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Phy 122
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Isotopes
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A problem in the Modern Physics problem set states
Set 57 Problem number 23
Problem
Which isotopes of which elements in the table below might possibly decay into which isotopes of which other elements by means of alpha emission?
Which isotopes of which elements in the table below might possibly decay into which isotopes of which other elements by means of beta decay consisting of emission of a negative electron?
Which isotopes of which elements in the table below might possibly decay into which isotopes of which other elements by means of beta decay consisting of electron capture?
From the table below, determine at least one series of alpha and beta decays that will lead from uranium-238 to lead-206. Specify the two isotopes involved in each decay, and tell whether the process is an alpha or a beta decay.
Nuclear Composition of Selected Heavy Nuclei
particle or atom
lead
bismuth
polonium
radon
radium
thorium
palladium
uranium
atomic number
82
83
84
86
88
90
91
92
nucleons
214
214
218
222
226
230
234
238
nucleons
210
210
214
234
235
nucleons
206
209
210
234
Solution
We are interested here in alpha and beta decays:
Alpha decay consists of the emission of an alpha particle. An alpha particle consists of two protons and two neutrons. Therefore, emission of an alpha particle will reduce the number of protons, and therefore the atomic number, by 2 and will reduce the number of nucleons by 4. Note that just because the numbers of protons and nucleons indicate a possibility of the decay does not ensure that the decay is possible--the masses of the particles involved must also be such that the total mass of the decay products is not less than that of the original atom.
Beta decay consists of the decay of a neutron in the nucleus into a proton and an electron, which increases the number of protons and therefore the atomic number by 1 without changing the number of nucleons (the original neutron and the proton that replaces it both count as nucleons).
Another possible mechanism of beta decay is electron capture, in which an electron is captured by a proton in the nucleus. With additional energy released from the resulting changes in the binding energy of the nucleus, the two particles 'reverse decay' into a neutron. This reduces the atomic number by 1 without changing the number of nucleons.
We look at possible alpha and beta decays among the listed elements. Recall again that just because a decay is possible by the table doesn't mean that it can actually happen. Decays happen only if the particles end up in a lower total-energy state.
Potential alpha decays include the following:
Uranium 238 to Thorium 234
Thorium 234 to radium 226, with the emission of two additional neutrons.
Radium 226 to radon 222.
Radon 222 to polonium 218.
Polonium 218 to lead 214.
Polonium 210 to lead 206.
Note that though bismuth 214 has four less nucleons than polonium 218, it has only one less proton and hence this decay cannot occur by alpha emission.
Potential beta decays include the following:
Thorium 234 could emit a negative electron as a neutron decays into a proton and thereby yield palladium 234.
Lead 214 could emit a negative electron as a neutron decays into a proton and thereby yield bismuth 214.
Lead 210 could emit a negative electron as a neutron decays into a proton and thereby yield bismuth 210.
Bismuth 210 could emit a negative electron as a neutron decays into a proton and thereby yield polonium 210.
Bismuth 209 could emit a negative electron as a neutron decays into a proton and thereby yield polonium 209.
Any of these decays could occur in reverse by the capture of an electron, which together with a proton could form a neutron.
A possible transition from U-238 to Pb-206 is:
U-238 alpha decay to Th-234.
Th-234 beta decay to Pa-234.
Pa-234 beta decay to U-234.
U-234 alpha decay to Th-230.
Th-230 alpha decay to Ra-226.
Ra-226 alpha decay to Rd-222.
Rd-222 alpha decay to Po-218.
Po-218 alpha decay to Pb-214.
Pb-214 beta decay to Bi-214.
Bi-214 beta decay to Po-214.
Po-214 alpha decay to Pb-210.
Pb-210 beta decay to Bi-210.
Bi-210 beta decay to Po-210.
Po-210 alpha decay to Pb-206.
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Would it be acceptable on test day to give one example for each category question rather than figuring out every possible combination?
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You won't get a question on a test that asks you to list all the possibilities; that would be too long for a test. I believe all the questions that arise on the test are of reasonable length.
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Phy 122
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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Magnitude
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A modern physics problem 17 states:
Problem
A hypothetical atom with negligible kinetic energy has a mass of 172 amu. It undergoes an alpha decay. The remaining atom has atomic mass 167.994 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
The change in atomic mass is approximately 172 amu - ( 167.994 amu + 4.001 amu) = 4.99677E-03 amu, or about 8.29463E-03 * 10^-27 kg.
This corresponds to an energy of E = m c^2 = 8.29463E-03 * 10^-27 kg ( 3 * 10^8 m/s) ^ 2 = 4.49709 * 10^-13 Joules.
A mole of these nuclei would constitute 6.02 * 10^23 nuclei, each releasing 4.49709 * 10^-13 Joules. The total energy released would therefore be
energy released per mole of decaying particles = (6.02 * 10^23 nuclei) * ( 4.49709 * 10^-13 Joules / nucleus) = 27.0724 + 10^10 Joules.
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I think the numbers might not be populating correctly or I have missed a step in my calculations:
I get
This corresponds to an energy of E = m c^2 = 8.29463E-03 * 10^-27 kg ( 3 * 10^8 m/s) ^ 2 = 7.43 * 10^-13 Joules. Instead of the 4.49... listed above.
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Yes, the 4.49 does not follow correctly whereas the 7.43 is reasonable.
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???Am I right there???
???Also the 8.29463E-03 * 10^-27 kg - am I right in assuming that I could just write that as 8.29 * 10^-30???
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Yes, and your form would be correct scientific notation.
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