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Unformatted text preview: PHYSICS TOPICAL: Atomic and Nuclear Structure Test 1
Time: 23 Minutes* Number of Questions: 18 * The timing restrictions for the science topical tests are optional. If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit. MCAT DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions. Study the passage, then select the single best answer to each question in the group. Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions. If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain. Indicate your selection by blackening the corresponding circle on your answer sheet. A periodic table is provided below for your use with the questions. PERIODIC TABLE OF THE ELEMENTS
1 H 1.0 3 Li 6.9 11 Na 23.0 19 K 39.1 37 Rb 85.5 55 Cs 132.9 87 Fr (223) 4 Be 9.0 12 Mg 24.3 20 Ca 40.1 38 Sr 87.6 56 Ba 137.3 88 Ra 226.0 21 Sc 45.0 39 Y 88.9 57 La * 138.9 89 Ac 227.0 22 Ti 47.9 40 Zr 91.2 72 Hf 178.5 104 Rf (261) 58 Ce 140.1 90 Th 232.0 23 V 50.9 41 Nb 92.9 73 Ta 180.9 105 Ha (262) 59 Pr 140.9 91 Pa (231) 24 Cr 52.0 42 Mo 95.9 74 W 183.9 106 Unh (263) 60 Nd 144.2 92 U 238.0 25 Mn 54.9 43 Tc (98) 75 Re 186.2 107 Uns (262) 61 Pm (145) 93 Np (237) 26 Fe 55.8 44 Ru 101.1 76 Os 190.2 108 Uno (265) 62 Sm 150.4 94 Pu (244) 27 Co 58.9 45 Rh 102.9 77 Ir 192.2 109 Une (267) 63 Eu 152.0 95 Am (243) 64 Gd 157.3 96 Cm (247) 65 Tb 158.9 97 Bk (247) 66 Dy 162.5 98 Cf (251) 67 Ho 164.9 99 Es (252) 68 Er 167.3 100 Fm (257) 69 Tm 168.9 101 Md (258) 70 Yb 173.0 102 No (259) 71 Lu 175.0 103 Lr (260) 28 Ni 58.7 46 Pd 106.4 78 Pt 195.1 29 Cu 63.5 47 Ag 107.9 79 Au 197.0 30 Zn 65.4 48 Cd 112.4 80 Hg 200.6 5 B 10.8 13 Al 27.0 31 Ga 69.7 49 In 114.8 81 Tl 204.4 6 C 12.0 14 Si 28.1 32 Ge 72.6 50 Sn 118.7 82 Pb 207.2 7 N 14.0 15 P 31.0 33 As 74.9 51 Sb 121.8 83 Bi 209.0 8 O 16.0 16 S 32.1 34 Se 79.0 52 Te 127.6 84 Po (209) 9 F 19.0 17 Cl 35.5 35 Br 79.9 53 I 126.9 85 At (210) 2 He 4.0 10 Ne 20.2 18 Ar 39.9 36 Kr 83.8 54 Xe 131.3 86 Rn (222) * GO ON TO THE NEXT PAGE. 2 as developed by Atomic and Nuclear Structure Test 1
Passage I (Questions 16) Semiconductor lasers typically produce light with a wavelength on the order of 1 m. The active region of the laser consists of a thin, narrow semiconductor crystal about 200 m long with partially reflecting facets at the ends through which light can enter or exit. In a semiconductor crystal, electron energies are confined to bands (collections of energy levels very close together in energy) rather than to the discrete energy levels found in a single atom. All semiconductors have a valence band, which is almost completely filled with electrons, and a higher energy conduction band. The smallest energy difference between these two bands is called the band gap E g. Figure 1 shows the energy of the bands, E, as a function of the electron wave momentum for a typical semiconductor. (k is the magnitude of the wave vector of the electron and plays a role analogous to that of a quantum number: the larger its value, the greater the momentum of the electron.) In the semiconductor laser, electrons are optically pumped from the valence band to the conduction band with a light source that produces photons of frequency vP and energy EP = hvP, where h is Planck's constant.
E(k) Conduction Band Eg Valence Band k 1 . Semiconductor lasers are often used as light sources for optical fiber communications. As a photon travels from the laser (index of refraction 3.5) into a glass fiber (index of refraction 1.4), which of the following changes? I. Its energy II. Its frequency III. Its wavelength A. B. C. D. I only III only I and II only I, II, and III 2 . Which of the following is true of the energy Ep of the photons produced by the optical pump? A . Ep must be smaller than the band gap Eg. B . Ep must equal the band gap Eg. C . Ep must be greater than or equal to the band gap E g. D . Ep can be chosen independently of the band gap E g. 3 . What value of Es will yield the most efficient laser? A. B. C. D. the value just before the vertical drop in Figure 2 the value just after the vertical drop in Figure 2 Eg infinity Figure 1 A second light source, having energy Es near the band gap, can now be amplified by the process of stimulated emission. In this process, a photon interacts with a conduction band electron causing it to drop to the valence band. This induces the emission of another photon of the same frequency, which then stimulates another conduction band electron to drop to the valence band. The number of photons increases, and the light is amplified as it travels back and forth through the active region. Figure 2 shows a graph of the natural log of the output intensity Iout divided by the input intensity Iin as a function of Es for a typical semiconductor laser. (Note: hc = 1.24 10 6 eVm, where c is the speed of light in a vacuum.) 4 . A photon emitted by a semiconductor laser is incident on a hydrogen atom. Assuming no ionization takes place, the photon will be absorbed by the atom: A . only if its energy equals the difference between two electron energy levels. B . only if its energy is an integer multiple of Planck's constant h. C . only if its energy is greater than 13.6 eV. D . independent of the value of its frequency. In (Iout/Iin) 0 Eg Es GO ON TO THE NEXT PAGE. Figure 2 KAPLAN 3 MCAT
5 . A particular semiconductor has a band gap of 2.48 eV. What is the wavelength of the photon that is emitted when an electron drops from the lowest energy state of the conduction band to the highest energy state of the valence band? A. B. C. D. 4.0 2.0 1.0 0.5 m m m m 6 . Examination of Figure 2 reveals that above a certain energy, ln (Iout/Iin) becomes less than 0. This implies that: A . Ep = Eg. B . Eg = 0. C . the input light is being absorbed instead of amplified. D . the process of stimulated emission is amplifying the input light. GO ON TO THE NEXT PAGE. 4 as developed by Atomic and Nuclear Structure Test 1
Passage II (Questions 713) Figure 1 shows a graph of the logarithm of the relative abundance of the elements in the universe as a function of mass number A. The dominant process for the synthesis of elements heavier than iron (Fe) is neutron capture. In neutron capture, energetic neutrons bombard a target nucleus which absorbs one or more of them to form an unstable isotope. The unstable nucleus then undergoes beta decay until it becomes stable. A nucleus produced by one or more beta decays is called a beta decay daughter. There are two situations in which neutron capture occurs. The first case is called the s-process and results when the neutron density of the environment is moderate. The capture process is slow, and the time that it takes the target nucleus to absorb a neutron is large compared to the time that it takes for the resulting unstable nucleus to beta decay. Certain nuclei with a "magic" number of neutrons have a small probability for capture. Therefore, beta decay daughters with magic neutron numbers N = 50, 82, and 126 account for the abundance peaks in Figure 1 at A = 90, 138, and 208. The second situation in which neutron capture occurs is called the r-process and results when the neutron density of the environment is very high. In this case the rate of capture is very rapid compared to the beta decay rate of the resulting unstable nucleus. Therefore, a nucleus captures many neutrons before undergoing beta decay. Since an unstable nucleus with a magic number of neutrons has a low probability of absorbing more neutrons, its beta decay daughter will have a relatively high abundance. This accounts for the second set of peaks in Figure 1 at A = 80, 130, and 194.
Logarithm of Relative Abundance 7. undergoes a beta decay in the following reaction:. 118 In X + e .What is the element X? 49 A. B. C. D.
119 49 In 118 50 Sn 118 48 Cd 1 1H 118 49 In 8 . Although the Sun is primarily composed of hydrogen, why is rapid neutron capture more common than rapid proton capture? A . Free neutrons rapidly decay to stable protons. B . Neutrons are in greater supply than protons. C . Free protons have less kinetic energy than free neutrons because they are heavier. D . Neutrons do not experience electrostatic repulsion from the nucleus. 9 . Which of the following nuclear reactions will maximize the production of heavy elements by neutron capture? A. B. C.
13 4 16 6 C + 2 He 8 O 1 2 3 1 H + 1 H 2 He + 8 8 5 B 4 Be + e +n - D. n p+e He 8 D 6 4 2 0 -2 Li-Be-B N= 50 N=82 N=126 1 0 . A 111 Cd atom is exposed to a high density of 48 neutrons and absorbs 7 of them before it beta decays. After the neutron flux discontinues, what is the final stable isotope produced? A.
118 48 Cd 122 50 Sn 118 50 Sn 119 49 In Iron Group B. C. D. 50 100 150 200 Atomic Weight Figure 1
Figure 1 adapted from Reviews of Modern Physics Journal #29, 1957. Reprinted courtesy of the American Physical Society. GO ON TO THE NEXT PAGE. KAPLAN 5 MCAT
1 1 . What is the ratio of the abundance of hydrogen in the universe to that of iron? A. B. C. D. 1:104 1:103 105:1 106:1 1 2 . An element undergoes beta decay, and 25% of the original sample remains after 16 hours. What is the half-life of the element? A. B. C. D. 1 hour 2 hours 4 hours 8 hours 1 3 . A stable nucleus resulting from an s-process neutron capture is likely to: A . have a magic number of neutrons. B . result from a single beta decay of a nucleus having a magic number of neutrons. C . beta decay into a nucleus having a magic number of neutrons. D . undergo an r-process neutron capture. GO ON TO THE NEXT PAGE. 6 as developed by Atomic and Nuclear Structure Test 1
Questions 14 through 18 are NOT based on a descriptive passage. 1 4 . The element 132 is formed as a result of 2 alpha 45 decays, 2 positron decays and a gamma decay. Which of the following is the parent element? A. B. C. D.
124 39 140 47 140 49 140 51 1 8 . The following is a table of the abundance of a radioactive element as a function of time. What is the half-life of the element? Time (years) number of radioactive particles (10 23 ) 4 1.5 0.55 0.20 0 5 10 15 1 5 . In demonstrating the photoelectric effect with zinc, ultraviolet light is used rather than visible light because ultraviolet light: A. B. C. D. moves faster. has a greater energy per photon. is more intense. is not visible. A. B. C. D. 2.5 years 3.5 years 5 years Cannot be determined from the information given. 1 6 . The four lowest energy levels of an electron in an atom are the ground state and the first three excited states. What is the maximum number of different lines in the emission spectrum of the atom that can be accounted for by transitions in these four levels? A. B. C. D. 3 4 6 12 1 7 . Four hydrogen atoms fuse to form a helium atom in a series of reactions. How much energy, in MeV, is released in this process? (Note: The atomic mass of hydrogen is 1.0080 amu. and the atomic mass of helium is 4.0026 amu. c2 = 932 MeV/amu.) A. B. C. D. 0 MeV 6.8 MeV 18.9 MeV 27.4 MeV END OF TEST KAPLAN 7 MCAT
ANSWER KEY: 1. B 6. C 2. C 7. B 3. A 8. D 4. A 9. A 5. D 10. C 11. 12. 13. 14. 15. C D A D B 16. C 17. D 18. B 8 as developed by Atomic and Nuclear Structure Test 1
EXPLANATIONS Passage I (Questions 1--6) 1. B The photon, being a quantum of light, exhibits also the wave behavior of light (and of waves in general) as it crosses the boundary from one medium to another. The frequency remains the same, and so does the energy since it is related to the frequency by the equation E = hf. The wavelength and the speed, on the other hand, do change. The speed of light in a medium with an index of refraction n ( 1) is given by v = c/n, where c is the speed in vacuum. This accounts for the refraction of light as described by Snell's law. Since the speed changes while the frequency remains constant, the wavelength will have to change to maintain the equality v = f. Only statement III is therefore correct, making B the correct choice. C In the passage we are told that electrons are pumped from the valence band to the higher energy conduction band. The band gap Eg is the smallest difference in energy between these two bands. When an electron is optically pumped, it absorbs a photon of a particular energy and makes a transition to a higher energy state. By conservation of energy, the difference between the final and initial energy states of the electron equals the energy of the photon absorbed. Within the band gap there are no states that the electron can occupy, so the electron must absorb enough energy to make a transition across the gap. The smallest energy that can bring about this transition is equal to the band gap. In this case an electron in the highest energy level of the valence band is excited to the lowest energy level of the conduction band. In general, an electron may start at a lower level in the valence band and/or end up at a higher energy level in the conduction band. The photon energy then will need to be greater. The photon absorbed by an electron must therefore have an energy at least as big as the band gap in general. 2. conduction band
transition E g (band gap) valence band A Even though efficiency is not formally defined for us in this context, it should not be difficult to reason that the higher the output intensity relative to the input intensity, the more efficient the system is in extracting energy. Either from this alone or from examining the answer choices, our attention should be directed to an examination of the graph in Figure 2. The x-axis corresponds to the energy of the second light source, while the y-axis plots the natural logarithm of Iout/Iin. The properties of logarithms are worth reviewing. In general, the larger a number, the larger its natural logarithm, but the two do not increase at the same rate. The natural logarithm of a negative number is undefined; the natural logarithm of a number between 0 and 1 is negative; while the natural logarithm of 1 is zero. From the graph, the y-value is zero when Es < Eg. So under such conditions, the quotient (Iout/Iin) is equal to one, since its natural logarithm is zero. Or, in other words, Iout = Iin. The y-value then increases until the vertical drop. Immediately before the drop, the y-value is at its maximum, and therefore so is Iout/Iin: the laser is at its most efficient. Immediately after the drop, the y-value takes on a negative value (and stays negative). A negative value for the natural logarithm means that the quotient (Iout/Iin) is less than one: i.e., the output intensity is less than the input intensity. In short, the laser is operating at its most efficient level (producing the most output intensity per unit of input intensity) when Es takes on the value right before the discontinuity occurs. 4. A When a hydrogen atom absorbs a photon, its electron makes a transition to a higher energy state. This can only occur if the difference in energy between the initial and final states equals the energy of the photon. Choice B states that the photon has to have an energy that is an integer multiple of h. We know that the energy of a photon is E = hf, where f is the frequency. Choice B then suggests that the photon's frequency has to be an integer in order for it to be absorbed. This is a gross distortion of the concept of quantization. The actual quantization requirement is that the energy of light comes in packets and has a minimum value of hf, i.e. it can be hf, 2hf, 3hf, etc. There is no restriction on the value of f itself. Besides, this requirement tells us nothing about the absorption of the photon. Choice C is the condition for ionization: A photon above this value will eject the electron from the atom, and the residual energy will be the kinetic energy of the electron. We are told, however, that ionization does not occur and besides, a photon can be absorbed without causing ionization. So choice C is incorrect. Choice D is also clearly wrong in that the energy of a photon is proportional to its frequency. 3. KAPLAN 9 MCAT
5. D The energy of the photon emitted is equal to the energy difference between the final and initial states of the electron. The final state of the electron in this case is the highest energy state of the valence band, while the initial state is the lowest energy state of the conduction band. The energy difference is the band gap. (See also explanation to #2 above.) The photon energy must therefore be 2.48 eV. To determine its wavelength, we use the relationship: E = hf = hc/ = 6. hc 1.24 10 6 eVm = = 0.5 106 m = 0.5 m E 2.48 eV C As discussed in the explanation to #3, ln (Iout/Iin) < 0 means that (Iout/Iin) < 1. This is a property of the natural logarithmic function. The output intensity is therefore less than the input intensity. Some of the input light has been absorbed. Choice A states that the energy E p of the photon that "pumps" electrons from the valence to the conduction band is equal to the band gap Eg. This cannot be deduced: pumping is needed to populate the conduction band with electrons so they can relax. As long as this is accomplished, the actual energy of photons used in pumping plays no role in the subsequent processes of stimulated emission and/or absorption. Choice B states that the band gap is zero. This need not be the case either. Choice D describes the region of the plot where ln (Iout/Iin) is greater than zero. Passage II (Questions 7--13) 7. B This question does not require any information from the passage and is a straightforward one dealing with beta-decay. In this process, a neutron decays into a proton and an electron that is ejected from the nucleus. As a result, the mass number A (number of neutrons + protons) does not change. The atomic number, Z, which is the number of protons in the nucleus, increases by 1. The general scheme for the beta-decay of an isotope Y is therefore:
A 0 A ZY Z +1X + 1 where the beta-particle is the ejected electron. Given the parent nucleus in the question, the daughter nucleus has to be one with atomic number 49 + 1 = 50 and mass number 118. 8. D Neutrons and protons both reside in the nucleus and have roughly the same mass. The most important difference is in the charge: The proton is positively-charged while the neutron carries no charge. How does this come into play? In order for a particle to be captured by a nucleus, it must penetrate the nucleus and get close enough to the nucleons so that the short range strong force can exert its effect and bind the particle to the nucleus. Since the protons in the original nucleus impart to it a positive charge, it will repel other protons that approach it. A neutron, however, experiences no electrostatic repulsion from the nucleus, making the initial approach much easier. 9. A We are told in the first paragraph of the passage that in neutron capture, the target nucleus absorbs one or more neutrons with which it is being bombarded, and then undergoes beta-decay. To initiate the process, then, a source of free neutrons is needed. Among all the reactions given in the choices, only choice A produces neutrons. 10. C Cadmium-111, upon absorption of 7 neutrons, will become cadmium-118. However, this is not a stable nucleus and, as the passage tells us, subsequently undergoes beta-decay. We are not told that how many times beta-decay occurs; but the mass number should remain at 118. Only choices A and C are possible. Choice A corresponds to no beta-decay, and therefore contradicts the information given. Choice C is the daughter nucleus after two beta-decay processes, and is the correct answer. Note that if one did not pay attention to the mass number, and assumed that beta-decay occurs only once, one might have leapt to the wrong choice of D. 10 as developed by Atomic and Nuclear Structure Test 1
11. C This question requires graphical analysis along with manipulation of logarithms. Figure 1 shows a plot of the logarithm of the natural abundance for each element. We are, however, not given any information as to what these abundances are relative to. This, as we shall see, poses no problem. Let us call Ai the abundance of element i, and A 0 the reference abundance. The graph in Figure 1 is then a plot of log (Ai/A0) for each element i. For hydrogen, log (AH/A0) is very close to 11. Similarly, for iron, log (AFe/A0) is about 6. We can then perform the following manipulations, keeping in mind the property of logarithms: log AH A Fe log = 11 6 = 5 A0 A0 log ( log ( AH A Fe )=5 A0 A0 AH A0 )=5 A0 AFe AH =5 AFe log AH 105 = 105 = AFe 1 AH:AFe = 105:1 12. D By definition, for each half-life that passes, 50% of the sample remains. After one half-life, then, 1 of the original 2 1 1 1 1 sample remains, and after two half-lives, of this , or of the starting sample, will have remained. 25% is equivalent to 2 2 4 4 , and therefore in this question, 16 hours must correspond to two half-lives. The half-life of the element is thus 8 hours. A The s-process is described in the second paragraph of the passage. In such a process, a nucleus captures a neutron, and before it has an opportunity to capture more, undergoes beta-decay. If a beta-decay daughter has a magic number of neutrons, it will be unlikely that it will capture more neutrons to undergo more nuclear reactions. Such daughter nuclei will therefore be stable. Choice A correctly describes the situation. Choice B is incorrect. A stable nucleus formed under the s-process has a magic number of neutrons after the beta-decay, not before. If a nucleus with a magic number of neutrons undergoes beta-decay, it will have one fewer neutron. Choice C is wrong because a stable nucleus will not undergo beta-decay. A stable nucleus formed under the s-process, in fact, is the product of a beta-decay into a nucleus having a magic number of electrons. Choice D is incorrect because there is no evidence to suggest that a stable nucleus formed by an s-process is likely to undergo an r-process. Independent Questions (Questions 1418) D Let Z and A be the atomic and mass numbers of the parent nucleus respectively. First let us consider the alpha decays. An alpha decay occurs when the parent nucleus emits a helium nucleus. The parent therefore loses 2 protons and 2 neutrons: in other words, its atomic number decreases by 2 and its mass number decreases by 4. In this question the parent undergoes two such decay reactions; Z thus decreases by 4 and A by 8. Now let us consider the effect of the positron decays. In a positron (or beta-plus) decay, a proton decays into a neutron and a positron that is emitted. The mass number is not affected, while the atomic number decreases by 1. The net result of two positron decays is therefore a net decrease of Z by 2. Finally, for the gamma decay process, neither the atomic number nor the mass number is affected since all that is "shedded" by the parent nucleus is excess energy in the form of a photon. The total effect of all the decay processes is that Z decreases by (4 + 2) = 6 and A decreases by 8. Note that regardless of the exact order in which these decay reactions take place (which is not given in the question stem), the overall effect would be the same: i.e. it does not matter whether the positron or the alpha decays occur first, etc. 14. 13. KAPLAN 11 MCAT
Now that we have identified the relation between the parent's and the daughter's mass and atomic numbers, identifying the correct choice should be straightforward. The daughter nucleus in the question has an atomic number of 45 and a mass number of 132. The parent, then, must have an atomic number of (45 + 6) = 51 and a mass number of (132 + 8) = 140. You may notice that once the atomic number has been determined, we can arrive at the correct choice. B We are asked to determine which answer choice best describes why ultraviolet light is better than visible light when demonstrating the photoelectric effect with zinc. We are not given any specific information about the photoelectric effect for zinc. However, ultraviolet light has a higher frequency than visible light. Since the energy of a photon is the product of its frequency and Planck's constant, a photon of UV light has a greater energy than a photon of visible light. Let us explore why this would be more desirable for the photoelectric effect. In the photoelectric effect, the incident photons must each have a certain minimum energy in order to liberate an electron from the metal. This minimum energy is known as the work function of the metal. If the energy of the photon is not high enough, the photoelectric effect will not occur: no electrons will be ejected. If zinc has a work function that is higher than the energy corresponding to the energy of a visible light photon, then light of higher frequency will be needed. We do not know for sure if it is indeed the case, but this would certainly constitute an explanation as to why an ultraviolet photon is more effective. Choice A is incorrect because all electromagnetic radiation travels at the same speed in vacuum (3 108 m/s), which is where the experiment is carried out. (Even in other media the speed is not too different from light of one frequency to another.) UV light, therefore, does not move faster than visible light. Intensity is the number of photons crossing a unit cross-sectional area per unit time, and this is independent of the frequency of the photons involved. Choice C is therefore incorrect. While UV light is certainly not visible (otherwise it would be visible light by definition), this has nothing to do with why it is more effective than visible light photons in inducing the photoelectric effect. 16. C An electron emits a photon when it makes a transition from a higher to a lower energy level. Such a process would correspond to a line in the emission spectrum. The number of lines in the emission spectrum is therefore dependent on the number of transitions to a lower energy level. This is most easily visualized with the following diagram:
n=4 n=3 n=2 15. n=1 An electron in the ground state (n = 1) cannot emit a photon since there is no lower-lying energy level. An electron in the n = 2 level can emit a photon and move to the ground state. An electron in the n = 3 level can move down to either the n = 2 or ground state, and the energy of these two transitions would be different. Finally, an electron in the n = 4 level can move down to any of the three levels beneath it, corresponding to three different transitions. There is thus a total of 6 transitions, and this would be the maximum number of lines one can observe in the emission spectrum of a sample of such atoms. (This number is a maximum because if, for example, the states are equally-spaced, then the (2 1), (3 2) and (4 3) transitions would emit photons of the same energy, and thus correspond to lines in the same position in the emission spectrum.) 17. D Each one of the four hydrogen atoms has an atomic mass of 1.0080 amu, and if mass were conserved, the product of the fusion would be expected to have a mass of 1.0080 4 = 4.0320 amu. The actual mass of the helium atom thus formed, however, is 4.0026 amu. This difference in mass, or mass defect, has been converted into energy and released. The mass defect is related to the amount of energy released via the equation E = mc2, where m is the mass defect and c the speed of light in vacuum. In this question, c2 is given in terms of MeV/amu. Straightforward multiplication of this with the mass defect, in terms of amu, would thus give an energy value in MeV: 12 as developed by Atomic and Nuclear Structure Test 1
E = (4.0320 4.0026) amu 932 MeV MeV = 0.0294 amu 932 amu amu Since the answer choices are far apart, we can safely round 932 off to 1000. The value of E is therefore 29.4 MeV. The only answer choice that is close to this value is 27.4 MeV. B The half-life is the time it takes for half of the radioactive atoms to decay. We are thus looking for the time when half of the substance is left. Looking at the values in the table, none of the values for the number of particles is one half that of another, so we cannot just read off the half-life. The first pair of values is 4 1023 and 1.5 1023 particles, respectively. This is more than a halving of the initial size of the sample, and so the half-life has to be less than 5 years. In order to choose between choices A and B, we need to do some interpolation: if the half-life is 2.5 years, then the sample size would decay to 2 1023 when t = 2.5 years. When t = 5 years, two half-lives will have passed and the sample would have been halved again, reduced to one-quarter its original size, i.e. 1 1023 particles. This, however, is not the case. The table clearly tells us that at t = 5 years, there are 1.5 1023 particles. The element therefore does not decay quite as rapidly, which means that it has a longer half-life than 2.5 years. Choice B is correct. Choice D is incorrect because there is enough information to determine the half-life. We could have obtained the value of the half-life by actual calculation. Since we know that radioactive decays are first-order reactions, we know the general form the rate law would take: N t = N 18. where N is the size of the sample and is some constant. The integrated rate law, relating the size of the sample at any time to the original sample size, is: N = N0et The values given in the table can be fitted to determine the value of , which is related to the half-life, 1/2, by: = ln 2 1/2 So for example, we could have taken the last entry of the table and substitute the values into the integrated rate law: 0.20 1023 = 4.0 1023 e15 0.2 e15 = = 0.05 4.0 15 = ln (0.05) = 3 = 0.2 ln 2 = 0.2 1/2 1/2 = ln 2 0.69 = = 3.45 yrs 0.2 0.2 Of course, we do not need to actually do this (it would also have been very difficult since we have needed to take the natural logarithm of different numbers). The point is simply that the values given in the table contain the information necessary to derive the complete relationship between the number of radioactive particles left and time elapsed. Once we know that the answer can be found somehow, the approach described at the beginning allows us to determine the right answer. KAPLAN 13 ...
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This note was uploaded on 04/08/2008 for the course MCAT 100 taught by Professor Wong during the Spring '08 term at HKU.
- Spring '08