Answers to Practice Test Questions 2 Atoms, Isotopes and Nuclear Chemistry
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1 Answers to Practice Test Questions 2 Atoms, Isotopes and Nuclear Chemistry. Fluine has only one stable isotope. Its mass number is _9_. A neutral atom of fluine has 9 protons, 0 neutrons and 9 electrons. Fluine can only fm one kind of ion. The charge of this ion is -, and it is fmed when fluine gains one electron(s). 2. Most elements exist as a mixture of isotopes (atoms with the same number of protons, but different numbers of neutrons) which have different masses. The mass repted on the periodic table is a weighted average of the isotopic masses. 3. Step : Wk out the percent abundance f each isotope If the percent abundance of 235 U is 80.00% and the percent abundance of 234 U is negligible (i.e. ~0%) then the percent abundance of 238 U must be 20.00% (since the sum of the percent abundances must be 00%). Step 2: Calculate the average atomic mass of the enriched uranium % U 235 % U 238 = % 00% 80.00% 20.00% = ( u) ( u) 00% 00% = 88.0u 47.6u = 235.6u Step 3: Check your wk Does your answer seem reasonable? Are sig. fig. crect? The maj isotope in this sample is 235 U so it makes sense that the average atomic mass is close to 235 u. The average atomic mass of the uranium will decrease, approaching u (the mass of 235 U).
2 4. Step : Look up average atomic mass of copper (Cu) on periodic table M Cu = u Step 2: Set up equations relating percent abundances to average atomic mass and to each other x y M Cu = M Cu 63 M Cu 65 00% 00% There are only two naturally occurring isotopes, so x y = 00% Step 3: Solve f one of the two percent abundance values (solving f x is shown) x 00% x M Cu = M Cu 63 M Cu 65 00% 00% x ( 00% x) u = ( u) ( u) 00% 00% % = x % x.9982x = 38.2% x = 69.5% Therefe, the natural abundance of 63 Cu is 69.5% Step 4: Solve f the other percent abundance value The natural abundance of 65 Cu is 00% % = 30.85% Step 5: Check your wk Does your answer seem reasonable? Are sig. fig. crect? The average atomic mass of Cu is less than 64 u (the halfway point between 63 and 65), so we expect the natural abundance of 63 Cu to be greater than that of 65 Cu. 5. (e) 00 40ZZZZ 00 4 NNNN 0 ββ 22 82PPPP BBBB 0 ββ 8 82PPPP HH 4 2 αα 7 NN 0 CC 6 pp NNNN nn NNNN pp 0 nn is a proton 0 is a neutron 6. (e) 3 26 AAAA 0ee 26 2MMMM It is also acceptable to write 0 ββ instead of 0 ee AAAA HHHH 0 ββ ZZZZ YY 0 ββ EEEE 4 2 HHHH 256 0MMMM 0 nn XXXX 95 SSSS 3 nn UU
3 7. 2 HH HH HH 0 0 ββ 0 vv You were only expected to include the neutrino 2 2 HH HH 0 0 ββ 0 vv because you were told it was produced and provided with the symbol HH HH 0 γγ 2 3 HHHH 3 2 HHHH 4 HH HH To balance the equation, you need a second product HHHH 2 HH to add 2 to the mass number and 2 to the atomic 2 3 HHHH 3 2 HHHH 4 2 pp pp number. 2HHHH is not a reasonable choice HHHH 2 pp Two HH ( two pp) is. You can answer this question with a picture like the one shown below (iginal image at ): Or you can answer by showing how all the equations add up to give the overall equation. If doing this, you will want to show the proton produced in part as HH instead of pp. Note that the reactions shown in parts and must each be done twice to generate 3 the two 2HHHH needed f part. 2 HH HH HH 0 0 ββ 0 vv 2 HH HH HH 0 0 ββ 0 vv HH HH 0 γγ HH HH 0 γγ HHHH 2HHHH HH HH 4 4 HH 2 0 ββ vv 2 0 γγ overall
4 PPPP 30ZZZZ 2CCCC 0 nn Step : Calculate the mass change f this reaction mm = (MM CCCC 277 MM nn ) (MM PPPP 208 MM ZZZZ 70 ) mm = mmmmmm mmmmmm mmmmmm mmmmmm mm = mmmmmm Step 2: Calculate the energy change f this reaction EE = mmcc 2 EE = mmmmmm mm ss 2 kkkk 000 JJ kkkk mm2 ss 2 3 JJ EE = mmmmmm Therefe, J/mol of energy is absbed. Step 3: Check your wk Does your answer seem reasonable? Are sig. fig. crect? 9. F each of the isotopes listed below, use the band of stability graph to determine whether not it is expected to be stable. F each unstable isotope, predict which mode(s) of decay are most likely. Z = 42 N = 53 stable (falls along band of stability i.e. the black dots) Z = 4 N = 3 unstable (underneath the band of stability = not enough neutrons; also, N/Z < which is very uncommon among stable isotopes) predict decay by either electron capture of positron emission (to increase N/Z); nuclide is too small f alpha emission actually decays by positron emission Z = 9 N = 28 unstable (above the band of stability = too many neutrons) predict beta emission (to decrease N/Z) actually decays by beta emission Z = 9 N = 6 unstable (underneath the band of stability = not enough neutrons; also, N/Z < which is very uncommon among stable isotopes) predict decay by either electron capture of positron emission (to increase N/Z); nuclide is too small f alpha emission actually decays by positron emission (e) Z = 99 N = 4 unstable (underneath the band of stability = not enough neutrons) predict decay by alpha emission (since nuclide is large) actually decays by alpha emission
5 0. Penetrating power decreases with increasing mass and charge. That is because particles with greater mass and charge are me likely to interact with the atoms/molecules in matter rather than travel past them through the material. Since it has neither mass n charge, gamma radiation should have the highest penetrating power. Neutrinos also have no charge and have very small mass, so they are also expected to have high penetrating power. In fact, their penetrating power is even higher than that of gamma radiation! The penetrating power of the types of radiation emitted by the radioactive isotopes affects the amount and type of safety equipment required to handle those isotopes. e.g. Alpha radiation can be stopped by cloth so, as long as the radioactive isotope(s) will not be inhaled ingested, proper clothing may suffice to protect the wker. On the other hand, if the radioactive isotope(s) emit beta radiation, they will likely be kept behind a protective shield since they would penetrate both clothing and skin. Note that higher penetrating power does not necessarily mean me dangerous. Neutrinos, f example, have such high penetrating power that they tend to pass right through matter (including humans) without interacting with it. Also note that there are other facts to consider beyond the penetrating power of the radiation how much radiation is there? how much energy is the radiation carrying?. Absbed dose (the amount of radiation absbed by a given mass of biological tissue) is repted using the gray. Gy = J/kg (energy absbed divided by mass of tissue) Equivalent dose (the amount of gamma radiation that would do the same amount of biological damage if absbed by the same mass of tissue) is measured using the sievert. Sv = J/kg (energy of gamma rays required to do same amount of damage divided by mass of tissue) 2. To compare the activities of the wooden artifact and of living wood, you need to convert them into the same units! Either convert the activity of the artifact into Bq/g use the activity of living wood to calculate the activity of 25.0 g of living wood (in Bq). Step : Calculate the activity of the artifact in Bq/g 4.65 BBBB BBBB AA 2 = = Step 2: Calculate the decay constant f 4 C llll(2) = kk tt /2 kk = llll(2) tt /2 = llll(2) 5730 yy = yy
6 Step 3: Organize your infmation kk = yy AA = BBBB AA 2 = 0.86 BBBB tt = 0 yy (no time passed) tt 2 =??? yy Make sure units f time and k match Step 4: Calculate time required f activity of 4 C to drop from 840 Bq/g to 748 Bq/g llll AA 2 AA = kk(tt 2 tt ) llll BBBB BBBB = ( yy )(tt 2 0 yy) llll(0.729) = yy tt 2 llll(0.729) tt 2 = = 2608 yy = yy 03 yy This makes the object made from wood chopped down in about 600 BC. Step 5: Check your wk Does your answer seem reasonable? Are sig. fig. crect? Less than half the 4 C activity has been lost so the age of the wood should be less than the half-life of 4 C. 3. Step : Organize your infmation kk =??? AA = 46. BBBB AA 2 = 39.8 BBBB tt = 0 mmmmmm (no time passed) tt 2 = 35 mmmmmm Step 2: Calculate the decay constant f 93 Tc llll AA 2 AA = kk(tt 2 tt ) llll 39.8 BBBB 46. BBBB = kk(35 mmmmmm 0 mmmmmm) llll(0.863) = 35 min kk kk = llll(0.863) = mmmmmm mmmmmm = mmmmmm Step 3: Calculate the half-life of 93 Tc llll(2) = kk tt /2 tt /2 = llll(2) kk = llll(2) mmmmmm = 65 mmmmmm =.7 02 min = 2.8 hoooooooo TTh αα UU tt /2 = yy dd yy h 60mmmmmm 60ss =.40 dd h mmmmmm 07 ss
7 ln(2) = kk tt /2 (e) (f) (g) (h) kk = ln(2) ln(2) = = 4.96 tt / ss 0 8 ss MM UU 238 = mmmmmm mmmmmm nn UU 238 = 3.5 = 0.05mmmmmm NN UU 238 = 0.05mmmmmm aaaaaaaaaa = aaaaaaaaaa mmmmmm AA = kkkk = ( ss )( ) = ss = BBBB aaaaaaaaaaaaaaaa dddddddd = (#pppppppppppppppppp) EE pppppppppppppppp mmmmmmmm wwwwwwwwwwww aaaaaaaaaaaaaaaa dddddddd = JJ 5 JJ = 5 0 = 5 85kkkk kk 0 5 GGGG eeeeeeeeeeeeeeeeeeee dddddddd = RRRRRR aaaaaaaaaaaaaaaa dddddddd = (20)(5 0 5 ) = 0 3 SSSS The following answer is *far* me in-depth than was required f the mark when this was used as a test question. If you eat food off a plate with a uranium glaze, you will ingest uranium. There are two mechanisms f this: First of all, you will scrape some of the glaze off the plate with your utensils. Secondly, some foods may leach uranium out of the plate. Uranium that you have ingested will emit alpha particles. This uranium will certainly move through your digestive tract, but some of it will also end up in your bloodstream. Most of the uranium in the bloodstream is eliminated by the kidneys within 24 hours, so the urinary system will also have significant radiation exposure. The RBE (WR) values we used in calculations of equivalent dose are whole-body averages, but some tissues have higher RBEs than others. It is thus very likely that some of the alpha particles emitted by the uranium you ingest will damage gans that are me sensitive to this fm of radiation than your skin. There are two other facts that make ingestion wse than external exposure. One is that, if you are sitting in front of a plate, even holding it while washing it, many of the alpha particles will travel in directions that do not intersect your body. On the other hand, every alpha particle emitted internally will be absbed by your body. This increases the absbed dose by a large fact in and of itself. The other fact has to do with the fact that alpha particles are easily stopped, and that you wear clothes, so external exposure if mitigated by the protection affded by your clothes, while you have no such protection against alpha particles emitted internally. The imptant thing here is to note the mechanism: By eating off of uranium-glazed plates, you will ingest some uranium. Decay of the uranium atoms results in alpha emission, and this emission occurs inside your body if the uranium was ingested. It is the alpha particles arising from uranium decay that cause the damage. Some students wrote answers that suested that the radiation did something to the food. Certainly, the radiation may damage the molecules of which your food is made, but so does digestion. Irradiating a material does not make it radioactive. The food you eat off a radioactive plate does not radiate alpha particles that it picked up earlier.
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