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Phy 232
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Problem
What is the approximate uncertainty in the velocity of an electron (mass 9.11 * 10^-31 kg) known to remain within a distance of 1.9 Angstroms of a proton?
What kinetic energy would the electron have at this velocity?
Approximately how many times would the electron 'orbit' the proton in a circular 'orbit' at this distance and velocity?
Compare the centripetal acceleration at this distance and velocity to the acceleration of the electron due to the electrostatic force between an electron and a proton at this distance.
&&&In this problem, I know you have to start by determining the position uncertainty, which is described as 2(1.9*10^-10 m)= 3.8*10^-10 m. Next, we find the momentum uncertainty which is dp = (6.63*10^-34 Js) / (3.8*10^-10 m) = 1.74*10^-24 kg*m/s. My question is about the momentum uncertainty. According to the formula, dp = h / (2*r). I calculated this above and came up with my answer, however in the introductory problem set, for this same problem, the calculated answer was 3.4*10^-24 kg*m/s, which is exactly double my answer. I was wondering if we were supposed to multiply by 2 at the end of the equation or if my answer is right and the intro problem had an error in it?
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Either way would be OK, since the uncertainty estimate is not to be regarded as very precise.
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Problem
A hypothetical atom with negligible kinetic energy has a mass of 230.889 amu. It undergoes an alpha decay. The remaining atom has atomic mass 226.8831 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
The change in atomic mass is approximately 230.889 amu - ( 226.8831 amu + 4.0026 amu) = 3.350735E-03 amu, or about 5.56222E-03 * 10^-27 kg.
This corresponds to an energy of E = m c^2 = 5.56222E-03 * 10^-27 kg ( 3 * 10^8 m/s) ^ 2 = 3.015662 * 10^-13 Joules.
A mole of these nuclei would constitute 6.02 * 10^23 nuclei, each releasing 3.015662 * 10^-13 Joules. The total energy released would therefore be
energy released per mole of decaying particles = (6.02 * 10^23 nuclei) * ( 3.015662 * 10^-13 Joules / nucleus) = 18.15429 + 10^10 Joules.
&&&I also had another question about this problem. I understand the process but I was wondering about the calculations. After calculating 230.889 amu -(226.8831 amu + 4.0026 amu )= .0033 amu = 5.478*10^-30 kg. This answer is close to the answer posted on the introductory problem set page which was 5.56*10^-3*10^-27 = 5.56*10^-30 kg. However, after calculating the energy, E = (5.478*10^-30 kg)*(3.0*10^8 m/s)^2 I get 4.93*10^-13 Joules instead of the 3.01*10^-13 Joules that was listed in the problem solution. Is my math wrong up to this point because the difference in my energy answer and the one listed causes a significant difference in the final answer of total energy released and I want to make sure that I'm not messing anything up.
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It appears that the random number generator was off. I think your result is the correct one.
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A hypothetical atom with negligible kinetic energy has a mass of 223 amu. It undergoes a gamma decay. The remaining atom has atomic mass which is less than that of the original by .0000174 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
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 .0000174 kg of mass lost:
photon energy = .0000174 amu * (3 * 10^8 m/s)^2 = 2.8884 * 10^-32 kg * (3 * 10^8 m/s)^2 = 25.9956 * 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 ) / ( 25.9956 * 10^-16 Joules ) = .7639755 * 10^-10 meters = .7639755 Angstroms.
A mole of these nuclei would constitute 6.02 * 10^23 nuclei, each releasing 25.9956 * 10^-16 Joules. The total energy released would therefore be
energy released per mole of decaying particles = (6.02 * 10^23 nuclei) * ( 25.9956 * 10^-16 Joules / nucleus) = 15.64935 * 10^8 Joules.
&&&My last question is about this problem with gamma decay. I noticed that the first line to the given solution talks beta decay, which is another problem completely. I just wanted to make sure that the given solution does indeed correspond to the question.
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The statement of this problem is also randomized, and the word should have been 'gamma'. The entire solution, fortunately, does address the situation of gamma decay.
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#$&*
Phy 232
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I have a question about a test that I previously took. I checked blackboard tonight and received my grade back from phy 232 test 3, which I recently took last Thursday, July 25th. However, I have not received a grade for phy 232 test 2, which I took back on July 11th. I just want to make sure that you have my phy 232 test 2 and that nothing went wrong during the delivery process. My test 3 was sent by the same person in the same fashion as test 2 was. It should have been sent electronically to your email from a Mrs. Custer from Danville Community College. I am leaving to go out of town this weekend and am trying to finish the course before I leave so I am just double checking to make sure that test 2 is in your hands. Thank you.
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I do have the test, which I printed when it came in but mislaid. I'll be posting the score tomorrow.
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