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Modern Physics Relativity Atomic structure Nuclear physics
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Postulates Einstein’s first postulate, called the principle of relativity, states: The laws of physics are the same in every inertial frame of reference. Einstein’s second postulate states: The speed of light in vacuum is the same in all inertial frames of reference and is independent of the motion of the source.
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Time Dilation
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Length Contraction
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Mass Energy
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Problems P1. A spacecraft is moving with respect to the earth. A flashing light on the spacecraft generate a flash every 1.5 s. A person based on earth measures that the time between the flashes is 2.5 s. How fast is the spacecraft moving relative to the earth? P2. What is the apparent length of a metre stick which is in the spacecraft to a person on earth? P3. The period of a pendulum is measured to be 3.00 s in the reference frame of the pendulum. What is the period when measured by an observer moving at a speed of 0.50c relative to the pendulum?
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Problems P1. The electron in a hydrogen atom makes a transition from the n = 2 energy level to the ground level (n = 1). Find the wavelength and frequency of the emitted photon. P2. What is the radius of the electron orbit for a Rydberg atom for which n = 273? How fast is the electron moving in a Rydberg atom for which n = 273? P3. What is the energy of a photon that, when absorbed by a hydrogen atom, could cause an electronic transition from (a) the n = 2 state to the n = 5 state and (b) the n = 4 state to the n = 6 state? P4. A photon is emitted when a hydrogen atom undergoes a transition from the n = 5 state to the n = 3 state. Calculate (a) the energy (in electron volts), (b) the wavelength, and (c) the frequency of the emitted photon.
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P5. The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n = 2 as shown in Figure. Consider the photon of longest wavelength corresponding to a transition shown in the figure. Determine (a) its energy and (b) its wavelength. Consider the spectral line of shortest wavelength corresponding to a transition shown in the figure. Find (c) its photon energy and (d) its wavelength. (e) What is the shortest possible wavelength in the Balmer series?
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Tro, Chemistry: A Molecular Approach 15 Facts About the Nucleus Every atom of an element has the same number of protons – atomic number (Z) Atoms of the same elements can have different numbers of neutrons – isotopes – different atomic masses Isotopes are identified by their mass number (A) – mass number = number of protons + neutrons
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16 Facts About the Nucleus The number of neutrons is calculated by subtracting the atomic number from the mass number The nucleus of an isotope is called a nuclide – less than 10% of the known nuclides are non- radioactive, most are radionuclides Each nuclide is identified by a symbol – Element -Mass Number = X-A
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17 Important Atomic Symbols ParticleSymbolNuclear Symbol protonp+p+ neutronn0n0 electrone-e- alpha beta positron
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18 Radioactivity Radioactive nuclei spontaneously decompose into smaller nuclei – Radioactive decay – We say that radioactive nuclei are unstable The parent nuclide is the nucleus that is undergoing radioactive decay, the daughter nuclide is the new nucleus that is made Decomposing involves the nuclide emitting a particle and/or energy All nuclides with 84 or more protons are radioactive
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Tro, Chemistry: A Molecular Approach 19 Other Properties of Radioactivity radioactive rays can ionize matter – cause uncharged matter to become charged – basis of Geiger Counter and electroscope radioactive rays have high energy radioactive rays can penetrate matter radioactive rays cause phosphorescent chemicals to glow – basis of scintillation counter
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20 Transmutation Rutherford discovered that during the radioactive process, atoms of one element are changed into atoms of a different element - transmutation – Dalton’s Atomic Theory statement 3 bites the dust in order for one element to change into another, the number of protons in the nucleus must change
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Tro, Chemistry: A Molecular Approach 21 Types of Radioactive Rays Rutherford discovered there were three types of radioactivity alpha rays ( ) – what we now know to be helium nucleus beta rays ( ) – negligible mass – electron-like gamma rays ( – form of light energy (not particle like and )
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Tro, Chemistry: A Molecular Approach 22 Alpha Emission an particle contains 2 protons and 2 neutrons – helium nucleus most ionizing, but least penetrating loss of an alpha particle means – atomic number decreases by 2 – mass number decreases by 4
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23 Beta Emission a particle is like an electron – moving much faster – produced from the nucleus when an atom loses a particle its – atomic number increases by 1 – mass number remains the same in beta decay, a neutron changes into a proton
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24 Gamma Emission gamma ( ) rays are high energy photons of light no loss of particles from the nucleus no change in the composition of the nucleus – Same atomic number and mass number least ionizing, but most penetrating generally occurs after the nucleus undergoes some other type of decay and the remaining particles rearrange
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25 Positron Emission positron has a charge of +1 c.u. and negligible mass – anti-electron when an atom loses a positron from the nucleus, its – mass number remains the same – atomic number decreases by 1 positrons appear to result from a proton changing into a neutron
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Tro, Chemistry: A Molecular Approach26 Practice - Write a nuclear equation for each of the following alpha emission from U-238 beta emission from Ne-24 positron emission from N-13 electron capture by Be-7
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27 Fundamental law of radioactive decay Each nucleus has a fixed probability of decaying per unit time. Nothing affects this probability (e.g., temperature, pressure, bonding environment, etc.) [exception: very high pressure promotes electron capture slightly] This is equivalent to saying that averaged over a large enough number of atoms the number of decays per unit time is proportional to the number of atoms present. Therefore in a closed system: (Equation 3.1) – N = number of parent nuclei at time t – = decay constant = probability of decay per unit time (units: s –1 ) To get time history of number of parent nuclei, integrate 3.1: (3.2) – N o = initial number of parent nuclei at time t = 0.
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28 Definitions The mean life of a parent nuclide is given by the number present divided by the removal rate (recall this later when we talk about residence time): (3.3) The half life t 1/2 of a nucleus is the time after which half the parent remains: –This is also the “e-folding” time of the decay: The activity is decays per unit time, denoted by parentheses: (3.4)
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1.How long does it take for 60 percent of a sample of radon to decay? Half life of a radon is 3.82 days. 2. Find the activity of 1 mg of radon, 222 Rn, whose atomic mass is 222u. What will be the activity of this sample be exactly one week later? 1u= 1.66 x 10 -27 kg
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4. A freshly prepared sample of a certain radioactive isotope has an activity of 10 mCi. After 4 hour, its activity is 8 mCi. (a) What is the decay constant? (b) Calculate the half life. (c) What is the sample’s activity 30 h after it is prepared? (d) How many atoms of the isotope were contained in the freshly prepared sample?
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33 In nuclear fission, a large nucleus is bombarded with a small particle the nucleus splits into smaller nuclei and several neutrons large amounts of energy are released Nuclear Fission
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34 Nuclear Fission When a neutron bombards U-235, an unstable nucleus of U-236 forms and undergoes fission (splits) smaller nuclei are produced such as Kr-91 and Ba-142 neutrons are released to bombard more 235 U
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35 Nuclear Fission Diagram and Equation
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36 Nuclear Fusion Nuclear fusion occurs at extremely high temperatures (100 000 000 °C) combines small nuclei into larger nuclei releases large amounts of energy occurs continuously in the sun and stars
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37 Fusion: small nuclei form larger nuclide, release energy This type of Fusion is being Examined as An alternative Energy source On Earth. Small nuclei come together to form larger nuclide, releasing energy
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Binding energy of the calculate the mass defect and the binding energy of an α-particle given the following date: m n = 1.0086665 u m p = 1.007825 u m α = 4.0026 u (1u = 931.5 MeV/c 2 ) Solution: The mass defect of an α-particle is given by Δm = (2 × 1.008665 + 2 × 1.007825) – 4.0026)u = 0.03038u The Binding energy is related to the mass defect by the reaction. B.E. = Δm.c 2 = 0.03038 × 931.5 MeV = 28.3 MeV (approximately)
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