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Tutorial 1: Universe and Astronomical Measures, Concept of Energy and Atomic Structure. About your tutor Name: CHING Chee Leong : S12-03-17, Phys. Depart. :65162534 : [email protected] Research interest: Gravitational physics and cosmology under Loop Gravity approach. Foundations of quantum mechanics, field theory and symmetries. Consultation: Tuesday: 4-5pm; Wednesday: 2-3pm; Friday:3-4pm or email me for other convenient timings. About tutorial Tutorial timing: Each tutorial class lasts for roughly 50-55 minutes. After that students are allowed to leave the tutorial room. Please kindly sign the attendance list. During the tutorial: • Discuss questions posted in tutorial assignment. • Clear up any doubts. If time is permit, I will conduct some physical demonstrations in the class to enhance your understanding. Question 1: [Astronomical Unit] Astronomical distances are so large compared to terrestrial ones that much larger units of length are needed. An Astronomical Unit (AU) is equal to the average distance from the Earth to the Sun, about 92.9 x 106 miles. A parsec (pc) is the distance at which 1 AU would subtend an angle of exactly 1 second of arc (see fig. below). A light year (LY) is the distance that light, traveling through a vacuum with a speed of 186,000miles/sec (or 300,000km/sec) would cover in 1.0 year. An angle of exactly 1 arc sec Earth 1 pc 1 AU Sun 1 pc pc stands for parallax seconds Recall on some trigonometry q r S Note: For small θ (in radians), Sinq 180 radian 1 o o s s q s rq r r 180 radian 1o 60 min 60 60 sec 60 60 arc sec rad 180 1 1 arc sec rad 180 3600 a) Express the distance from the Earth to the Sun in parsecs and in light years. 1 AU q 1 arc sec 1 AU r 1 arc sec 1 par sec 1 arc sec 1 1AU pc 4.85 10 6 pc 180 3600 r since s = rθ s Given that 1AU= 92.9x106 miles and 1 light-years (LY)= speed of light x one year time =186,000miles/sec x (365 days x24hrs x60mins x60secs) = 5.87x1012 miles. Hence, distance from Earth to Sun is 1 AU 92.9 106 miles 5 1 . 58 10 1LY 5.87 1012 miles 1AU 1.58 10 5 LY 1b) To express 1ly and 1pc in miles, 1 light-years (LY)=186,000miles/sec x (365 days x24hrs x60mins x60seconds) = 5.87x1012 miles 1 AU 1 par sec 1 arc sec 92.9 106 miles 1 par sec 1.911013 miles 1 180 3600 2). [Stellar Parallax] There are many ways to measure the distance of a, say, star from the earth, and one way to do that is through this technique, called stellar parallax. a) Explain how this is done? Parallax is the apparent shift of a foreground object relative to some distant background as the observer’s point of view changes. Astronomers make use this trick as a measurement method. January July Stellar Parallax Measure the angular shift θ θ of the foreground star with respect to background stars, and compute P P P= 1/ Once we know the value of parallax angle P, we can compute D easily parallax angle which implies 2θ D = …? since sin P ≈ P, if P is very small Baseline = 1 AU January July Note: 1 AU = 1 Astronomycal Unit, which is the sunearth distance ≈150 million kilometers Stellar Parallax (cont’d) b) What are the advantages and (c) disadvantages of this technique? It’s arguably the most straightforward and simplest measurement technique… well, except that it only works for relatively nearby stars >_< The parallax shift is getting smaller and smaller as the distance increases (click on figure beside), making it more and more difficult to do measurement accurately. Only few hundreds stars whose distance can be measured this way Ancient astronomers believed in geocentric theory for they could not detect the parallax shift of stars with their naked-eye. Thus, they concluded that either the stars had to be very farrrrrr away, or earth was not moving… Well… they chose the wrong conclusion for thousands years apparently >_< 3) [Hubble’s Law and Cepheids] a) What is Hubble’s Law and how it changes our view on universe? Discuss. Hubble’s Law: Mathematically expresses the idea that more distant galaxies move away form us faster; its formula is ν=Ho x d, where ν is the galaxy’s speed away from us, d is its distance, and Ho is Hubble’s constant, 72.6 ± 3.1 (km/s)/Mpc. Our Universe is expanding and NOT static! Main Scientific Consequences: i) There is an origin for our universe (standard referred as “Big Bang”). ii) Hubble’s Law deduces that our universe has a finite age (not equal to finite space!). 3b) The Andromeda (nearest galaxy to our Milky Way) is approaching with a speed of about 300km/sec. What will be the detected wavelength of the stationary Hydrogen 656nm line? Since the source velocity is small compare to velocity of light, vr/c << 1, vr zc; convention: ' z vr 0 source is approaching vr 0 source is receding vr ' c 300 km / s '656 nm 300,000 km / s 656 nm ' 655.3 nm blue shifted 3c) . If we have a type-I Cepheid with the period of luminosity being approximately 140 hours and an apparent luminosity of 5 x 10-12 watt. What is it's distance from us in light years? From Figure 1.10 in the text book, we get the absolute luminosity (L) for period 140 hrs = 5.04 x 105 s, L ≈1000 Ls =1000 x (3.86 x 1026)W = 3.86x 1029 W The apparent luminosity is L LA 4d 2 L 3.86 10 29 19 d 7 . 84 10 m 12 4 LA 4 (5 10 ) 2.54 kilo pc 8287 light years Method used to measure distances in universe Method Distance (pc) Object Time(yr) Hubble's Law 109-1010 Quasars 1011 Apparent luminosity - Galaxies 108 Virgo Cluster 108 106-107 Andromeda Cluster 107 105 Andromeda Cluster 105 Cepheid Variables - Type I 101-104 Center of Milky Way 103 Parallax 100 Alpha Centauri 101 Apparent luminosity - Super Giants Cepheid Variables - Type II A big question (worth billion dollar) 4) [Mathematic and Group Theory] Do all the even numbers form a group? Explain your answer. Also, why the theory of group is so important in nature? In order to form a group, four conditions must behold: (i) Closure, (ii) Associativity, (iii) Identity exist and (IV) Inverse. Examples: Group action: Group action: Addition Multiplication (NOT a group). (YES). We can check 4 four criteria (i) 2 x 2=4 (still even number) are satisfied. (ii)(2x4)x10=2x(4x10) (i) 2+2=4 (still even number) (iii)Identity= 1 such that 2x1=1x2=2 (but (ii)(2+4)+10=2+(4+10) the identity is not in the set (even (iii)Identity= 0 such that number) 2+0=0+2=2 (iv)Inverse 2 x 1/2=1 (1/2 is not even (iv)Inverse 2-2=0 number) Nature prefers to manifest herself with a lot of symmetries. Group theory is the natural language to study the symmetries of the systems, i.e. rotation, translation inertial frames idea (in classical physics), or gauge theory, supersymmetries (in quantum theory), etc. Symmetries naturally lead to the Conservation laws, i.e energy, momentum and electric charges conservation. 5) [Mechanical energy] We are on top of a 100m high tower with our friend “Frednotsoclever” and would like to investigate the laws of nature. a) What is the potential energy of an iron ball with a mass of 1kg exactly at the height of the tower? Potential energy has no absolute meaning, only the difference (between point considered and reference point) in potential energy carries physical significant. U mgh 1kg (9.81ms 2 ) (100m) 981J b) If we drop the ball. What is the kinetic energy of the iron ball an infinitely short time before hitting the ground? Assume initially velocity of ball is zero (i.e. at rest), and due to energy conservation, KE U 981J c) “Frednosoclever” wants to determine the impact velocity by dividing the height of the building by the time it takes to fall from the top to the bottom. Of course this is wrong. How much will his calculation differ from the correct result? From energy conservation, 1 2 KE mv 981J 2 v 44.3ms 1 correct result From kinematics, 44.3ms1 v u at t 4.52s 2 9.81ms Frednosoclever’s 100m v' 22.12ms 1 false result calculation: 4.52s The discrepancy in impact velocity is around 50%. Fred result is wrong because his calculation is the average velocity and not the instantaneous one (in fact, the velocity is varying along the trajectory). 6) [Conservation of Energy] a) If the asteroid that hit the Yucatan Peninsula was travelling at 10km/s when it released an energy of 1023J, estimate its mass. 1 2 1 mv m (10 103 ) 2 1023 J 2 2 m 2 1015 kg KE b) One kilogram of the explosive TNT releases about 4.184x106J of energy. Estimate theTNT equivalent of the energy released in part (a) above. 10 23 J 16 2 . 39 10 kg of TNT 6 4.184 10 J 2.39 1013Tons of TNT c) The atomic bombs that destroyed the cities of Hiroshima and Nagasaki during World War II were about the equivalent of 15 kilotons of TNT each. Estimate the energy released by the asteroid in terms of the equivalent number of atomic bombs of this size. 1023 J 9 1 . 59 10 atomic bombs 6 6 15 10 4.184 10 J 7) [Atomics Size] Arrange the following objects according to their size in decreasing order: Chlorine Nucleus, Chlorine Molecule (Cl2), Chlorine Atom (Cl), Chlorine Ions (Cl - ). Find out the quantitative values for the above objects. Some terminologies in chemistry: Atomic radius: Distance from the atomic nucleus to the outermost stable electron orbital in an atom that is at equilibrium. (commonly known as size of atom) Covalent radius: Corresponds to half of the distance between two identical atomic nuclei bound by covalent bond. Cl Cl ratom; cl rcov; cl2 From experiment rcov; cl2≈0.099 nm Cl Let us approximate size of Cl2 := 2 x rcov; cl2 = 0.198 nm rcov; cl2 Negative ions are larger than their atoms! How about positive ion? Smaller. - Cl Ionic radius of Chlorine, rion;cl= 0.181 nm + Cl From experiment ratom; cl≈0.100 nm Na Na ratom; cl Cl From nucleus physics, Hence, r nucleus; cl = 1.2 x 10-15 x (35.5)1/3 rnucleus= ro A1/3 = 3.944 fm where A= mass number; Acl= 35.5 and ro= 1.2 x 10-15 m (empirical constant) 8) [Nuclear Radioactivity] The radioactive isotope 57Co decays by electron capture with a half-life of 272 days. a) Find the decay constant and the lifetime. Half Life 272days 2.35107 s Life time (or mean life-time) is defined as, Tmean T1/ 2 2.35 107 ln 2 0.693 1 3.39 107 s decay constant 1 Tmean 2.95 10 8 s 1 half life is not mean life time! b) If you have a radiation source containing 57Co, with activity 2.00µCi (1Ci≡3.70x1010 decays/sec), how many radioactive nuclei does it contain? dN (t ) A 2.00 Ci dt 7.40 104 decays / s 7.40 10 4 s 1 N (t ) 2.95 10 8 s 1 2.511012 Nuclei dN (t ) / dt