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21.6 Binding energy

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  • Define and discuss binding energy.
  • Calculate the binding energy per nucleon of a particle.

The more tightly bound a system is, the stronger the forces that hold it together and the greater the energy required to pull it apart. We can therefore learn about nuclear forces by examining how tightly bound the nuclei are. We define the binding energy    (BE) of a nucleus to be the energy required to completely disassemble it into separate protons and neutrons . We can determine the BE of a nucleus from its rest mass. The two are connected through Einstein’s famous relationship E = ( Δ m ) c 2 size 12{E= \( Δm \) c rSup { size 8{2} } } {} . A bound system has a smaller mass than its separate constituents; the more tightly the nucleons are bound together, the smaller the mass of the nucleus.

Imagine pulling a nuclide apart as illustrated in [link] . Work done to overcome the nuclear forces holding the nucleus together puts energy into the system. By definition, the energy input equals the binding energy BE. The pieces are at rest when separated, and so the energy put into them increases their total rest mass compared with what it was when they were glued together as a nucleus. That mass increase is thus Δ m = BE / c 2 size 12{Δm="BE"/c rSup { size 8{2} } } {} . This difference in mass is known as mass defect . It implies that the mass of the nucleus is less than the sum of the masses of its constituent protons and neutrons. A nuclide A X size 12{"" lSup { size 8{A} } X} {} has Z size 12{Z} {} protons and N size 12{N} {} neutrons, so that the difference in mass is

Δ m = ( Zm p + Nm n ) − m tot . size 12{Δm= \( ital "Zm" rSub { size 8{p} } + ital "Nm" rSub { size 8{n} } \) - m rSub { size 8{"tot"} } } {}

Thus,

BE = ( Δ m ) c 2 = [ ( Zm p + Nm n ) − m tot ] c 2 , size 12{"BE"= \( Δm \) c rSup { size 8{2} } = \[ \( ital "Zm" rSub { size 8{p} } + ital "Nm" rSub { size 8{n} } \) - m rSub { size 8{"tot"} } \]c rSup { size 8{2} } } {}

where m tot size 12{m rSub { size 8{"tot"} } } {} is the mass of the nuclide A X , m p is the mass of a proton, and m n is the mass of a neutron. Traditionally, we deal with the masses of neutral atoms. To get atomic masses into the last equation, we first add Z size 12{Z} {} electrons to m tot size 12{m rSub { size 8{"tot"} } } {} , which gives m A X size 12{m left ("" lSup { size 8{A} } X right )} {} , the atomic mass of the nuclide. We then add Z size 12{Z} {} electrons to the Z size 12{Z} {} protons, which gives Zm 1 H size 12{ ital "Zm" left ("" lSup { size 8{1} } H right )} {} , or Z size 12{Z} {} times the mass of a hydrogen atom. Thus the binding energy of a nuclide A X size 12{"" lSup { size 8{A} } X} {} is

BE = { [ Zm ( 1 H ) + Nm n ] − m ( A X ) } c 2 .

The atomic masses can be found in Appendix A , most conveniently expressed in unified atomic mass units u ( 1 u = 931 . 5 MeV / c 2 size 12{1" u"="931" "." 5 "MeV"/c rSup { size 8{2} } } {} ). BE is thus calculated from known atomic masses.

The image shows some spherical protons and neutrons pulled out from a nucleus. The work done to pull them apart is binding energy.
Work done to pull a nucleus apart into its constituent protons and neutrons increases the mass of the system. The work to disassemble the nucleus equals its binding energy BE. A bound system has less mass than the sum of its parts, especially noticeable in the nuclei, where forces and energies are very large.

Things great and small

Nuclear Decay Helps Explain Earth’s Hot Interior

A puzzle created by radioactive dating of rocks is resolved by radioactive heating of Earth’s interior. This intriguing story is another example of how small-scale physics can explain large-scale phenomena.

Radioactive dating plays a role in determining the approximate age of the Earth. The oldest rocks on Earth solidified about 3 . 5 × 10 9 size 12{3 "." 5 times "10" rSup { size 8{9} } } {} years ago—a number determined by uranium-238 dating. These rocks could only have solidified once the surface of the Earth had cooled sufficiently. The temperature of the Earth at formation can be estimated based on gravitational potential energy of the assemblage of pieces being converted to thermal energy. Using heat transfer concepts discussed in Thermodynamics it is then possible to calculate how long it would take for the surface to cool to rock-formation temperatures. The result is about 10 9 size 12{"10" rSup { size 8{9} } } {} years. The first rocks formed have been solid for 3 . 5 × 10 9 size 12{3 "." 5 times "10" rSup { size 8{9} } } {} years, so that the age of the Earth is approximately 4 . 5 × 10 9 size 12{4 "." 5 times "10" rSup { size 8{9} } } {} years. There is a large body of other types of evidence (both Earth-bound and solar system characteristics are used) that supports this age. The puzzle is that, given its age and initial temperature, the center of the Earth should be much cooler than it is today (see [link] ).

The figure shows that the center of the Earth cools by three heat transfer methods. Convection heat transfer in the center region, then conduction heat transfer moves thermal energy to the surface, and finally radiation heat transfer from the surface to space.
The center of the Earth cools by well-known heat transfer methods. Convection in the liquid regions and conduction move thermal energy to the surface, where it radiates into cold, dark space. Given the age of the Earth and its initial temperature, it should have cooled to a lower temperature by now. The blowup shows that nuclear decay releases energy in the Earth’s interior. This energy has slowed the cooling process and is responsible for the interior still being molten.

We know from seismic waves produced by earthquakes that parts of the interior of the Earth are liquid. Shear or transverse waves cannot travel through a liquid and are not transmitted through the Earth’s core. Yet compression or longitudinal waves can pass through a liquid and do go through the core. From this information, the temperature of the interior can be estimated. As noticed, the interior should have cooled more from its initial temperature in the 4 . 5 × 10 9 size 12{4 "." 5 times "10" rSup { size 8{9} } } {} years since its formation. In fact, it should have taken no more than about 10 9 size 12{4 "." 5 times "10" rSup { size 8{9} } } {} years to cool to its present temperature. What is keeping it hot? The answer seems to be radioactive decay of primordial elements that were part of the material that formed the Earth (see the blowup in [link] ).

Nuclides such as 238 U and 40 K have half-lives similar to or longer than the age of the Earth, and their decay still contributes energy to the interior. Some of the primordial radioactive nuclides have unstable decay products that also release energy— 238 U size 12{"" lSup { size 8{"238"} } U} {} has a long decay chain of these. Further, there were more of these primordial radioactive nuclides early in the life of the Earth, and thus the activity and energy contributed were greater then (perhaps by an order of magnitude). The amount of power created by these decays per cubic meter is very small. However, since a huge volume of material lies deep below the surface, this relatively small amount of energy cannot escape quickly. The power produced near the surface has much less distance to go to escape and has a negligible effect on surface temperatures.

A final effect of this trapped radiation merits mention. Alpha decay produces helium nuclei, which form helium atoms when they are stopped and capture electrons. Most of the helium on Earth is obtained from wells and is produced in this manner. Any helium in the atmosphere will escape in geologically short times because of its high thermal velocity.

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?
Kate Reply
what is the change in momentum of a body?
Eunice Reply
what is a capacitor?
Raymond Reply
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.
Gautam
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
Maria Reply
please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
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.
Mehmet
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
Saheed Reply
50 m/s due south east
Someone
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
Ramon Reply
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.
Someone
Scratch that
Someone
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.....
Someone
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
Someone
definitely of physics
Haryormhidey Reply
how many start and codon
Esrael Reply
what is field
Felix Reply
physics, biology and chemistry this is my Field
ALIYU
field is a region of space under the influence of some physical properties
Collete
what is ogarnic chemistry
WISDOM Reply
determine the slope giving that 3y+ 2x-14=0
WISDOM
Another formula for Acceleration
Belty Reply
a=v/t. a=f/m a
IHUMA
innocent
Adah
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?
Nassze Reply
how do lnternal energy measures
Esrael
Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
JALLAH Reply
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.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
Raymond
What is specific heat capacity
Destiny Reply
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
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, Physics 101. OpenStax CNX. Jan 07, 2013 Download for free at http://legacy.cnx.org/content/col11479/1.1
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