\,\text{MeV}$ of the alpha decay of uranium-238.
- Exercise 10-8
The mass defect of the oxygen-16 nucleus is $0.137005\,\text{u}$. Calculate the binding energy of the nucleus in joules by first converting the mass defect to kilograms. Use $1\,\text{u} = 1.66054 \cdot 10^{-27}\,\text{kg}$ and $c = 2.998 \cdot 10^{8}\,\text{m/s}$. Give the answer in the unit $10^{-11}\,\text{J}$.
Answer
$2.045 \cdot 10^{-11}\,\text{J}$
Show the solution
Given
mass defect $\Delta m = 0.137005\,\text{u}$, $1\,\text{u} = 1.66054 \cdot 10^{-27}\,\text{kg}$, $c = 2.998 \cdot 10^{8}\,\text{m/s}$
Asked
the binding energy $E_\text{B}$ in the unit $10^{-11}\,\text{J}$
Principle
The binding energy is the energy corresponding to the mass defect. In SI units the mass is in kilograms and the speed in metres per second, so the energy is obtained in joules.
Formula
$E_\text{B} = \Delta m\,c^2$
Unit conversions
$\Delta m$$= 0.137005\,\text{u}$$= 0.137005 \cdot 1.66054 \cdot 10^{-27}\,\text{kg}$$= 2.27502 \cdot 10^{-28}\,\text{kg}$Substitution
$E_\text{B}$$= 2.27502 \cdot 10^{-28}\,\text{kg} \cdot (2.998 \cdot 10^{8}\,\text{m/s})^2$$= 2.27502 \cdot 10^{-28}\,\text{kg} \cdot 8.98800 \cdot 10^{16}\,\tfrac{\text{m}^2}{\text{s}^2}$$= 2.04479 \cdot 10^{-11}\,\text{J}$Answer
$2.045 \cdot 10^{-11}\,\text{J}$
Four significant figures like the speed of light $2.998 \cdot 10^{8}\,\text{m/s}$.
Reasonableness
Let us convert to electronvolts by putting in $1\,\text{J} = \tfrac{1}{1.602 \cdot 10^{-19}}\,\text{eV}$:
$2.045 \cdot 10^{-11} : 1.602 \cdot 10^{-19}\,\text{eV}$$\approx 1.277 \cdot 10^{8}\,\text{eV}$$= 127.7\,\text{MeV}$almost the same as the $127.6\,\text{MeV}$ calculated with the factor $931.5\,\tfrac{\text{MeV}}{c^2}$; the difference comes from the rounding of the constants.
- Exercise 10-9
The Sun radiates a power of $3.83 \cdot 10^{26}\,\text{W}$ into space. The energy of the radiation comes from nuclear reactions in the core of the Sun, so the mass of the Sun decreases.
By how much does the mass of the Sun decrease in one day because of the radiation? The speed of light is $c = 2.998 \cdot 10^{8}\,\text{m/s}$. Give the answer in the unit $10^{14}\,\text{kg}$.
Answer
$3.68 \cdot 10^{14}\,\text{kg}$
Show the solution
Given
the radiated power of the Sun $P = 3.83 \cdot 10^{26}\,\text{W}$, time $t = 1\,\text{d}$, $c = 2.998 \cdot 10^{8}\,\text{m/s}$
Asked
the decrease $\Delta m$ of the mass of the Sun in one day, in the unit $10^{14}\,\text{kg}$
Principle
The radiation takes energy $E = Pt$ away from the Sun. By the equivalence of mass and energy, a decrease of energy decreases the mass by $\tfrac{E}{c^2}$. We assume the power constant and ignore the matter carried away by the solar wind.
Formula
$E = Pt$ and $E = \Delta m\,c^2$, so $\Delta m = \dfrac{Pt}{c^2}$
Unit conversions
$t$$= 1\,\text{d}$$= 24 \cdot 3600\,\text{s}$$= 86\,400\,\text{s}$and $1\,\text{W} = 1\,\tfrac{\text{J}}{\text{s}}$.
Substitution
$\Delta m$$= \dfrac{3.83 \cdot 10^{26}\,\tfrac{\text{J}}{\text{s}} \cdot 86\,400\,\text{s}}{(2.998 \cdot 10^{8}\,\text{m/s})^2}$$= \dfrac{3.3091 \cdot 10^{31}\,\text{J}}{8.988 \cdot 10^{16}\,\tfrac{\text{m}^2}{\text{s}^2}}$$= 3.6817 \cdot 10^{14}\,\text{kg}$Answer
$3.68 \cdot 10^{14}\,\text{kg}$
Three significant figures like the radiated power $3.83 \cdot 10^{26}\,\text{W}$.
Reasonableness
In a second the Sun loses about $4.3 \cdot 10^{9}\,\text{kg}$, and in $86\,400$ seconds about
$4 \cdot 10^{9}\,\text{kg} \cdot 9 \cdot 10^{4}$$\approx 4 \cdot 10^{14}\,\text{kg}$Compared with the mass of the Sun, about $2 \cdot 10^{30}\,\text{kg}$, the amount is negligible.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 10-10
One neutron is to be removed from a helium-4 nucleus, leaving a helium-3 nucleus: ${}^{4}_{2}\text{He} \to {}^{3}_{2}\text{He} + \text{n}$. The atomic masses are $m({}^{4}\text{He}) = 4.002603\,\text{u}$ and $m({}^{3}\text{He}) = 3.016029\,\text{u}$, and the mass of the neutron is $m_\text{n} = 1.008665\,\text{u}$.
How much energy does removing the neutron require? Use $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$. Give the answer in megaelectronvolts.
Answer
$20.58\,\text{MeV}$
Show the solution
Given
$m({}^{4}\text{He}) = 4.002603\,\text{u}$, $m({}^{3}\text{He}) = 3.016029\,\text{u}$, $m_\text{n} = 1.008665\,\text{u}$, $1\,\text{u} = 931.5\,\tfrac{\text{MeV}}{c^2}$, reference value $\tfrac{E_\text{B}}{A} = 7.07\,\text{MeV}$
Asked
the energy $E$ needed to remove a neutron, in megaelectronvolts
Principle
The particles of the final state, helium-3 and a free neutron, are together heavier than helium-4. So the reaction does not release energy; an amount of energy corresponding to the difference of masses must be supplied, that is, the reaction energy is negative. Atomic masses can be used, because both helium atoms have two electrons.
Formula
$Q = (m_\text{initial} - m_\text{final})\,c^2$ and the energy needed is $E = -Q$, so $E = (m_\text{final} - m_\text{initial})\,c^2$, where $m_\text{final} = m({}^{3}\text{He}) + m_\text{n}$ and $m_\text{initial} = m({}^{4}\text{He})$
Unit conversions
$931.5\,\tfrac{\text{MeV}}{c^2}$ is put in place of the unit $\text{u}$, so $c^2$ cancels.
Substitution
$m_\text{final} - m_\text{initial}$$= 3.016029\,\text{u} + 1.008665\,\text{u} - 4.002603\,\text{u}$$= 0.022091\,\text{u}$so
$E$$= 0.022091 \cdot 931.5\,\text{MeV}$$= 20.578\,\text{MeV}$Answer
$20.58\,\text{MeV}$
Four significant figures like the conversion factor. This is almost three times the binding energy per nucleon of helium-4, $7.07\,\text{MeV}$.
Reasonableness
The energy must be positive, because a tightly bound nucleus is being broken up. It is greater than the average $7.07\,\text{MeV}$, because the helium-3 that is left is much more loosely bound: its binding energy is only about $7.7\,\text{MeV}$, that is, $2.6\,\text{MeV}$ per nucleon, so removing the neutron takes most of the binding energy $28.3\,\text{MeV}$ of helium-4. This is why helium-4 is exceptionally stable.
- Exercise 10-11
Two deuterium nuclei join to form a helium-4 nucleus: ${}^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \to {}^{4}_{2}\text{He}$. The binding energy per nucleon is $1.112\,\text{MeV}$ for deuterium and $7.074\,\text{MeV}$ for helium-4.
How much energy is released in the reaction? Calculate it from the binding energies without atomic masses. Give the answer in megaelectronvolts.
Answer
$23.85\,\text{MeV}$
Show the solution
Given
for deuterium $\tfrac{E_\text{B}}{A} = 1.112\,\text{MeV}$ ($A = 2$), for helium-4 $\tfrac{E_\text{B}}{A} = 7.074\,\text{MeV}$ ($A = 4$)
Asked
the reaction energy $Q$ in megaelectronvolts
Principle
Let us imagine the reaction in two steps: the deuterium nuclei are first broken into nucleons, which requires their binding energy, and the nucleons are then assembled into a helium-4 nucleus, which releases its binding energy. The number of nucleons is conserved, so the energy released is the increase of the binding energy. By the conservation of energy, the intermediate steps do not matter.
Formula
$Q = E_\text{B}({}^{4}\text{He}) - 2\,E_\text{B}({}^{2}\text{H})$, where $E_\text{B} = A \cdot \tfrac{E_\text{B}}{A}$, so
$Q$$= 4 \cdot \tfrac{E_\text{B}}{A}({}^{4}\text{He}) - 2 \cdot 2 \cdot \tfrac{E_\text{B}}{A}({}^{2}\text{H})$Unit conversions
None needed: all energies are in megaelectronvolts.
Substitution
$Q$$= 4 \cdot 7.074\,\text{MeV} - 2 \cdot 2 \cdot 1.112\,\text{MeV}$$= 28.296\,\text{MeV} - 4.448\,\text{MeV}$$= 23.848\,\text{MeV}$Answer
$23.85\,\text{MeV}$
Four significant figures like the given binding energies $1.112\,\text{MeV}$ and $7.074\,\text{MeV}$.
Reasonableness
From atomic masses the same result:
$(2 \cdot 2.014102\,\text{u} - 4.002603\,\text{u})\,c^2$$= 0.025601 \cdot 931.5\,\text{MeV}$$\approx 23.85\,\text{MeV}$Energy is released because helium-4 is much higher in the graph than deuterium: this is the source of the energy of fusion.
- Exercise 10-12
A pot contains $1.00\,\text{kg}$ of water that cools from a temperature of $80.0\,{}^\circ\text{C}$ to a temperature of $20.0\,{}^\circ\text{C}$. The specific heat capacity of water is $4.19\,\tfrac{\text{kJ}}{\text{kg} \cdot \text{K}}$ and the speed of light $2.998 \cdot 10^{8}\,\text{m/s}$.
By how much does the mass of the water decrease as it cools? Give the answer in the unit $10^{-12}\,\text{kg}$.
Answer
$2.80 \cdot 10^{-12}\,\text{kg}$
Show the solution
Given
mass of the water $m = 1.00\,\text{kg}$, initial temperature $80.0\,{}^\circ\text{C}$, final temperature $20.0\,{}^\circ\text{C}$, specific heat capacity $c_\text{v} = 4.19\,\tfrac{\text{kJ}}{\text{kg} \cdot \text{K}}$, speed of light $c = 2.998 \cdot 10^{8}\,\text{m/s}$
Asked
the decrease $\Delta m$ of the mass of the water, in the unit $10^{-12}\,\text{kg}$
Principle
As it cools, the water gives off heat $Q = c_\text{v} m \Delta T$, and its internal energy decreases by the same amount, because the state of matter does not change. The equivalence of mass and energy applies to every change of energy, so the mass decreases by $\tfrac{Q}{c^2}$.
Formula
$Q = c_\text{v} m \Delta T$ and $Q = \Delta m\,c^2$, so
$\Delta m$$= \dfrac{c_\text{v} m \Delta T}{c^2}$Unit conversions
$c_\text{v}$$= 4.19\,\tfrac{\text{kJ}}{\text{kg} \cdot \text{K}}$$= 4190\,\tfrac{\text{J}}{\text{kg} \cdot \text{K}}$the change of temperature is the same in degrees Celsius and in kelvins:
$\Delta T$$= 80.0\,{}^\circ\text{C} - 20.0\,{}^\circ\text{C}$$= 60.0\,\text{K}$Substitution
$\Delta m$$= \dfrac{4190\,\tfrac{\text{J}}{\text{kg} \cdot \text{K}} \cdot 1.00\,\text{kg} \cdot 60.0\,\text{K}}{(2.998 \cdot 10^{8}\,\text{m/s})^2}$$= \dfrac{2.514 \cdot 10^{5}\,\text{J}}{8.988 \cdot 10^{16}\,\tfrac{\text{m}^2}{\text{s}^2}}$$= 2.797 \cdot 10^{-12}\,\text{kg}$Answer
$2.80 \cdot 10^{-12}\,\text{kg}$
Three significant figures like the mass, the specific heat capacity and the change of temperature.
Reasonableness
The change of mass is about three trillionths of the mass of the water, far below the precision of any balance. This is why the equivalence of mass and energy shows in practice only in nuclear reactions, where the mass defect is almost one per cent.