Chapter 2: Principal Constituents of Matter PDF
Document Details
Uploaded by Deleted User
Tags
Related
- ATAR Physics Unit 4 Topic 3: The Standard Model PDF
- Lecture Notes 1 (Nature of Electricity) - DONE PDF
- Generalization of Klein-Gordon Equation in Curved Space (PDF)
- San Beda College Alabang NSTP Waiver (PDF)
- Classical Mechanics 1 (PHY6001) - UMM AL-QURA UNIVERSITY PDF
- Fundamental Nuclear Physics PDF
Summary
This document provides a detailed explanation of the principal constituents of matter, including atoms and subatomic particles. It delves into historical discoveries and experimental findings. The chapter also covers the work of key scientists such as Democritus, Dalton, and Faraday and includes multiple figures describing scientific processes.
Full Transcript
Chapter 2 : Principal contituents of matter 1 Introduction In the fifth century B.C. the Greek philosopher Democritus expressed the belief that all matter consists of very small, indivisible particles, which he named atomos (meaning uncuttable or indivisible). It was in 1808 that John Dalton, formul...
Chapter 2 : Principal contituents of matter 1 Introduction In the fifth century B.C. the Greek philosopher Democritus expressed the belief that all matter consists of very small, indivisible particles, which he named atomos (meaning uncuttable or indivisible). It was in 1808 that John Dalton, formulated a precise definition of the indivisible building blocks of matter that we call atoms. Dalton provided the basic theory: all matter—whether element, compound, or mixture—is composed of small particles called atoms. If matter were composed of atoms, what were atoms composed of? Were they the smallest particles, or was there something smaller? A series of experiments that began in the 1850s and extended into the twentieth century clearly demonstrated that atoms actually possess internal structure; that is, they are made up of even smaller particles, which are called subatomic particles. This research led to the discovery of three such particles—electrons, protons, and neutrons. Discoveries of elementary particles are in this sequence: -Electron by Crookes -Proton by Goldstein -Neutron by Chadwick 2. The discovery of elementary particles 2.1 Electron 2.1.1 Highlighting the existence of electrons a) Faraday’s Law: the big merit of Faraday law on electrolysis was to suggest a relationship between matter and electricity. In 1830, Michael Faraday demonstrated that if electricity is passed through a solution of an electrolyte, it results in chemical reactions at electrodes. The reaction result in the liberation and deposition of matter at electrodes. These results suggest the particulate nature of electricity. Figure 1 Electrolytic cell The two laws of Faraday may be summarized as follows: -The mass formed on an electrode is proportional to the applied charge (quantity of electricity) -If q quantity of electricity gives rise to one atom, 2q,3q, 4q,…nq gives rise to 1,2,3,4…n atoms -Masses of produced substances are proportional to their atomic masses. Later it was proved that the produced mass (m) at electrode is given by: 𝑚 = M ∗ 𝑄 /𝑛 ∗ 𝐹 𝑁 ∗ 𝑒− = 1 𝐹𝑎𝑟𝑎𝑑𝑎𝑦 = 96500 C𝑜u𝑙𝑜𝑚𝑏. ( e* N = 6.022x1023 x 1.6 10 -19 ) 𝑚 = M ∗ 𝐼.𝑡 / 𝑛 ∗ 𝐹 Q: charge, M : atomic mass, n : number of electrons, exchanged in the redox reaction , F : Faraday’s number, t :time, I : current intensity, N: Avogadro’s number é : charge of one electron b) Crookes tube Based on previous works on discharge tube, William Crooks carried out in 1878 discharge tube experiments and discovered new radiations and called them cathode rays, since these rays travel from the cathode towards anode. Crookes tube consists of sealed glass tube (60cm), equipped with two metallic electrodes (cathode, negative electrode and anode, positive electrode). (Figure 2 and S1). The discharge tube contains gas (air, helium or neon) at low pressure (10-6 atm). (The air has been evacuated with vacuum pump). Figure 2. Crookes tube When a high voltage of about 15000 V is applied between the electrodes, the glass screen in opposite side of cathode glows. Evidence is brought that florescence is due to the catholic ray formed by negatively charged particles named electrons. These are the main conclusion of Crookes experiment: - If an object is placed in the path of cathodic ray, it projects a shadow of the object on the screen. the shadow caused by the object indicates that particles were being blocked on their way from the cathode to the anode. And the ray from cathode is propagated in straight line. (rectilinear propagation). - When a small rotor blades are placed in the path of the cathode rays, they seem to rotate (see S2). This proves that the cathode rays are made up of particles of a certain mass so that they have some kinetic energy. - The ray is deflected by electric field towards positive plate (Figure 3), the direction of deflection suggested that they were negatively charged particles Crookes' work opened the door to a number of important discoveries. Other scientists were able to demonstrate that the "cathode ray" was actually a stream of electrons. The term “electron” was coined in 1891 by Irish physicist George Stoney, from “electric ion.” Figure 3. Deflection of rays by electric field. 2.1.2 Characteristics of electron a) J.J Thomson experiment: measure of e/m In 1897, JJ Thomson uses vacuum tube, inside which cathode emits electrons. Electrons are accelerated in electrostatic field created by collimating anodes (through a hole). When electrons leave the anodes they made a very narrow beam. The beam travels two metallic plates with opposite charges. Electrons are under the effect of electric field and are deflected from their path. More importantly, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths using electromagnetism laws (figure 4 and S5), J. J. Thomson was able to calculate the mass-to-charge ratio (e/m) of the electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms. Thomson found the charge-to-mass ratio (e/m) of electrons. (|e|/m)= 1.7588 × 1011 C/Kg. C: Coulomb, is the unit of electric charge Figure 4. J.J Thomson experiment for the measure of e/m Calculation of the mass to charge ratio e/m In J.J. Thomson experiment, in the first step, a beam of electrons is deflected in the presence of an electric field E and the deviation Y is measured. In the second step, the deviation of the beam is cancelled by imposing magnetic field B in the same space of electric field. The first step: The study of the motion of electrons in the capacitor domain ( 0 to L) gives: The forces affecting the electron are electric force and gravity force (neglected) ∑ 𝐹𝑖 = 𝑚𝑎 Fe + mg = Fe= ma , Fe = qE = eE qE= ma (1) a is the acceleration -On OX axis: the movement is uniform and rectilinear The acceleration ax= 0, vx=v0, and thus x= v0t (2) -On OY axis: the movement is uniform accelerated a = ay # 0, implies that : y = ½ a t2 (3) from (1), by replacing a by its value, we get y = ½ (eE/me) t2 arc parabolic From (2) , t = x/v0 Y = ½ (eE/me) (x/v0)2 When x= L, y= y0 y0 = ½(eE/me) (L/v0)2 (4) The measure of y0 gives the value of the mass to charge ratio e/me The second step : electric field ( E) and magnetic field ( B) are both applied and regulated so that the path is rectilinear Uniform rectilinear motion : ∑ 𝐹𝑖 = 𝐹𝑒 − 𝐹𝐵 = 0, Fe= FB Fe= eE FB= Bev0 Thus E=Bv0 and v0= E/B By inserting this value of v0 in equation (4) the value of e/me is deduced. However, the experiment of Thomson was unable to provide separately the charge of electron e and its mass me. b) Millikan experiment: Determination of the charge é In 1909, more information about the electron was uncovered by Robert A. Millikan via his “oil drop” experiments. Millikan created microscopic oil droplets, (by an atomizer, or spray bottle) which could be electrically charged by friction as they formed or by using X-rays. The experimenter follows the motion of one drop with a microscope. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure.6) Millikan concluded that this value must therefore be a fundamental charge—the charge of a single electron Figure 6 Millikan’s experiment measured the charge of individual oil drops. The tabulated data are examples of a few possible values. (table below). One can see how an electron charge is measured by Millikan. Millikan found that all drops had charges that were 1.6x 10-19 C multiples. Since the charge of an electron was now known due to Millikan’s research, and the charge-to- mass ratio was already known due to Thomson’s research (1.759 × 1011 C/kg), it only required a simple calculation to determine the mass of the electron as well. Mass of electron = 1.602 × 10-19 C / 1.759 × 1011 C× 1 kg = 9.107 × 10-31 kg 2.2. The nucleus and nucleons 2.2.1 Nucleus : a) Goldstein experiment – nucleus charge In 1886 Eugen Goldstein discovered evidence for the existence of this positively charged particle. He noted that cathode-ray tubes with a perforated cathode emit a glow from the end of the tube near the cathode. Goldstein concluded that in addition to the electrons, or cathode rays, that travel from the negatively charged cathode toward the positively charged anode, there is another ray that travels in the opposite direction, from the anode toward the cathode. Because these rays pass through the holes, or channels, in the cathode, Goldstein called them canal rays. He called these canal rays and showed that they were composed of positively charged particles Figure 7. Goldstein experiment cathode Figure 8. Ionization of gaseous atoms b) Rutherford experiment Ernest Rutherford put forth the idea of the nuclear model of the atom in 1911, based on experiments done in his laboratory by Hans Geiger and Ernest Marsden. These scientists observed the effect of bombarding thin gold foil (and other metal foils) of 4.10-7m thickness with alpha radiation from radioactive ( 𝟒𝟐𝑯𝒆2+ ) substances such as uranium (Figure 9). Figure 9. Rutherford experiment: Alpha-particle scattering from metal foils Alpha radiation is produced by a radioactive source and formed into a beam by a lead plate with a hole in it. (Lead absorbs the radiation.) Scattered alpha particles are made visible by a zinc sulfide screen, which emits tiny flashes where particles strike it. A movable microscope is used for viewing the flashes. It was found that most of the alpha particles passed through a metal foil as though nothing were there, but a few (about 1 in 8000 figure 10) were scattered at large angles and sometimes almost backward. (Alpha particles are helium nuclei.) Most of the alpha particles pass through the metal atoms of the foil, undeflected by the lightweight electrons. When an alpha particle does happen to hit a metal-atom nucleus, however, it is scattered at a wide angle because it is deflected by the massive, positively charged nucleus Figure 10. Description of alpha beam passage in gold foil According to Rutherford’s model, most of the mass of the atom (99.95% or more) is concentrated in a positively charged center, or nucleus, around which the negatively charged electrons move. Although most of the mass of an atom is in its nucleus, the nucleus occupies only a very small portion of the space of the atom. Nuclei have diameters of about 10-15 m (103 pm), whereas atomic diameters are about 10 -10 m , a hundred thousand times larger. 2.2.2 Nucleons a) Proton In 1918, Rutherford bombarded nitrogen atoms by a beam of highly accelerated alpha particles. A light particle positively charged appears which was proved to be the nucleus of hydrogen and was named proton. The mass of proton mp was determined later to be mp = 1.6726 x 10-24 g= 1.6726 x 10-27 kg It was 1838 times greater than the mass of electron b) Neutron It was not until 1932 that James Chadwick found evidence of neutrons, uncharged, subatomic particles with a mass approximately the same as that of protons. Chadwick bombarded light atom beryllium (and B, Li) by alpha particles. However, the neutrons produced this way were initially hypothesised to be high-energy gamma radiation because they were unaffected by electric and magnetic fields. Chadwick conducted the same experiment using beryllium and paraffin block but provided a different interpretation. He claimed that this unknown radiation was actually neutral particles – neutrons. The neutrons are formed according to the following reaction: Figure 11. Chadwick experiment The mass of neutron was found 1838 times higher than that of electron mn= 1,6747.10-27 Kg The following table summarized the properties of the three subatomic particles. 3. Constitution of the atom 3.1 Symbolic representation of an atom –planetary model Rutherford envisioned the atom as a miniature solar system, with electrons orbiting around a massive nucleus, and as mostly empty space, with the nucleus occupying only a very small part of the atom. The nucleus comprised protons and neutrons. Figure 12. Structure of the atom 3.2 Mass and atomic number The number of protons in the nucleus of each atom of an element is called the atomic number (Z). In a neutral atom the number of protons is equal to the number of electrons, so the atomic number also indicates the number of electrons present in the atom. The chemical identity of an atom can be determined solely by its atomic number. The mass number (A) is the total number of neutrons and protons present in the nucleus of an atom of an element. Except for the most common form of hydrogen, which has one proton and no neutrons, all atomic nuclei contain both protons and neutrons. A nuclide X is an atomic entity characterized by number of the protons and neutrons that compose the nucleus. A= Z + N N= Number of neutrons Example: The mass of atom ma ma = Z * mp + N * mn + Z * me- The mass of electrons is neglected compared to protons and neutrons mass, thus, the nucleus represents almost the totality of the atom mass. ma = Z *mp + N *mn Atomic charge Atomic charge = number of protons - number of electrons. 3.3 Isotopes Atoms or nuclides that have the same atomic number but different mass numbers are called isotopes of the same element. -An element may have one or more isotopes. - Isotopes have same chemical properties and the same place in the periodic table. - Isotopes may only be separated by mass spectrometry Example: Hydrogen isotopes: Hydrogen , Deuteriul and tritium. Example : Cl : 𝟑𝟓 𝟑𝟕 𝟏𝟕𝑪𝒍 , 𝟏𝟕𝑪𝒍 O : 𝟏𝟔𝟖𝑶 , 𝟏𝟕𝟖𝑶 , 𝟏𝟖𝟖𝑶 Relative abundance of isotopes The percentage in number of atoms of each isotope present in natural mixture in called natural abundance. 𝟏𝟐 𝟔𝑪 : 98.89 % 𝟏𝟑 𝟔𝑪 : 1.1 % 𝟏𝟒 𝟔𝑪 : traces (1/100000) The molar mass of element = the relative atomic mass: Maverage Maverage =∑ Xi *Mi /100 With : Xi : designate the natural abundance of the isotope i with molar mass Mi. Example: X( 𝟏𝟐𝟔𝑪) =98.89 %, M ( 𝟏𝟐𝟔𝑪 )= 12g/mol X( 𝟏𝟑𝟔𝑪) =1.1 %, M ( 𝟏𝟑𝟔𝑪 )= 13g/mol Mmoy= (12*98.89 +13*1.1)/100= 12.02 g/mol 3.4. Binding energy in nucleus Mass defect ∆m: The mass of nucleus is lower than the sum of particles composing it taken in free state. The loss in mass accompanying the formation of an atom from protons, neutrons, and electrons is due to the conversion of that mass into energy that is evolved as the atom forms. It is called binding energy. The conversion between mass and energy is most identifiably represented by the mass-energy equivalence equation as stated by Albert Einstein: ∆E = ∆mc2 where ∆E is energy, ∆m is mass of the matter being converted, and c is the speed of light in a vacuum (c=3.108 m/s). The unit in International System is Joule (J). A variety of units are, however, commonly used for nuclear binding energies, including electron volts (eV). By definition one electron volt, 1 eV equalling the amount of energy necessary to the move the charge of an electron across an electric potential difference of 1 volt. The energy E, E = q*U q: the charge in Coulomb, q= I.t, I current intensity in A , t in seconds and U the difference of potential in volt. E= q*U (C*V) = I*t*U ( A*V*s) = (J) 1 eV = = 1,6.10-19 C.1V 1 eV = 1.602 × 10–19 J Also used, mega eV: MeV : 1 MeV = 106 eV = 1.602 × 10–13 J Energy of 1 amu ∆E = ∆mc2 = (3*108)2 ∆m= 1 amu= 1.66*10-27 kg = 1/NA * g= 10-3 /NA * kg (NA Avogadro’s number) ∆E=1.66. 10-27*(3*108)2 =14.94x10-11/ 1.6x10-19= 933 MeV