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course Phy 202
8/3 at 9:15
ProblemTwo identical hypothetical nuclei, each with mass ( 10 - .0086) amu, fuse to form a nucleus with mass ( 19 - .00848) amu and a neutron, whose mass is about 1.00867 amu. How much energy would be given off if .33 kg of these nuclei fused?
Solution
Mass defect= 2(10-0.0086)-((19-0.00848)-(1+0.00867))
=12.68x10^-30
Energy= 13.7x10^-30 (3x10^8)^2= 123.1x10^-14
Nuclei= 0.33kg/33.2x10^-27
=0.99x10^-25
Energy released= 0.99x10^25 x 123.1x10^-14
= 122.3x10^11 J
The total mass of the original particles is 2 ( 10 - .0086 ) amu, while the total mass of the resulting nucleus and neutron is ( 19 - .00848) amu + (1 + .00867 ) amu.
The total mass defect is therefore
• mass defect = 2 ( 10 - .0086 ) amu - [ ( 19 - .00848 ) amu - ( 1 + .00867 ) amu ] = .00855 amu = 13.68 * 10^-30 kg.
The corresponding energy is therefore
• energy = 13.68 * 10^-30 kg ( 3 * 10^8 m/s) ^ 2 = 123.1 * 10^-14 Joules.
The mass of each of the resulting particles is very close to 2 * 10 amu = 33.2 * 10^-27 kg, so in .33 kg there will be approximately
• number of nuclei = .33 kg / ( 33.2 * 10^-27 kg) = .9939 * 10^25.
These nuclei would therefore give off
• energy released = .9939 * 10^25 nuclei * ( 123.1 * 10^-14 J / nucleus ) = 122.3 * 10^11 Joules.
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Problem
The table below shows masses, in atomic units, of various neutral isotopes of several elements. If an atomic mass unit is 1.660502 * 10^-27 kg, if a proton has mass 1.007276 amu and a neutron mass 1.008665 amu, and if a neutral hydrogen atom has mass 1.007825 amu, what is the mass defect of the isotope in the third row of column # 10? What is the mass defect per nucleon for this isotope?
What is the binding energy of this atom, in electron volts? What is the binding energy per nucleon?
Properties of Selected Nuclei
particle or atom cobalt nickel iron gold mercury lead bismuth polonium radium thorium
atomic number 27 28 26 79 80 82 83 84 88 90
mass (amu) 59.933822 59.93079 56.9354 197.9682 197.9668 205.9744 208.980383 208.9824 226.0254 230.0331
mass (amu) 56.936292 209.984105
Solution
You told me this problem is beyond the course of Physics 202
The element in Column 10 has atomic number 88, which is equal to the number of charged particles in the nucleus--i.e., to the number of protons. We therefore have 88 in the nucleus.
The atomic mass is 226.0254, representing the number of atomic mass units in a neutral atom. This mass is very close to a whole number of atomic mass units. If we round this number off we obtain the number of nucleons--i.e., the total number of protons and neutrons--in the nucleus.
So we see that there are 226 nucleons in the nucleus. Since 88 of these are protons, it is easy to see that we have 226 - 88 = 138 neutrons in the nucleus.
The atom consists of its neutrons, protons and electrons, one electron for each proton. A basic hydrogen atom consists of a single proton and a single electron. Therefore if we add the masses of all the neutrons, plus the mass of a hydrogen atom for each proton, we will get the total unbound mass of all the particles that make up the atom:
• total mass of constituent particles = 88 * 1.007825 amu + 138 * 1.008665 = 227.8844 amu.
This is somewhat greater than the mass of the actual atom, which is 226.0254 amu. The difference is
• 227.8844 amu - 226.0254 amu = 1.858963 amu,
and represents the energy that binds the particles of the atom together.
Since there are 226 nucleons in the nucleus, the mass defect per nucleon is
• Mass Defect Per Nucleon = 1.858963 amu / 226 = 8.225 * 10^-3 amu.
The binding energy is the energy equivalent of the mass defect:
• binding energy = (mass defect) * c^2 = 1.858963 amu * (3 * 10^8 m/s) ^ 2 = 1.858963 amu * 1.660502 * 10^-27 kg / amu * (3 * 10^8 m/s) ^ 2 = 17.35 * 10^-10 Joules.
The binding energy per nucleon is therefore
• binding energy per nucleon = 17.35 * 10^-10 J / ( 226 nucleons) = 76.76 * 10^-13 J / nucleon.
This binding energy is easiest to understand in units of electron volts:
• binding energy per nucleon (electron volts)
= ( 76.76 * 10^-13 J / nucleon) * (1 eV / {1.6 * 10^-19 eV/Joule} )
= 47.98 * 10^6 eV
= 47.98 * 10^6 MeV.
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Same for problems 22 and 23
I see here my given solutions. If there are questions, please insert them using #### to mark your insertions.
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