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A graph is shown where the x-axis is labeled “Number of neutrons, open parenthesis, n, close parenthesis” and has values of 0 to 180 in increments of 10. The y-axis is labeled “Number of protons, open parenthesis, Z, close parenthesis” and has values of 0 to 120 in increments of 10. A green shaded band of varying width, labeled “Radioactive,” extends from point 0 on both axes to 178 on the y-axis and 118 on the x-axis in a linear manner. The width of this band varies from 8 to 18 units in width according to the x-axis measurements. A blue line in a roughly zig-zag pattern runs through the middle of the shaded band and stops at 128 on the y-axis and 82 on the x-axis. This line is labeled “Nonradioactive.” An unlabeled, black, solid line extends from point 0, 0 to 120, 120 in a linear manner.
This plot shows the nuclides that are known to exist and those that are stable. The stable nuclides are indicated in blue, and the unstable nuclides are indicated in green. Note that all isotopes of elements with atomic numbers greater than 83 are unstable. The solid line is the line where n = Z.

The nuclei that are to the left or to the right of the band of stability are unstable and exhibit radioactivity    . They change spontaneously (decay) into other nuclei that are either in, or closer to, the band of stability. These nuclear decay reactions convert one unstable isotope (or radioisotope    ) into another, more stable, isotope. We will discuss the nature and products of this radioactive decay in subsequent sections of this chapter.

Several observations may be made regarding the relationship between the stability of a nucleus and its structure. Nuclei with even numbers of protons, neutrons, or both are more likely to be stable (see [link] ). Nuclei with certain numbers of nucleons, known as magic numbers , are stable against nuclear decay. These numbers of protons or neutrons (2, 8, 20, 28, 50, 82, and 126) make complete shells in the nucleus. These are similar in concept to the stable electron shells observed for the noble gases. Nuclei that have magic numbers of both protons and neutrons, such as 2 4 He , 8 16 O , 20 40 Ca , and 82 208 Pb , are called “double magic” and are particularly stable. These trends in nuclear stability may be rationalized by considering a quantum mechanical model of nuclear energy states analogous to that used to describe electronic states earlier in this textbook. The details of this model are beyond the scope of this chapter.

Stable Nuclear Isotopes
Number of Stable Isotopes Proton Number Neutron Number
157 even even
53 even odd
50 odd even
5 odd odd

The relative stability of a nucleus is correlated with its binding energy per nucleon    , the total binding energy for the nucleus divided by the number or nucleons in the nucleus. For instance, we saw in [link] that the binding energy for a 2 4 He nucleus is 28.4 MeV. The binding energy per nucleon for a 2 4 He nucleus is therefore:

28.4 MeV 4 nucleons = 7.10 MeV/nucleon

In [link] , we learn how to calculate the binding energy per nucleon of a nuclide on the curve shown in [link] .

A graph is shown where the x-axis is labeled “binding energy per nucleon, open parenthesis, M e V, close parenthesis” and has values of 0 to 10 in increments of 1. The y-axis is labeled “Mass number” and has values of 0 to 260 in increments of 20. A line of best fit beginning at point 0, 0 is drawn through points “8, 5.5; 9, 7.3; 18, 7.1; 20, 7.5; 19, 7.9; 27, 7.8; 21, 8.1; 25, 8.4; 37, 8.6; 43, 8.8; 57, 8.6; 60, 8.9; 70, 9; 88, 8.8; 102, 8.9; 108, 8.5; 126, 8.7; 133, 8.8; 143, 8.2; 157, 8.1; 167, 8.2; 195, 7.9; 205, 7.9; 241, 7.3 and 255, 75. An upward-facing arrow near the bottom left of the graph is labeled “Fusion” while a left-facing arrow near the top right is labeled “Fission.”
The binding energy per nucleon is largest for nuclides with mass number of approximately 56.

Calculation of binding energy per nucleon

The iron nuclide 26 56 Fe lies near the top of the binding energy curve ( [link] ) and is one of the most stable nuclides. What is the binding energy per nucleon (in MeV) for the nuclide 26 56 Fe (atomic mass of 55.9349 amu)?

Solution

As in [link] , we first determine the mass defect of the nuclide, which is the difference between the mass of 26 protons, 30 neutrons, and 26 electrons, and the observed mass of an 26 56 Fe atom:

Mass defect = [ ( 26 × 1.0073 amu ) + ( 30 × 1.0087 amu ) + ( 26 × 0.00055 amu ) ] 55.9349 amu = 56.4651 amu 55.9349 amu = 0.5302 amu

We next calculate the binding energy for one nucleus from the mass defect using the mass-energy equivalence equation:

E = m c 2 = 0.5302 amu × 1.6605 × 10 −27 kg 1 amu × ( 2.998 × 10 8 m/s ) 2 = 7.913 × 10 −11 kg m/s 2 = 7.913 × 10 −11 J

We then convert the binding energy in joules per nucleus into units of MeV per nuclide:

7.913 × 10 −11 J × 1 MeV 1.602 × 10 −13 J = 493.9 MeV

Finally, we determine the binding energy per nucleon by dividing the total nuclear binding energy by the number of nucleons in the atom:

Binding energy per nucleon = 493.9 MeV 56 = 8.820 MeV/nucleon

Note that this is almost 25% larger than the binding energy per nucleon for 2 4 He .

(Note also that this is the same process as in [link] , but with the additional step of dividing the total nuclear binding energy by the number of nucleons.)

Check your learning

What is the binding energy per nucleon in 9 19 F (atomic mass, 18.9984 amu)?

Answer:

7.810 MeV/nucleon

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Key concepts and summary

An atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force. A nucleus has less mass than the total mass of its constituent nucleons. This “missing” mass is the mass defect, which has been converted into the binding energy that holds the nucleus together according to Einstein’s mass-energy equivalence equation, E = mc 2 . Of the many nuclides that exist, only a small number are stable. Nuclides with even numbers of protons or neutrons, or those with magic numbers of nucleons, are especially likely to be stable. These stable nuclides occupy a narrow band of stability on a graph of number of protons versus number of neutrons. The binding energy per nucleon is largest for the elements with mass numbers near 56; these are the most stable nuclei.

Key equations

  • E = mc 2

Chemistry end of chapter exercises

Write the following isotopes in hyphenated form (e.g., “carbon-14”)

(a) 11 24 Na

(b) 13 29 Al

(c) 36 73 Kr

(d) 77 194 Ir

(a) sodium-24; (b) aluminum-29; (c) krypton-73; (d) iridium-194

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Write the following isotopes in nuclide notation (e.g., " 6 14 C " )

(a) oxygen-14

(b) copper-70

(c) tantalum-175

(d) francium-217

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For the following isotopes that have missing information, fill in the missing information to complete the notation

(a) 14 34 X

(b) X 36 P

(c) X 57 Mn

(d) 56 121 X

(a) 14 34 Si ; (b) 15 36 P ; (c) 25 57 Mn ; (d) 56 121 Ba

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For each of the isotopes in [link] , determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.

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Write the nuclide notation, including charge if applicable, for atoms with the following characteristics:

(a) 25 protons, 20 neutrons, 24 electrons

(b) 45 protons, 24 neutrons, 43 electrons

(c) 53 protons, 89 neutrons, 54 electrons

(d) 97 protons, 146 neutrons, 97 electrons

(a) 25 45 Mn +1 ; (b) 45 69 Rh +2 ; (c) 53 142 I −1 ; (d) 97 243 Bk

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Calculate the density of the 12 24 Mg nucleus in g/mL, assuming that it has the typical nuclear diameter of 1 × 10 –13 cm and is spherical in shape.

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What are the two principal differences between nuclear reactions and ordinary chemical changes?

Nuclear reactions usually change one type of nucleus into another; chemical changes rearrange atoms. Nuclear reactions involve much larger energies than chemical reactions and have measureable mass changes.

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The mass of the atom 11 23 Na is 22.9898 amu.

(a) Calculate its binding energy per atom in millions of electron volts.

(b) Calculate its binding energy per nucleon.

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Which of the following nuclei lie within the band of stability shown in [link] ?

(a) chlorine-37

(b) calcium-40

(c) 204 Bi

(d) 56 Fe

(e) 206 Pb

(f) 211 Pb

(g) 222 Rn

(h) carbon-14

(a), (b), (c), (d), and (e)

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Which of the following nuclei lie within the band of stability shown in [link] ?

(a) argon-40

(b) oxygen-16

(c) 122 Ba

(d) 58 Ni

(e) 205 Tl

(f) 210 Tl

(g) 226 Ra

(h) magnesium-24

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Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
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what is the change in momentum of a body?
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what is a capacitor?
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Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
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A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
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please solve
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8m/s²
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What is Thermodynamics
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velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
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A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
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50 m/s due south east
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which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
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I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
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Scratch that
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temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
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about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
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definitely of physics
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what is field
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physics, biology and chemistry this is my Field
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field is a region of space under the influence of some physical properties
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pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
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No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
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Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
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like charges repel while unlike charges atttact
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What is specific heat capacity
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specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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