#$&*
course Phy 202
5/4 10 pmI didn't know how to approach number 5. You'll probably get more problems tomorrow night.
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Problem Number 1
A hypothetical atom with negligible kinetic energy has a mass of 245.863 amu. It undergoes an alpha decay. The remaining atom has atomic mass 241.8544 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.
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E=mc^2
Start 245.863 amu End-241.8544 amu
m=245.863-(241.8544 amu+ 4.0026 amu)=.006 amu
E= .006 amu (1.66 *10^-27 kg) (3.8 *10^8 m/s)^2=1.44*10^-12 J
1 mole=6.02 *10^23 nuclei * 1.44 *10^-12 J/nucleus=8.66*10 ^11 J
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Problem Number 2
A photon of 156.0689 nm electromagnetic radiation encounters an electron in the n = 2 orbital of a hydrogen atom, and causes it to 'jump' to the n = 3 orbital. What will be the wavelength of a photon which carries away any excess energy from the collision?
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wavelength is 1.56 x 10^-7 meters
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I looked for a problem set to address this one, but I can’t find it if it’s out there.
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@& How much energy does it take to jump from the n = 2 to the n = 3 orbital? Remember that E = - 1 / n^2 * 13.6 eV*@
@& What is the energy of the photon?
How much is left after the photon supplies the necessary energy to the electron?
What therefore is the wavelength of the photon that comes away from the interaction?*@
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Problem Number 3
Find the velocity an electron in a circular orbit would require at a distance of 1.722 Angstroms from a proton. Find the deBroglie wavelength of this electron. Determine whether the corresponding probability wave would 'fit' the circumference of the orbit without undergoing destructive interference. [ The mass of an electron is 9.11 * 10^-31 kg; the proton has a much greater mass; Planck's Constant is 6.62 * 10^-34 J s; k = 9 * 10^9 N m^2 / C^2 ]
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deBroglie wavelength=h/mv
v=sqrt(k*qE^2/mr)
r=1.722x10^-10 m
v=sqrt (9 * 10^9 N m^2 / C^2 *(1.6 * 10^-19 C)^2/(9.11 * 10^-31 kg*1.722x10^-10 m))
v=sqrt(1.47 x10 ^12 m^2/s^2)
v=1.21*10^6 m/s
deBroglie wavelength=6.62 * 10^-34 J s/(9.11 * 10^-31 kg*1.21*10^6 m/s)=6.006*10^-10 m
circumference= 2`pi r=2 `pi 1.722 x10^10 m=1.08 * 10^11 m
@& The circumference is on the order of 10^-10 m, not 10_10 m.*@
@& The circumference turns out to be about 10 * 10^-10 m = 10^-9 m.*@
1.08*10^11m/6.006*10^-10 m=1.8032 NO, it will not fit the circumference without undergoing destructive interference
@& 10^11 / 10^-10 is 10^21*@
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Problem Number 4
What is the momentum of a photon of light whose wavelength is .0095 nm? How much momentum would be lost in 1 minute by a 98000-watt beam of this light?
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p=E/c
E=h*f f=c/`lambda
c=3.8 *10^8 m/s
wavelength=9500 m
F=3.8 *10^8 m/s/9500 m=4.0*10^4 cycles/second
@& nm stands for nanometer, 10^-9 meter
frequency is on the order of 10^15 Ha.*@
E=4.0*10 ^4 cycles/s*6.62 * 10^-34 J s=2.65 *10^-29 J
p=2.65 *10^-29 J/ 3.8 * 10^8 m/s=6.97 *10^-38 kg m/s
Momentum lost in 1 minute by a 98000 watt beam of light (J/s/ m^2)
98000 J/s/m^2/(60 seconds*6.97 x10^-34 kg m/s)=2.34*10^36 kg m/s
@& You're using the right relationships but watch your powers of 10, and interpretation of things like 'nm'.*@
@& These errors are throwing your results off by quite a lot. Fortunately not difficult to correct.*@
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Problem Number 5
A beam consisting of 100 eV electrons (electron mass 9.11 * 10^-31 kg) is incident on a thin wafer of a crystal with layer spacing 3.5 Angstroms. Surprisingly we find that the electrons, which are particles, are scattered in such a way as to form an interference pattern identical to that of a wave. What will be the distance between central interference maxima at a distance of 20 cm from the wafer?
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@& Start by finding the angle with respect to the perpendicular bisector at which the path difference is 1 wavelength.
The separation of the two sources is the 3.5 Angstrom spacing of the layers in the crystal.*@
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Problem Number 6
What is the approximate uncertainty in the velocity of a proton known to remain within a nucleus of diameter 2.8 * 10^-15 m? What kinetic energy would the proton (mass approximately 1.6 * 10^-27 kg) have at this velocity?
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.Uncertainty velocity=`dp/m
@& The electron is confined to within `dx = 2.8 * 10^-15 m.
The uncertainty principle gives you the uncertainty in momentum.
From that you can find the uncertainty in the velocity.*@
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Problem Number 7
A hypothetical atom with negligible kinetic energy has a mass of 166 amu. It undergoes a gamma decay. The remaining atom has atomic mass which is less than that of the original by .0000135 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?
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@& What is a gamma decay?
What is the mass defect?
What is the energy equivalent of the mass defect?
What therefore is the wavelength of the photon that carries away this energy?*@
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Problem Number 8
The table below depicts properties of the isotopes of the eight lightest elements. For the element of column 10, is it possible from an energy standpoint for the isotope in the fourth line to emit a neutron and change to the isotope in the third line?
Properties of Selected Particles and Light Nuclei (most common isotope listed first)
particle or atom proton neutron hydrogen helium lithium beryllium boron carbon nitrogen oxygen
atomic number 1 0 1 2 3 4 5 6 7 8
mass (amu) 1.007276 1.008665 1.007825 4.002603 6.015122 9.012182 10.01294 12 14.00307 15.99492
mass (amu) 2.014102 3.016029 7.016004 11.00931 13.00336 15.00011 16.99913
mass (amu) 3.016049 17.99916
electron mass (amu) atomic mass unit
0.000549 1.660502 * 10^-27 kgw
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Nitrogen emit a neutron and change to isotope in 3rd line?
Mass=15.00011 amu Neutron mass is 1.6749 * 10^-27 kg=1.008 amu
The nucleus and neuron will have a combined mass of 14.00307 amu + 1.008 amu=15.011 amu
It is not possible because the mass of the resulting atom is greater than the initial mass.
@& Good.*@
@& Check my notes. Some of my notes break the problem down into smaller questions, so feel free to insert responses (mark with &&&&).*@