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course Phy 202
8/3 at 1:00
ProblemHow many electron volts of energy are associated with a single atomic mass unit, or amu (an amu is approximately 1.66 * 10^-27 kg)? How many electron volts are associated with the mass of an electron (electron mass approximately 9.11 * 10^-31 kg)? What would be the wavelength of a single photon with this energy? Compare each wavelength with the diameter of a typical atom, which is 10^-10 m or 1 Angstrom, and with the diameter of a typical proton, which is about 10^-15 m or 1/100,000 Angstrom; this distance is also called a Fermion.
Solution
Energy equivalent 1 amu= 1.66x10^-27x(3x10^8)^2=1.5x10^-10
1.5x10^-10/(1.6x10^-19)=9.3x10^8
Energy equivalent of an electron=9.11x10^-31x(3x10^8)^2= 8.2x10^-14
8.2x10^-14/(1.6x10^-19)=510x10^8
Wavelength= 6.62x10^-23x(3x10^8)/(8.2x10-14)=3x10^-12
Wavelength= (6.62x10^-34x(3x10^8)/(1.5x10^-10)=1.3x10^-15
The energy equivalent of 1 amu is
• energy equivalent of 1 amu: E = mc^2 = 1.66 * 10^-27 kg * (3 * 10^8 m/s) ^ 2 = 1.5 * 10^-10 Joules, or
1.5 * 10^-10 J / ( 1.6 * 10^-19 J / eV) = 9.3 * 10^8 eV, or 930 million electron volts (930 meV).
The energy equivalent of an electron mass is
• energy equivalent of 1 electron mass: E = mc^2 = 9.11 * 10^-31 kg * (3 * 10^8 m/s) ^ 2 = 8.2 * 10^-14 Joules, or
8.2 * 10^-14 J / ( 1.6 * 10^-19 J / eV) = 510 * 10^8 eV, or 510 thousand electron volts (510 keV).
The photon wavelengths will are found from E = hf = h c / `lambda to be `lambda = h c / E for the corresponding energies:
• wavelength of 510 keV photon: `lambda = ( 6.62 * 10^-34 J s ) * ( 3 * 10^8 m/s) / ( 8.2 * 10^-14 J ) = 3 * 10^-12 meters, about 1000 times the diameter of a proton and 1/100 the diameter of a typical atom
and
• wavelength of 931 meV photon: `lambda = ( 6.62 * 10^-34 J s ) * ( 3 * 10^8 m/s) / ( 1.5 * 10^-10 J ) = 1.3 * 10^-15 meters, on the order of the diameter of a proton.
Photons of these wavelengths can occur in particle accelerators and in cosmic rays. The energy of such photons can in various interactions convert its form into a combination of mass and energy, with the energy manifesting itself in the form of kinetic energy of massive particles and/or in the energy of lower-frequency photons.
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Problem
A certain hypothetical atom contains 51 protons and 66 neutrons in its nucleus and has an atomic mass of ( 117 - .051 ) atomic mass units, or amu (an amu is approximately 1.66 * 10^-27 kg; a proton mass is about 1 + .00073 amu, neutron mass about 1 + .00087 amu). What is the mass defect of this atom? What is the energy equivalent of the mass defect of a mole of these atoms?
Solution
Total mass= 51x(1+0.00073)+ 6x(1+0.00087)=(117+0.09465)
Mass defect per mole= 0.1455 x (6.02x10^23)= 1.46x10^-4
Energy= 1.46x10^-4(3x10^8)^2= 1.3x10^13
The total mass of 51 protons and 66 neutrons is 51 * (1 + .00073) amu + 66 * ( 1 + .00087) amu = ( 117 + .09465 ) amu.
This exceeds the actual mass of the atom by ( 117 + .09465 ) amu - ( 117 - .051 ) amu = .14565 amu.
A mole of these atoms would contain an Avagadro's Number of nuclei with total mass defect
• mass defect per mole = mass defect per atom * Avagadro's Number = .14565 amu * (6.02 * 10^23) = ( .14565 * (1.66 * 10^-27 kg) ) * (6.02 * 10^23) = 1.455509 * 10^-4 kg.
The energy in this mass would be
• energy per mole = m c^2 = 1.455509 * 10^-4 kg ( 3 * 10^8 m/s ) ^ 2 = 1.309958 * 10^13 Joules.
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Problem
A certain hypothetical atom contains 54 protons and 69 neutrons in its nucleus and has an atomic mass of 122.3 atomic mass units, or amu (an amu is approximately 1.66 * 10^-27 kg). How many protons and how many neutrons will it end up with if it undergoes an alpha decay? How many if it undergoes a beta decay? How many if it undergoes a gamma decay?
Solution
?? There are no numbers in this solution to answer the question
In alpha decay an atomic nucleus emits an alpha particle, which is a common helium nucleus containing 2 protons and 2 neutrons. The atom therefore loses 2 protons and 2 neutrons, leaving a total of 52 protons and 67 neutrons. The total mass of the remaining nucleus and the alpha particle will be less than that of the original nucleus; the 'lost' mass is converted to energy and goes into the kinetic energy of the remaining nucleus and the alpha particle. Since in this case the remaining nucleus is much more massive than the alpha particle conservation of momentum ensures that most of the the energy will be carried by the alpha particle, with a negligible amount carried by the nucleus.
In beta decay a neutron effectively changes to a proton. It is necessary in any interaction for the total charge of the particles involved to remain constant, so the system emits an electron to balance the charge. The number of neutrons therefore decreases by 1 while the number of proton increases by 1, resulting in a nucleus with 55 protons and 68 neutrons. The total mass of the remaining nucleus and the electron will be less than that of the original nucleus; the 'lost' mass is converted to energy and goes into the kinetic energy of the remaining nucleus and the electron. Since in this case the remaining nucleus is much, much more massive than the electron, conservation of momentum ensures that practically all the energy will be carried by the alpha particle.
In gamma decay the number of protons and neutrons remains the same because gamma decay consists of the emission of a high-energy photon from within the nucleus. The resulting nucleus will have less mass than the original, by an amount equivalent to the energy carried by the photon.
Problem
A hypothetical atom with negligible kinetic energy has a mass of 229.741 amu. It undergoes an alpha decay. The remaining atom has atomic mass 225.733 amu. What is the kinetic energy and/or wavelength (whichever is more appropriate) of the emitted particle, assuming that the kinetic energy of the remaining atom is negligible? How much energy would be released by the decay of a mole of these atoms? Note that the mass of a helium atom is about 4.0026 amu, where an amu is approximately 1.66 * 10^-27 kg.
Solution
Mass change= 229.7-(225.7+4)= 5.4x10^-3 amu
Energy= 8.9x10^-3x10^-27(3x10^8)^1= 4.8x10^-13 J
Energy released= 6.02x10^23x4.8x10^-13= 29.1+10^10 J
The change in atomic mass is approximately 229.741 amu - ( 225.733 amu + 4.0026 amu) = 5.364895E-03 amu, or about 8.905727E-03 * 10^-27 kg.
This corresponds to an energy of E = m c^2 = 8.905727E-03 * 10^-27 kg ( 3 * 10^8 m/s) ^ 2 = 4.828406 * 10^-13 Joules.
A mole of these nuclei would constitute 6.02 * 10^23 nuclei, each releasing 4.828406 * 10^-13 Joules. The total energy released would therefore be
• energy released per mole of decaying particles = (6.02 * 10^23 nuclei) * ( 4.828406 * 10^-13 Joules / nucleus) = 29.06701 + 10^10 Joules.
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Problem
A hypothetical atom with negligible kinetic energy has a mass of 161 amu. It undergoes an alpha decay. The remaining atom has atomic mass 156.5 amu. What is the kinetic energy and/or wavelength (whichever is more appropriate) of the emitted particle, assuming that the kinetic energy of the remaining atom is negligible? How much energy would be released by the decay of a mole of these atoms? Note that the mass of a helium nucleus is about 4.001 amu and the mass of an electron about .00055 amu, where an amu is approximately 1.66 * 10^-27 kg.
Solution
161-(156.5+4)= 0.83x10^-27
Energy- 0.83x10^-27 x(3x10^8)^2= 449x10^-13
Energy released= 6.02x10^23x449x10^-13= 2703+10^10
The change in atomic mass is approximately 161 amu - ( 156.5 amu + 4.001 amu) = .4990001 amu, or about .8283401 * 10^-27 kg.
This corresponds to an energy of E = m c^2 = .8283401 * 10^-27 kg ( 3 * 10^8 m/s) ^ 2 = 449.1001 * 10^-13 Joules.
A mole of these nuclei would constitute 6.02 * 10^23 nuclei, each releasing 449.1001 * 10^-13 Joules. The total energy released would therefore be
• energy released per mole of decaying particles = (6.02 * 10^23 nuclei) * ( 449.1001 * 10^-13 Joules / nucleus) = 2703.582 + 10^10 Joules.
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Problem
A hypothetical atom with negligible kinetic energy has a mass of 194 amu. It undergoes a beta decay. The remaining atom has atomic mass 193.6 amu. What is the kinetic energy and/or wavelength (whichever is more appropriate) of the emitted particle, assuming that the kinetic energy of the remaining atom is negligible? How much energy would be released by the decay of a mole of these atoms? Note that the mass of a helium nucleus is about 4.001 amu and the mass of an electron about .00055 amu, where an amu is approximately 1.66 * 10^-27 kg?
Solution
194-(194+0.0055)= 0.66x10^-27
0.66x10^-27 (3x10^8)^2= 6.0x10^-13
Energy released=
6.02x10^23 x 6.0x10^-13= 35+10^10 J
The change in atomic mass is approximately 194 amu - ( 193.6 amu + .00055 amu) = .3994439 amu, or about .6630769 * 10^-27 kg.
This corresponds to an energy of
• E = m c^2 = .6630769 * 10^-27 kg ( 3 * 10^8 m/s) ^ 2 = 5.967692 * 10^-13 Joules.
A mole of these nuclei would constitute 6.02 * 10^23 nuclei, each releasing 5.967692 * 10^-13 Joules. The total energy released would therefore be
• energy released per mole of decaying particles = (6.02 * 10^23 nuclei) * ( 5.967692 * 10^-13 Joules / nucleus) = 35.92551 + 10^10 Joules.
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Problem
A hypothetical atom with negligible kinetic energy has a mass of 216 amu. It undergoes a gamma decay. The remaining atom has atomic mass which is less than that of the original by .0000115 amu. What is the kinetic energy and/or wavelength (whichever is more appropriate) of the emitted particle, assuming that the kinetic energy of the remaining atom is negligible? How much energy would be released by the decay of a mole of these atoms? Note that the mass of a helium nucleus is about 4.001 amu and the mass of an electron about .00055 amu, where an amu is approximately 1.66 * 10^-27 kg?
Solution
Energy- 0.000115x(3x10^8)^2=117.181x10^-16
Wavelength- 6.62x10^-34x3x10^8/17x10^-16=1.156x10^-10
Energy released= 6.02x10^23x17x10^-16=0.3x10^8
In a beta decay all the energy released is carried away in a photon--there is no massive particle involved. This energy corresponds to a change in the 'orbit' of a particle in the nucleus, and is analogous to the effect of a change in the orbital of an electron.
The energy therefore corresponds to the entire .0000115 kg of mass lost:
• photon energy = .0000115 amu * (3 * 10^8 m/s)^2 = 1.909 * 10^-32 kg * (3 * 10^8 m/s)^2 = 17.181 * 10^-16 Joules.
The wavelength of this photon is found from the relationship E = h f = h c / `lambda to be
• wavelength = h c / E = ( 6.62 * 10^-34 J s ) ( 3 * 10^8 m/s ) / ( 17.181 * 10^-16 Joules ) = 1.155928 * 10^-10 meters = 1.155928 Angstroms.
A mole of these nuclei would constitute 6.02 * 10^23 nuclei, each releasing 17.181 * 10^-16 Joules. The total energy released would therefore be
• energy released per mole of decaying particles = (6.02 * 10^23 nuclei) * ( 17.181 * 10^-16 Joules / nucleus) = 10.34296 * 10^8 Joules.
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