放射性同位素地球化学1ppt课件

合集下载

放射性同位素地球化学

Nd同位素的演化(1) －整体地球(CHUR)，地幔和地壳的分异
岩浆演化：超基性基性中性酸性， Sm/Nd(147Sm/144Nd)比值降低
Nd同位素的演化(2)-模式年龄
亏损地幔模式年龄
eNd(0)
TDM ←亏损地幔模式年龄
TCHUR
Nd同位素亏损地幔模式年龄的计算
1) 143Nd/144Nd = (143Nd/144Nd )DM + (147Sm / 144Nd) (elTDM – 1)
=
1 137.88
(el235T – el235t) (el238T – el238t)
U-Pb等时线的形成
207Pb 204Pb
( ) 207Pb 204Pbo
等时线
t2
c2
增长曲线
b2
c1
t1
a2 a1
b1
a
b
c
to
235U 204Pb
低Pb高U的体系 - 锆石 U-Pb体系的演化，谐和线
谐和线
作用、幔源岩浆发生结晶分异作用和富集REE的矿物发生分
选作用的时间等。T2DM的计算还需知道地幔物质进入地壳后，并在发生Sm/Nd比值变化前的147Sm/144Nd比值，即地壳的
147Sm/144Nd比值。对于沉积岩类，往往用上地壳的平均比值
来代替：0.1180.017(540个全球沉积岩平均值)，但对于中下
Sample/Chondrite
100
Chondrite N-MORB E-MORB OIB Continental Crust Upper Crust Lower Crust
10 1 La Ce Pr Nd Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

《放射性同位素》课件

特性
放射性同位素具有不稳定性和不稳定性，会自发地发生核反应，释放出能量和射线。
放射性同位素的应用领域
医学诊断和治疗
01
放射性同位素在医学领域中广泛应用于诊断和治疗，如放射性
核素显像、放射性核素治疗等。
工业检测和控制
02
放射性同位素可以用于工业检测和控制，如厚度测量、金属探
伤等。
科学研究
03
放射性同位素在科学研究领域中广泛应用于核物理、化学、生
03
放射性同位素还可以用于治疗肿瘤，通过向肿瘤组织发射高能射线，杀死癌细胞并抑制其生长。
04
放射性同位素在医学领域的应用需要严格控制剂量和安全性，以避免对健康造成损害。
工业检测与控制
放射性同位素在工业领域的应用主要包括检测和控制工艺流程。
输标02入题
通过使用放射性同位素标记物质，可以检测产品的质量和纯度，例如在石油工业中检测油品的纯度和在食品工业中检测食品的成分。
安全与环保的挑战
放射性废物的处理与处置
放射性同位素在生产、使用过程中产生的废物需要妥善处理和处置，以避免对环境和人类健康造成危害。
VS
辐射防护与安全监管
在使用放射性同位素的过程中，应加强辐射防护措施，确保工作人员和公众的安全。同时，需要建立完善的监管体系，确保放射性同位素的安全使用。
国际合作与政策法规的完善
有电离本领，穿透能力最强。
不同的辐射类型具有不同的能量和穿透能力，适用于不同的应用领域，例如医学影像技术、工业无损检测、核能利用等。
稳定性
稳定性是指放射性同位素原子核保持稳定状态的能力。
有些放射性同位素原子核不稳定，会发生衰变，释放出能量和射线；有些放射性同位素原子核稳定或半稳定，不会发生衰变或发生衰变但释放的能量较低。

放射性同位素地球化学

0
Ha wai i
160 ¡
180 ¡
160 ¡
放射性同位素地球化学
140 ¡
North America
120 ¡
放射性同位素地球化学
Jason Morgan‘s Plume Model
• Upwelling from thermal boundary layer at the base of the mantle
放射性同位素地球化学
2.1 地球的圈层结构（1），地幔的基本组成和结构
放射性同位素地球化学
放射性同位素地球化学
类地行星的形成
放射性同位素地球化学
主要陨石类型的相对含量
普通球粒陨石
普通球粒陨石
放射性同位素地球化学
球粒陨石类的主要特征
放射性同位素地球化学
碳质球粒陨石组成与太阳光球的组成基本一致
通用二元混合方程
• Vollmer（1976）和Langmuir等（1978）先后给出了二元混合体系微量元素浓度的通用表达式。该式理论上可适用于任何元素和同位素。对任何一个二组份混合体系，其方程为
Ax＋Bxy＋Cy＋D＝0 （5.62） • 其中x，y是横坐标、纵坐标的变量，可以是
元素或元素的比值。当端元1和端元2上的坐标即比值为（x1，y1）（x2，y2）时系数可表示为：
放射性同位素地球化学
亏损地幔的贡献－大洋地壳的形成
拉斑玄武岩
放射性同位素地球化学
放射性同位素地球化学
富集地幔的贡献－大洋岛的形成
碱性玄武岩
放射性同位素地球化学
Kamtschatka
60 ¡ Alaska
70
PACIFIC
40 ¡

放射性同位素地球化学1(共86张PPT)

1.4 U-Th-Pb体系
地球化学性质(一)
• U和Th均属锕系元素，常为+4价，但在地球表层条件下，U呈+6价；
• 由于较大的离子半径和高电价，U和Th 均表现为强不相容元素；
• +4价U、Th较稳定，但+6价的U可呈 UO22-溶于水而发生迁移；
地球化学性质(二)
• 除极少数情况下以沥青铀矿(uraninite， UO2)和硅酸钍矿(thorite)形式成独立矿物外，多数条件下U和Th呈分散状分布于造岩矿物中或集中于副矿物中(锆石、独居石、磷灰石、榍石)；
e 2) Nd= 0.25*T2－3T＋8.5 (T in Ga)
3) eNd=
143Nd/ 144Nd - (143Nd/144Nd) CHUR (143Nd/144Nd) CHUR
×１０4
(143Nd/144Nd )CHUR = 0.512638；(147Sm/144Nd) CHUR = 0.1967
根本的数学关系与参数
206Pb = 206Pbi + 238U (el238t – 1) 207Pb = 207Pbi + 235U (el235t – 1)
208Pb = 208Pbi + 232Th (el232t – 1)
(1)对于低Pb高U的体系〔如锆石〕
(206Pb/ 238U)* = (el238t – 1) (207Pb/ 235U)* = (el235t – 1)
Rb-Sr等时线的形成
87Sr 86Sr
( ) 87Sr 86Sr o
a
b
87Rb 86Sr
c
to
Rb-Sr等时线的形成
87Sr 86Sr

Ar-Ar——同位素地球化学课件PPT (1)

1. 如果样品不均匀，将带来误差； 2. 测试方法不同，测试精度不同，影响总体年龄精度； 3. 样品用量较大
• Ar-Ar法克服了这些缺点，同一份样品测试，同种方法，大大减小了样品量。
优点二：质谱测量
• 质谱仪的测试精度很高，而钾含量的测试误差相对较大，成为影响测试精度的主要因素；
• Ar-Ar法只需要质谱仪测试，而且测同位素比值，不测绝对含量，因此不需要加稀释剂。
缺点
收取。 • 石英瓶用镉箔屏蔽热中子，放入核反应堆。 • 照射过程中样品罐旋转，并置于冷水中。 • 快中子通量，累计通量，照射时间。
J值有关问题
• 标样在样品罐中的位置 • 多个标样，纵向和横向上 • 分别测试每个标样，看J值的变化规律
关于照射
• 1989年，《放射性同位素与射线装置放射防护条例》 • 2003 年以前，管理不严，国内发生多起事故。 • 2003 年，《放射性污染防治法》 • 2005年，《放射性同位素与射线装置安全和防护条例》
• 具备上述条件，照射前申请，由公安局、卫生局和环保局审批
放射性废物的处理
• 2003年，《中华人民共和国放射性污染防治法》
– 低、中水平放射性固体废物在符合国家规定的区域实行近地表处置。
– 高水平放射性固体废物实行集中的深地质处置。
• 2009年，《放射性物品运输安全管理条例》 • 2011年，《放射性废物安全管理条例》 • 法规前后放射性废物的处理方式很大变化。
39K为原子数，△T为照射的时间长度，φ（ε）为能量为ε的中子通量密度，ó（ε）为39K 对具有ε能量的中子的俘获截面，该积分的积分区间是中子的整个能谱。
• 39Ar不稳定，以269年的半衰期通过β辐射衰变成39K。

同位素地球化学PPT课件

32
1）轻稳定同位素
A. 原子量小，同一元素的各同位素间
的相对质量差异较大（ΔA/A≧5%）；
B. 轻同位素组成变化的主要原因是同
位素分馏作用造成的，其反应是可逆的。
2019/7/3
第五章同位素地球化学Ⅰ
33
2）重稳定同位素
A. 原子量大，同一元素的各同位素间的相
对质量差异小（ΔA/A＝0.7~1.2%），环境的物理和化学条件的变化通常不导致重稳定同位素组成的改变；
526262621放射性同位素衰变定律及同位素地质年代学原理622kar法及40ar39ar法年龄测定623rbsr法年龄测定624smnd法年龄测定625upb法年龄测定53621621同位素地质年代学的基本原理前提及分类541放射性原子释放出粒子和能量的现象即所谓的放2放射性衰变元素的原子核自发地发出粒子和释放能量而变成另一种原子核的过程
2019/7/3
第五章同位素地球化学Ⅰ
11
5. 同位素地球化学发展现状
同位素地球化学发展迅速，已渗透到地球科学的各个研究领域，如：大地构造学、岩石学、矿床学、海洋学、环境科学、空间科学等。
主要表现在以下方面：
♣ 实验测试技术不断完善和提高； ♣ 多元同位素体系的综合研究； ♣ 研究领域不断扩大； ♣ 各种新方法的出现。
28
② 类型
1）放射性同位素(unstable or radioactive isotope)
其原子核是不稳定的，它们能自发地放出粒子并衰变成另一种同位素。
2）稳定同位素(stable isotope)
原子核是稳定的，或者其原子核的变化不能被觉察。元素周期表中，原子序数相同，原子质量不同，化学性

放射性同位素 ppt课件

要经过筛选才能培育出优良品种．用 γ 射线治疗肿瘤对人体肯定有副
作用，因此要科学地控制剂量．本题正确选项为 D.
.
学习探究区
学案3
针对训练 2 正电子发射计算机断层显像(PET)的基本原理是：将放射性
同位素 15O 注入人体，参与人体的代谢过程.15O 在人体内衰变放出正
电子，与人体内负电子相遇而湮灭转化为一对光子，被探测器探测到，
目开
D．放射性同位素容易制造解析人工放射性同位素用作示踪原子，主要是用人工放射性同
位素代替没有放射性的同位素参与正常的物理、化学、生物过程，
既要利用化学性质相同，也要利用衰变规律不受物理、化学变化
的影响，同时还要考虑放射性废料容易处理，因此选项 A、C 正
确，选项 B、D 错误．
.
自我检测区
栏其中3105P 是3115P 的一种放射性同位素．
目开
5．放射性同位素的应用
(1) 利用放射性同位素放出的射线可进行 ___γ___ 探伤、消除
__静___电___、培育___良__种___、治疗癌症． (2)放射性同位素可以作为__示__踪____原子．
(1)1919 年卢瑟福发现质子的核反应：
目 174N＋42He→________＋11H
开 (2)1932 年查德威克发现中子的核反应：
94Be＋42He→________＋10n (3)1934 年约里奥—居里夫妇发现放射性同位素和正电子的核反应：
2173Al＋42He→________＋01n；3105P→3104Si＋________＋ν.
本 (2)示踪原子的应用：利用放射性元素能放出某种射线，可用探测器

《放射性同位素》课件 (2)

4 性质和用途
放射性同位素可以用于研究和治疗，还广泛应用于能源生产和工业生产中。
应用领域
医学
放射性同位素在肿瘤治疗和医学诊断中发挥着重要作用。
环境
放射性同位素用于环境监测和放射性废物处理。
能源
核能发电利用放射性同位素产生热能，驱动涡轮发电机产生电力。
考古学
利用放射性同位素碳-14对古代遗物进行年代测定。
核能与核电
核能
核能是指核反应过程中释放的能量，用来产生核电和核武器。
核电
核电是通过核反应产生热能，进而转化为电能的过程。
危害与防护
1 辐射影响
放射性同位素可能对生物体造成辐射伤害，导致放射病、癌症等健康问题。
2 防护措施
合理使用防护设备、隔离措施和安全操作符合核安全规范。
3 管理与监控
放射性同位素
放射性同位素是具有放射性的原子核形态，具有独特的特点和广泛的应用领域。
定义和特点
1 无稳定同位素
放射性同位素没有稳定的核结构，会通过放射性衰变释放能量和粒子。
2 半衰期
每种放射性同位素都有其特定的半衰期，即在该时间内，总数减少一半。
3 放射性衰变
放射性同位素会衰变为其他同位素或元素，释放出粒子、γ射线和能量。
采用监测措施和管理制度确保放射性同位素的安全和合规。
结论和展望
放射性同位素的特性和应用使其在科学、医学、能源等领域具有重要地位。未来的研究

同位素地球化学1

I NTRODUCTION AND P HYSICS OF THE N UCLEUSI NTRODUCTIONIsotope geochemistry has grown over the last 40 years to become one of the most important fields in the Earth Sciences as well as in geochemistry. It has two broad subdivisions: radiogenic isotope geochemistry and stable isotope geochemistry. These subdivisions reflect the two primary reasons why the isotopes of some elements vary in nature: radioactive decay and chemical fractionation. One might recognize a third subdivision: cosmogenic isotope geochemistry, but this subdivision is perhaps better considered a part of radiogenic isotope geochemistry, since although cosmogenically produced isotopes are not products of radioactive decay, they are products of nuclear reactions.The growth in the importance of isotope geochemistry reflects its remarkable success in attacking fundamental problems of Earth Science, as well as problems in astrophysics and physics. Isotope geo-chemistry has played an important role in transforming geology from a qualitative, observational science to a modern quantitative one. To appreciate the point, consider the Ice Ages, a phenomenon that has fascinated geologist and layman alike for the past 150 years. The idea that much of the northern hemisphere was once covered by glaciers was first advanced by Swiss zoologist Louis Agas-siz in 1837. His theory was based on observations of geomorphology and modern glaciers. Over the next 100 years, this theory advanced very little, other than the discovery that there had been more than one ice advance. Isotopic studies in the last 40 years (and primarily in the past 20) have deter-mined the exact times of these ice ages and the exact extent of temperature change (about 3° or so cooler in temperate latitudes). Knowing the timing of these glaciations has allowed us to conclude that variations in the Earth’s orbital parameters (the Milankovitch parameters) and resulting changes in insolation have been the direct cause of these ice ages. Comparing isotopically deter-mined temperatures with CO2 concentrations in bubbles in carefully dated ice cores leads to the hy-pothesis that atmospheric CO2 plays and important role in amplifying changes in insolation. Care-ful U-Th dating of corals is now revealing the detailed timing of the melting of the ice sheet. Com-paring this with stable isotope geothermometry shows that melting lagged warming (not too surpris-ingly). Other recent isotopic studies have revealed changes in the ocean circulation system as the last ice age ended. Changes in ocean circulation may also be an important feedback mechanism af-fecting climate. Twenty years ago, all this was very interesting, but not very relevant. Today, it pro-vides us with critical insights into how the planet’s climate system works. With the current concern over potential global warming and greenhouse gases, this information is extremely ‘relevant’.Other examples of the impact of isotope geochemistry could be listed. The list would include such diverse topics as ore genesis, mantle dynamics, hydrology, and hydrocarbon migration, monitors of the cosmic ray flux, crustal evolution, volcanology, oceanic circulation, archeology and anthropology, environmental protection and monitoring, and paleontology. Indeed, there are few, if any, areas of geological inquiry where isotopic studies have not had a significant impact.One of the first applications of isotope geochemistry remains one of the most important: geochro-nology and cosmochronology: the determination of the timing of events in the history of the Earth and the Universe. The first ‘date’ was obtained by Boltwood in 1907, who determined the age of a uranium ore sample by measuring the amount of the radiogenic daughter of U, namely lead, present. Other early applications include determining the abundance of isotopes in nature to constrain models of the nature of the nucleus and models of nucleosynthesis (the origin of the elements). Work on the latter problem still proceeds. The usefulness of stable isotope variations as indicators of the condi-tions of natural processes was recognized by Harold Urey in the 1940’s.This course will touch on many, though not all, of these applications. Before discussing applica-tions, however, we must build a firm basis in the physical and chemical fundamentals.P HYSICS OF THE N UCLEUSEarly Development of Atomic and the Nuclear TheoryThat all matter consists of atoms was first proposed by John Dalton, an English school teacher, in 1806. Prout showed in 1815 that atomic weights were integral multiples of the mass of hydrogen.This observation was strong support for the atomic theory, though it was subsequently shown not to hold for all elements. J. J. Thomson developed the first mass spectrograph in 1906 and provided the answer as to why the Law of Constant Proportions did not always hold: those elements not having in-teger weights had several isotopes, each of which had mass that was an integral multiple of the mass of H. In the mean time, Rutherford had made another important observation: that atoms con-sisted mostly of empty space. This led to Bohr’s model of the atom, proposed in 1910, which stated that the atom consisted of a nucleus, which contained most of the mass, and electrons in orbit about it.It was nevertheless unclear why some atoms had different mass than other atoms of the same element. The answer was provided by W. Bothe and H. Becker of Germany and James Chadwick of England: the neutron. Bothe and Becker discovered the particle, but mistook it for radiation.Chadwick won the Nobel Prize for determining the mass of the neutron in 1932. Various other experiments showed the neutron could be emitted and absorbed by nuclei, so it became clear t h a tdiffering numbers of neutrons caused some atoms to be heavier than other atoms of the same element.This bit of history leads to our first basic observation about the nucleus:it consists of protons and neutrons.Some Definitions and UnitsBefore we consider the nucleus in more detail, let’s set out some defi-nitions:N : the number of neutrons, Z : the number of protons (this is the same as atomic number, since the number of protons dictates the chemical properties of the atom), A : Mass number (N+Z), M : Atomic Mass, I :Neutron excess number (I=N-Z). Iso-topes have the same number of pro-tons, but different numbers of neutrons; isobars have the same mass number (N+Z); isotones have the same number of neutrons but different number of protons.The basic unit of nuclear mass is the dalton (formerly known as the amu, or atomic mass unit), which is based on the mass 12C ≡12, that is,the mass of 12C is 12 daltons. The masses of atomic particles are:proton: 1.007593 daltons (oramu, atomic mass units) =1.6726231 x 10-27 kg neutron 1.008982 daltonsFigure 1.1 Neutron number vs. proton number for stable nu-clides.electron 0.000548756 daltons = 9.10093897 x 10-31 kgNucleons, Nuclei, and Nuclear ForcesFigure 1.1 is a plot of N vs. Z showing which nuclides are stable. A key observation in understand-ing the nucleus is that not all nuclides (combinations of N and Z) are stable. In other words, we cannot simply throw protons and neutrons (collectively termed nucleons) together and expect them to necessarily to form a nucleus. For some combinations of N and Z, a nucleus forms, but is unstable, with half-lives from >1015yrs to <10–12sec.An interesting observation from Figure 1.1 is that N ≈Z for stable nuclei. Thus a significant portion of the nucleus consists of protons, which obviously tend to repel each other by electrostatic force. Notice also that for small A, N=Z, for large A, N>Z. This is another important observation t h a t will lead to the first model of the nucleus.From the observation that nuclei exist at all, it is apparent that another force must exist that is stronger than coulomb repulsion at short distances. It must be negligible at larger distances, otherwise all matter would collapse into a single nucleus. This force, called the nuclear force, is a manifestation of one of the fundamental forces of nature (or a manifestation of the single force in nature if you prefer unifying theories), called the strong force. If this force is assigned a strength of 1, then the strengths of other forces are: electromagnetic 10-2; weak force 10-5; gravity 10-39 (we’ll discuss the weak nuclear force later). Just as electromagnetic forces are mediated by a particle, the photon, the nuclear force is mediated by the pion. The photon carries one quantum of electromagnetic force field; the pion carries one quantum of nuclear force field. A comparison of the relative strengths of the nuclear and electromagnetic forces as a function of distance is shown in Figure 1.2.Atomic Masses and Binding EnergiesThe carbon 12 atom consists of 6 neutrons, 6 protons and 6 electrons. But using the masses listed above, we find that the masses of these 18 particles do not add to 12 daltons, the mass of 12C. There is no mistake, they do not add up. What has happened to the extra mass? The mass has been converted to the energy binding the nucleons.It is a general physical principle that thelowest energy configuration is the most stable. We would expect that if 4He is stable relative to two free neutrons and two free protons, 4He must be a lower energy state compared to the free particles. If this is the case, then we can predict from Einstein's mass-energy equiva-lence:E = mc2 1.1that the 4He nucleus has less mass that 2 free neutrons and protons. It does in fact have less mass. From the principle that the lowest energy configurations are the most stable and the mass-energy equivalence, we should be able to predict the relative stability of various nuclei from their masses alone.We define the mass decrement of an atom as:d = W – M 1.210–410–510–610–710–810–910–10Distance, 10-12 cmV,ergsFigure 1.2. The nuclear and electromagnetic potential of a proton as a function of distance from the proton.where W is the sum of the mass of the constituent particles and M is the actual mass of the atom. For example, W for 4He is W = 2m p +2m n + 2m e = 4.034248 daltons. The mass of 4He is 4.003873 daltons, so d = 0.030375 daltons. Converting this to energy using Equ. 1.1 yields 28.28 MeV. This energy is known as the binding energy . Dividing by A,the mass number, or number of nu-cleons, gives the binding energy per nucleon , E b :E b =W –M Ac21.10This is a measure of nuclear stabil-ity: those nuclei with the largestbinding energy per nucleon are the most stable. Figure 1.3 shows E b as a function of mass. Note that the nucleons of intermediate mass tend to be the most stable. This distribution of binding energy is important to the life history of stars, the abundances of the elements, and radioactive decay, as we shall see.Some indication of the relative strength of the nuclear binding force can be obtained by comparing the mass decrement associated with it to that associated with binding an electron to a proton in a hydrogen atom. The mass decrement we calculated above for He is of theorder of 1%, 1 part in 102. The massdecrement associated with bindingan electron to a nucleus of the orderof 1 part in 108. So bonds between nucleons are about 106 times stronger than bonds between electrons and nuclei.The Liquid Drop ModelWhy are some combinations of N and Z more stable than others? The answer has to do with the forces between nucleons and how nucleons are organized within the nucleus. The structure and organi-zation of the nucleus are questions still being actively researched in physics, and full treatment is cer-tainly beyond the scope of this class, but we can gain some valuable insight to nuclear stability by considering two of the simplest models of nuclear structure. The simplest model of the nucleus is the liquid-drop model , proposed by Niels Bohr in 1936. This model assumes all nucleons in a nucleus have equivalent states. As its name suggests, the model treats the binding between nucleons as similar to the binding between molecules in a liquid drop. According to the liquid-drop model, the total bindingFigure 1.3 Binding energy per nucleon vs. mass number.Figure 1.4 Variation of surface, coulomb, and volume energy per nucleon vs. mass number.of nucleons is influenced by 4 effects: a volume energy, a surface energy, an excess neutron energy, and a coulomb energy. The variation of three of these forces with mass number and their total effect is shown in Figure 1.4.In the liquid drop model, the binding energy is given by the equation:B(A,I)= a 1A – a 2A 2/3 – a 3I 2/4A – a 4Z 2/4A 3 + d1.11where:a 1: heat of condensation (volume energy µA) = 14 MeV a 2: surface tension energy = 13 MeV a 3: excess neutron energy = 18.1 MeV a 4: coulomb energy = 0.58 MeVd : even-odd fudge factor. Binding energy greatest for even-even and smallest for odd-odd.Some of the nuclear stability rules above can be deduced from equation 1.11. Solutions for equation 1.11 at constant A, that is for isobars, result in a hyperbolic function of I, as illustrated in Figure 1.5.For odd A, one nuclei will lie at or near the bottom of this function (energy well). For even A,twoFigure 1.5 Graphical illustration of total binding energies of the isobars of mass number A= 81 (left) and A=80 (right). Energy values lie on parabolas, a single parabola for odd A and two parabolas for even A. Binding energies of the 'last'proton and 'last' neutrons are approximated by the straight lines in the lower part of the figure. After Suess (1987).curves result, one for odd-odd, and one for even-even. The even-even curve will be the one with the lower (more stable) one.Odd-Even Effects and Magic NumbersSomething that we have alluded to and which the liquid drop model does not explain well is the even-odd effect. This effect is illustrated in Table 1.1. Clearly, even combinations of nuclides are much more likely to be stable than odd ones. This is the first indication that the liquid drop model does not provide a complete description of nuclear stability. Another observation not explained by the liquid drop model is the so-called Magic Numbers. The Magic Numbers are 2, 8, 20, 28, 50, 82, and 126. Some observations about magic numbers:1. Isotopes and isotones with magic numbers are unusually common (i.e., there are a lot of differentnuclides in cases where N or Z equals a magic number).2. Magic number nuclides are unusually abundant in nature (high concentration of the nuclides).3. Delayed neutron emission in fission occurs in nuclei containing N*+1 (where N* denotes a magicnumber) neutrons.4. Heaviest stable nuclides occur at N=126 (and Z=83).5. Binding energy of last neutron or proton drops for N*+1.6. Neutron-capture cross sections for magic numbers are anomalously low.7. Nuclear properties (spin, magnetic moment, electrical quadrupole moment, metastable isomericstates) change when a magic number is reached.Table 1.1. Numbers of stable nuclei for odd and even Z and NZ N A number of stable nuclei number of very long-lived nuclei(Z + N)odd odd even45odd even odd503even odd odd553even even even16511The Shell Model of the NucleusThe state of the nucleus may be investigated in a number of ways. The electromagnetic spectra emitted by electrons is the principal means of investigating the electronic structure of the atom. B y analogy, we would expect that the electromagnetic spectra of the nucleus should yield clues to its structure, and indeed it does. However, the g spectra of nuclei are so complex that not much progress has been made interpreting it. Observations of magnetic moment and spin of the nucleus have been more useful (nuclear magnetic moment is also the basis of the nuclear magnetic resonance, or NMR, technique, used to investigate relations between atoms in lattices and the medical diagnostic technique nuclear magnetic imaging).Nuclei with magic numbers of protons or neutrons are particularly stable or ‘unreactive’. This is clearly analogous to chemical properties of atoms: atoms with filled electronic shells (the noble gases) are particularly unreactive. In addition, just as the chemical properties of an atom are largely dictated by the ‘last’ valence electron, properties such as the nucleus’s angular momentum and mag-netic moment can often be accounted for primarily by the ‘last’ odd nucleon. These observations sug-gest the nucleus may have a shell structure similar to the electronic shell structure of atoms, and leads to the shell model of the nucleus.In the shell model of the nucleus, the same general principles apply as to the shell model of the atom: possible states for particles are given by solutions to the Schrödinger Equation. Solutions to this equation, together with the Pauli Exclusion principle, which states that no two particles can have exactly the same set of quantum numbers, determine how many nucleons may occur in each shell. In the shell model, there are separate systems of shells for neutrons and protons. As do electrons, pro-tons and neutrons have intrinsic angular momentum, called spin, which is equal to 1/2h (h =h/2π, where h is Planck's constant and has units of momentum, h = 6.626 x 10-34 joule-sec). The total nuclear angular momentum, somewhat misleadingly called the nuclear spin, is the sum of (1) the intrinsic an-gular momentum of protons, (2) the intrinsic angular momentum of neutrons,and (3) the orbital angular momentum ofnucleons arising from their motion in thenucleus. Possible values for orbitalangular momentum are given by l , theorbital quantum number, which mayhave integral values. The total angular momentum of a nucleon in the nucleus is thus the sum of its orbital angular momentum plus its intrinsic angular momentum or spin: j = l ± 1/2. The plus or minus results because the spin angular momentum vector can be either in the same direction or opposite direction of the orbital angular momentum vector. Thus nuclear spin is related to the constituent nucleons in the manner shown in Table 1.2.Let’s now return to magic numbers and see how they relate to the shell model. The magic numbers belong to two different arithmetic series:N = 2, 8, 20, 40, 70, 112...N = 2, 6, 14, 28, 50, 82, 126...The lower magic numbers are part of the first series, the higher ones part of the second. The numbers in each series are related by their third differences (the differences between the differences between the differences). For example, for the first of the above series:28204070112Difference 612203042Difference 681012Difference 222This series turns out to be solutions to the Schrödinger equation for a three-dimensional harmonic os-cillator (Table 1.3). (This solution is different from the solution for particles in an isotropic Coulomb field, which describes electron shells).Table 1.3. Particles in a Three-Dimensional Harmonic Oscillator (Solution of Schrödinger Equation)Nl 010213j1/21/23/21/23/25/21/23/25/27/2State s +p -p +s +d -d +p -p +f -f +No.2242462468S 261220Total(2)(8)(20) (40)N is the shell number; No. gives the number of particles in the orbit, which is equal to 2j +1; S gives the number ofparticles in the shell or state, and total is the total of particles in all shells filled. Magic number fail to follow the progression of the first series because only the f state is available in the fourth shell.Magnetic MomentA rotating charged particle produces a magnetic field. A magnetic field also arises from the orbital motion of charged particles. Thus electrons in orbit around the nucleus, and also spinning about an in-ternal axis, produce magnetic fields, much as a bar magnet. The strength of a bar magnet may be mea-sured by its magnetic moment, which is defined as the energy needed to turn the magnet from a posi-tion parallel to an external magnetic field to a perpendicular position. For the electron, the spin magnetic moment is equal to 1 Bohr magneton (µe ) = 5.8 ¥ 10-9 ev/gauss. The spin magnetic moment of the proton is 2.79 nuclear magnetons, which is about three orders of magnitude less than the Bohr magneton (hence nuclear magnetic fields do not contribute significantly to atomic ones). Surprisingly,in 1936 the neutron was also found to have an intrinsic magnetic moment, equal to -1.91 nuclear magne-tons. Because magnetism always involves motion of charges, this result suggested there is a non-uni-Table 1.2. Nuclear Spin and Odd-Even Nuclides Number of Nucleons Nuclear Spin Even-Even 0Even-Odd 1/2, 3/2, 5/2, 7/2 ...Odd-Odd 1,3form distribution of charge on the neutron, which was an early hint that neutrons, and protons, werecomposite particles rather than elementary ones.Total angular momentum andmagnetic moment of pairs of protons cancel because the vectors of each member of the pair are aligned in opposite directions. The same holds true for neutrons. Hence even-even nuclei have 0 angular momentum and magnetic moment.Angular momentum, or nuclear spin,of odd-even nuclides can havevalues of 1/2, 3/2, 5/2, and non-zero magnetic moment (Table 1.2). Odd-odd nuclei have integer value of angular momentum or 'nuclear spin'.From this we can see that the angu-lar momentum and magnetic mo-ment of a nuclear are determined by the last nucleon added to the nu-cleus. For example, 18O has eight protons and 10 neutrons, and hence 0angular momentum and magnetic moment. Adding one proton to this nucleus transforms it to 19F, which has angular momentum of 1/2 and magnetic moment of ~2.79. For thisreason, the shell model is also sometimes called the single-parti-cle model, since the structure can be recognized from the quantum-mechanical state of the “last”particle (usually). This is a little surprising since particles are assumed to interact.The three-dimensional harmonic oscillator solution explains only the first three magic numbers;magic numbers above that belong to another series. This difference may be explained by assuming there is a strong spin-orbit interaction, resulting from the orbital magnetic field acting upon the spin magnetic moment. This effect is called the Mayer-Jensen coupling. The concept is that the energy state of the nucleon depends strongly on the orientation of the spin of the particle relative to the orbit, and that parallel spin-orbit orientations are energetically favored, i.e., states with higher values of j tend to be the lowest energy states. This leads to filling of the orbits in a somewhat differ-ent order; i.e., such that high spin values are energetically favored. Spin-orbit interaction also occurs in the electron structure, but it is less important.Pairing EffectsIn the liquid-drop model, it was necessary to add a term d , the even-odd effect. This arises from a 'pairing energy' that exists between two nucleons of the same kind. When proton-proton and neutron-neutron pairing energies are equal, the binding energy defines a single hyperbola as a function of I (e.g., Figure 1.4). When they are not, as is often the case in the vicinity of magic numbers, the hyper-bola for odd A splits into two curves, one for even Z, the other for even N. An example is shown in Figure 1.6. The empirical rule is: Whenever the number of one kind of nucleon is somewhat larger than a magic number, the pairing energy of this kind of nucleon will be smaller than the other kind.neutron excess number in the vicinity of N=50.Capture Cross-SectionsInformation about the structure and stability of nuclei can also be obtained from observations of the probability that a nucleus will capture an additional nucleon. This probability is termed the cap-ture-cross section, and has units of area. Neutron capture cross sections are generally of greater use than proton capture cross sections, mainly because they are much larger. The reason for this is simply that a proton must overcome the repulsive coulomb forces to be captured, whereas a neutron, being neu-tral, does not feel the electrostatic forces. Neutron-capture cross sections are measured in barns, which have units if 10-24 cm2, and are denoted by s. The physical cross-section of a typical nucleus (e.g., Ca) is of the order of 5 x 10-25 cm2, and increases somewhat with mass number (more precisely, R = r0A1/3, where A is mass number and r0 is the nuclear force radius, 1.4 x 10-13 cm). While many neutron capture cross sections are of the order of 1 barn, they vary from 0 (for 4He) to 105 for 157Gd, and are not simple functions of nuclear mass (or size). They depend on nuclear structure, being for example, gen-erally low at magic numbers of N. Capture cross-sections also dependent on the energy of the neutron, the dependence varying from nuclide to nuclide.Collective ModelA slightly more complex model is called the collective model. It is intermediate between the liq-uid-drop and the shell models. It emphasizes the collective motion of nuclear matter, particularly the vibrations and rotations, both quantized in energy, in which large groups of nucleons can partici-pate. Even-even nuclides with Z or N close to magic numbers are particularly stable with nearly per-fect spherical symmetry. Spherical nuclides cannot rotate because of a dictum of quantum mechanics that a rotation about an axis of symmetry is undetectable, and in a sphere every axis is a symmetry axis. The excitation of such nuclei (that is, when their energy rises to some quantum level above the ground state) may be ascribed to the vibration of the nucleus as a whole. On the other hand, even-even nuclides far from magic numbers depart substantially from spherical symmetry and the excita-tion energies of their excited states may be ascribed to rotation of the nucleus as a whole.R EFERENCES AND S UGGESTIONS FOR F URTHER R EADINGDickin, A. 1995. Radiogenic Isotope Geochemistry. Cambridge: Cambridge University Press.Faure, G. 1986. Principles of Isotope Geology. New York: Wiley & Sons.Suess, H. E. 1987. Chemistry of the Solar System. New York: John Wiley and Sons.。

高中物理第4章第3节放射性同位素ppt课件

为2，月球的土壤中这种质量数为3的氦应表示为 .
放射性同位素的应用及防护
应用人工放射性同位素有哪些优势？提示：(1)种类多．天然放射性同位素只有60多种，而人工放射性同位素有1 000多种． (2)放射强度容易控制 . (3)可制成各种所需的形状． (4)半衰期短，废料易处理．
(1)射线的应用． ①利用放出的γ射线的穿透本领检查金属内部是否存在砂眼裂痕等，即所谓无损的γ探伤． ②利用射线的穿透本领与物质厚度的关系，来检查各种产品的厚度和密封容器中液体的高度等，从而实现自动控制． ③利用射线的电离本领使空气电离而把空气变成导电气体，以消除化纤、纺织品上的静电． ④利用射线的物理化学作用照射植物，引起植物变异而培育良种，也可以利用它杀菌、治病(如放疗)等．
(2)核反应的实质：以基本粒子(α粒子、质子、中子等)为“ 炮
弹” 去轰击原子核(靶核X)，从而促使原子核发生变化，生成
了新原子核(Y)，并放出某一粒子．
除了天然放射性元素会产生自发核衰变外，还可以利用天然放
射性的高速粒子或利用人工加速的粒子去轰击原子核，以产生
新的原子核，这个过程叫做核反应．
在核反应过程中，原子核的质量数和电荷数会发生变化，同时
(2)1932年查德威克发现中子的核反应： (3)1934年约里奥—居里夫妇发现放射性同位素和正电子的核反应：
1．同位素．具有相同质子数而中子数不同的原子，在元素周期表中处于同一位置，因而互称同位素．
2．放射性同位素．
(1)定义：具有放射性的同位素，叫做放射性同位素．
(2)分类：可分为天然放射性同位素和人工放射性同位素．
有些原子核发生变化，形成这种非放射性元素的同位素，但
这种同位素可能不稳定，会发生衰变，因此称这种同位素为

放射性同位素的应用PPT课件

40
2、在人工合成牛胰岛素的过程中掺入放
射性碳14的用途是（C ）
A.作催化剂
Байду номын сангаас
B.作介质
C.作示踪原子 D.作组成元素
3、请写出放射性同位素的三种应用。
点拨
答：放射性示踪技术、射线探伤和测厚、射线治疗肿瘤、射线育种、射线消毒、射线电离技术。（任选3种即可）
41
教材习题解答
本节习题属于自由发挥性习题请同学们自己解答
第三节
5
教学目标
1、知识与能力
了解放射性同位素应用的几个方面。理解放射性同位素各方面应用的原理。对于每个应用能举例说明。
6
2、过程与方法
了解放射线示踪的原理，能列举它在各方面的应用。知道射线探伤和测厚技术原理及应用。初步了解射线治疗的原理，知道放射线治疗的三种方式。初步了解射线育种、保存的原理。了解射线电离技术应用实例。
又如，肿瘤细胞比正常细胞能吸收更多的放射性元素，由此可以诊断肿瘤形成的部位。
再如，锝99可用来做脑部扫描，帮助医生诊断脑部疾病。
18
③ 工业技术
放射新性示踪在工业技术中同样有广泛的应用。
19
例如，将放射性同位素注入输油和输气管中，用探测器可以挖掘管道破裂和泄漏的位置，从而避免大规模的挖掘（如上页的图片）。
想一想
棉花在结桃、开花的时候需要较多的磷肥，但是什么时候的吸收率最高、磷在作物体内能停留多长时间、磷在作物体内分布情况如何呢？
15
我们知道，把磷肥喷在棉花叶子上，磷肥也能被吸收。那么我们可以用放射性同位素磷30制成肥料，喷洒在棉花页面上，然后每隔一定时间用探测器测量棉株各部位的放射性强度，就能研究棉花对肥料的吸收情况。

放射性同位素-核辐射的主角课件PPT

05
案例分析
核电站事故案例
三里岛核事故
1979年，美国宾夕法尼亚州三里岛核电站发生事故，导致反应堆芯部分熔化，大量放射性物质泄漏。事故后，核电站周围设立了隔离区，居民疏散，并对环境进行了长期的监测和清理。
切尔诺贝利核事故
1986年，乌克兰切尔诺贝利核电站发生事故，导致反应堆爆炸，大量放射性物质泄漏。事故造成大量人员伤亡和环境污染，影响范围波及整个欧洲。事故后，切尔诺贝利核电站周围设立了隔离区，并对受损反应堆进行了长期的处理和清理。
科学研究
在科学研究中，放射性同位素可用于核物理、化学、生物学等领域的研究。例如，研究原子核的结构和性质、化学键的性质、生物大分子的结构和功能等。
02
核辐射基础知识
核辐射的定义与种类
核辐射定义
核辐射是指由放射性同位素衰变或核反应过程中释放出的放射性物质所造成的各种电离辐射的总称。
核辐射种类
主要包括α射线、β射线、γ射线、X射线和中子射线等。
放射性同位素在科研领域的应用案例
放射性同位素标记化合物
放射性同位素标记化合物在科研领域中广泛应用于示踪研究。通过标记化合物中的放射性同位素，可以追踪化合物的代谢过程和反应机理，为科学研究提供重要的数据支持。
放射性同位素在地质学研究中的应用
放射性同位素在地质学研究中用于测定岩石和矿物的年龄。通过测量放射性同位素的衰变规律，可以推算出岩石和矿物的形成时间
放射性同位素是核辐射的主要来源之一，它们通过发射出各种射线（如α、β、γ射线）来释放能
量。
放射性同位素在核辐射中的作用是提供能量和信息，这些能量和信息可用于医疗、工业、科研等
领域。
在医疗领域，放射性同位素常用于诊断和治疗肿瘤等疾病，通过向病变组织发射射线来杀死癌细

Rb-Sr——同位素地球化学课件PPT

• 利用等时线既可求出年龄，又可以得到锶初始值
• 数据点拟合程度的好坏，也是检验所有样品是否一直保持封闭的一个尺度
0.710 0.708
t>0
87Sr/86Sr
0.706
t=0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
87Rb/86Sr
同源岩浆岩石Rb-Sr同位素演化和等时线图
等时线的拟合
• 最常用的是双误差最小二乘回归法进行拟合。
• 实际观测也表明，矿物中的Rb—Sr衰变图确实受到影响。
基本假设
• 假设在区域变质或接触变质过程中， 87Sr的变化仅仅是由于放射性同位素87Rb的衰变导致的，而矿物中铷和锶的浓度基本上保持不变
• 如果在热变质过程中，矿物发生了同位素均匀化，一般手标本大小的全岩样品都保持着封闭体系，则可以计算矿物等时年龄（内等时线），代表变质年龄，而全岩年龄代表成岩年龄
未受变质的沉积岩的年龄测定
• 沉积岩中的含铷矿物，既可以是自生的，也可以是它地生成的和碎屑的
• 自生的有海绿石、钾盐、光卤石等 • 海绿石较好，可以反映沉积年龄 • 蒸发岩矿物不够稳定，所测年龄不可靠
小结
• 铷和锶是分散元素，它们在火成岩、沉积岩和变质岩中的浓度从小于几ppm到大于几百ppm。
挑选单矿物时，除了含钾矿物以外，通常还会挑选一种富钙贫钾矿物，如磷灰石、榍石等。它投点基本落在纵轴上。
变质岩的年龄测定
变质岩的年龄测定
• 变质作用可以是热变质，也可以有流体的参与，结果使岩石的总体化学组成和微量元素成分都可能发生变化。
• 可以预料，变质作用对岩石中存在的天然放射性元素的母体和子体的关系将产生深刻影响。