C 14 I 131 Mendeteksi fungsi kelenjar gondok. Na 24 Pb 210

Transkripsi

1 NUCLEAR PHYSICS

2 OVERVIEW Properties of Nuclei Mass defect (mass difference) Nuclear binding energy Radioactivity a) Alpha decay b) Beta decay c) Gamma decay Nuclear Reaction The Decay Processes

3 Why we study this chapter? Berbagai manfaat radioisotopes: Co 60 Membunuh sel kanker Mendeteksi usia fosil C 14 I 131 Mendeteksi fungsi kelenjar gondok Mengetahui efektivitas kerja jantung Na 24 Pb 210 Mendeteksi pemalsuan keramik

4 Examples (UN) Radiasi dari radioisotop Co-60 dimanfaatkan untuk... A. Penghancuran batu ginjal B. Detektor Asap C. Menentukan umur fosil D. Terapi pada kelenjar gondok E. Membunuh sel kanker

5 Examples (UN) Zat radioisotop C-14 dapat digunakan untuk... A. Mendeteksi fungsi kelenjar gondok B. Mengetahui efektifitas kerja jantung C. Membunuh sel kanker D. Mendeteksi pemalsuan keramik E. Menentukan usia fosil

6 The Nucleus Only percent of atom's volume It contains 99.9% of the total mass of the atom. Nucleons Proton (p) Neutron (n)

7 The Nucleus Since the mass of nucleons is far too small to measure in kg, other standard unit is used: Mass of proton Atomic Mass Unit (amu) 1 amu = 1.66 x kg Mass of electron m p = amu m e = x 10-4 amu m n = amu Mass of neutron Extremely small 1836 times smaller than a proton

8 Atomic Number and Mass Number The number of protons in an atom is called The Atomic Number Z A The total number of protons and neutrons in an atom is called The Atomic Mass Number

9 Atomic Number and Mass Number 7 3 Li The number of Neutrons (N) Atomic Number = 3 Atomic Mass Number = 7 = A - Z N = 7-3 = 4 The difference between Z and A Will give us The number of neutrons (N) in the nucleus

10 Examples Calculate the number of proton and neutron in an atom notated by: (a) Au and (b) Bi (a) Atomic Number = 79 (protons) Jumlah Neutron = = 121 (b) Atomic Number = 83 (protons) Jumlah Neutron = = 126

11 Nuclear Force There must be a massive repulsive force between the protons, however they remain in the nucleus together. There must be another force that attracts the nucleons. This force is called Nuclear Attractive Force, or Nuclear Force or Strong Force. Dari mana datangnya gaya inti itu? O.o?

12 Nuclear Binding Energy Source of energy that holds protons and neutrons together in a nucleus arises from the transformation of mass to energy. For any atoms, the total mass in measured experimentally is always smaller than the sum of the mass of protons and neutrons. The amount of mass that is transformed into energy is called Mass Defect. Masa inti < ( Z.m p + N.m n ) Jumlah Proton Jumlah Neutron

13 Nuclear Binding Energy This mass difference between the individual nucleons and the fully formed nucleus of any atom can be calculated by: Defek Massa m =( Z.m p + N.m n ) - m inti The energy of this mass defect' is the same energy preventing the nucleus from breaking up. The energy which binds the protons and neutrons together in the nucleus is called the nuclear binding energy.

14 Example Massa inti atom Ca adalah 40,078 sma. Jika massa proton = 1,0078 sma dan neutron = 1,0087 sma, defek massa pembentukan Ca adalah. m =( Z.m p + N.m n ) - m inti m = [ 20(1.0078) + 20 (1.0087) ] sma m = 20(2.0165) sma sma m = sma

15 Nuclear Binding Energy From Einstein mass energy equation, The Energi Ikat Atom can be calculated: E ikat = m c 2 Untuk atom yang defek massanya adalah 1 sma, Maka energi ikat atomnya adalah: E ikat = (1 sma) (3x10 8 ) 2 E ikat = (1.66x10-27 kg)(3x10 8 m/s) 2 E ikat = 1.49x10-10 J = 931 MeV

16 Example (UN) Massa inti atom 12 Mg 24 adalah 23,993 sma. Massa nukleon bebas inti tersebut adalah massa proton 1,007 sma dan massa neutron 1,008 sma. Energi ikat inti 12 Mg 24 sebesar. (1 sma =931 MeV) E ikat = m 931 m =( Z.m p + N.m n ) - m inti m = [ 12(1.007) + 12 (1.008) ] sma m = 12(2.015) sma sma = sma

17 Example (UN) Massa inti atom 12 Mg 24 adalah 23,993 sma. Massa nukleon bebas inti tersebut adalah massa proton 1,007 sma dan massa neutron 1,008 sma. Energi ikat inti 12 Mg 24 sebesar. (1 sma =931 MeV) E ikat = m 931 MeV E ikat = m 931 E ikat = (0.187 sma) (931 MeV) E ikat = 174,1 MeV

18 Isotopes Isotopes merupakan kumpulan atom unsur yang memiliki sifat kimia sama, tetapi berbeda sifat fisikanya. Isotop yang tidak stabil dinamakan radioisotop. Isotop yang tidak stabil ini cenderung memancarkan partikel partikel radioaktif untuk mencapai kestabilannya. 12 C 13 C 14 C 15 C

19 Radioactivity

20 Radioactivity Most elements have stable isotopes. However, the nuclei of some isotopes are unstable. therefore they decompose and emit highly energetic particles Radioactive Example: radium, uranium, Emitted nuclear radiation plutonium only a small number are found naturally, most of them are artificially produced in laboratories.

21 Types of Radiation The radioisotope is located in a chamber where its radiation can pass through a magnetic field to strike a photographic plate 3 different darkened spots appear α alpha radiation β beta radiation γ gamma radiation

22 Penetration Power

23 α alpha radiation Alpha particles are identical to 4 α 2 neutrons and 2 protons 2 the nuclei of helium atoms (+2) Charged particle When an element undergoing an alpha decay, α - particle it emits 2 protons and 2 neutrons. There is a decrease of 2 in the atomic number and a decrease of 4 in the atomic mass number of the nucleus.

24 α alpha decay 238 U Th 90

25 β beta decay -1β 0 β - particle β - beta particles are identical to electrons, negatively charged (-1) and negligible mass. When an element undergoing an beta decay, it emits 1 electron (coming from the split of neutron) 1 n 1 p β 1 There is an increase of 1 in the atomic number and no change in the atomic mass number of the nucleus.

26 α and β particle decay 14 C 0 β N 7 Example, 11Na 24 undergoes beta decay. What is the resulting nucleus element? 24 Na 0 β X 12 X is Magnesium (Mg)

27 γ gamma rays Gamma (γ) rays are a type of electromagnetic radiation of very high energy. 0 0 γ γ - rays The emission of γ -rays from a nucleus is similar to the emission of photons from an excited atom. The mass and charge of gamma rays are both zero. There is no change in the atomic number and no change in the atomic mass number of the nucleus.

28 γ gamma rays 248 Pu 94 0 γ Pu 94 The unstable nucleus (*) then loses excess energy in the form of gamma rays to become a more stable nucleus of lower energy.

29 Nuclear Reaction

30 Nuclear Reaction Any process that involves a change in the nucleus of an atom is called nuclear reaction Tend to be unstable and combine with other nuclei (FUSION) Tend to be unstable and split into two or more nuclei (FISSION)

31 Reaksi Fisi

32 Reaksi Fusi

33 Nuclear Reaction Reaksi nuklir menghasilkankan energi yang sangat besar, lebih dari jutaan kali lipat energi yang diperoleh dari reaksi kimia biasa. Energi yang sangat besar itu diperoleh karena massa inti yang dihasilkan lebih kecil daripada jumlah total massa inti yang bereaksi. Jadi, sekali lagi, persamaan Einstein E =mc 2 menjelaskan bahwa massa yang hilang telah dikonversi menjadi energi pada hasil reaksinya.

34 Nuclear Reaction Reaksi Fusi m = m pereaksi - m hasil reaksi 1 2 H He H 0 n pereaksi Reaksi Fisi Hasil reaksi + E Energi pada reaksi nuklir: E = m 931 MeV 0 1n U 140 Cs Rb n + E pereaksi Hasil reaksi

35 Example (UN) Nilai E (energi yang dihasilkan) pada reaksi fusi tersebut adalah.. m = m pereaksi - m hasil reaksi m = ( ) - ( ) m = sma sma = sma m (931 MeV) ( sma) (931 MeV) = 0.88 MeV E = E =

36 Nuclear Decay

37 Nuclear Decay Imagine that you are studying a sample of radioactive material. You know that the atoms in the material are decaying into other types of atoms. How many of the unstable parent atoms remain after a certain amount of time? λ Decay constant of a radioactive material Large λ decays quickly Small λ decays slowly N 0 Initial Number of Nuclei Activity = λ N 0 The unit of Activity is Becquerel (Bq) 1 Curie (Ci) = 3.7 x Bq

38 Nuclear Decay N Undecayed Number of Nuclei N 0 Initial Number of Nuclei N = N 0 e λt The rate of radioactive decay is generally expressed in terms of half-life. Half-life is the period of time over which the number of radioactive nuclei decreases by half. Half-Life N = N 0 2

39 Nuclear Decay Half-life ln N = N 0 e λt N 0 2 = N 0 e λt 1 = ln e λt 2 ln 2 = λt ln 2 = λt ln 2 is equal to We can calculate the decay constant based on the information of half-life of a radioactive material by: λ = t 1 2

40 Half-life Example 222 A radioactive isotope, Radon-222 ( 86Rn) has a half-life of 4 days. An initial number of 2 x nuclei is released during radioactive decay. a) What is the activity of the sample? b) How many nuclei remain undecayed at the end of this period? ANSWER a) First, convert the unit of days into second! 4 days = 4 x 24 x 60 x 60 = seconds = 3.5 x 10 5 s Then, before we find the activity, we have to calculate the decay constant of Radon-222

41 Half-life Example 4 days = 4 x 24 x 60 x 60 = seconds = 3.5 x 10 5 s Then, before we find the activity, we have to calculate the decay constant of Radon-222 λ = = t x 10 5 = 0.2 x10 5 Activity = λ N 0 = ( 2 x10 6 ) ( 2 x ) Activity = 4 x 10 4 Bq

42 Half-life Example b) How many nuclei remain undecayed at the end of this period? N = N 0 e λt = (2 x ) e (2x10 6 )(3.5x10 5 ) = (2 x ) e (0.7) = (2 x ) 0.5 = Nuclei

43 Half-life Example 2 How long will it take a sample of polonium-210 with a half-life of 140 days to decay to one-sixteenth its original strength?

44 Half-life Example 3 The half-life of the radioactive Radium ( 226 Ra) nucleus is 5.0 x seconds. A sample contains 3.0 x nuclei. (a) What is the decay constant for this radioactive? (b) How many radium nuclei, in curies, will decay per second?

45 Half-life Example 3 A radioactive sample consists of 5.3 x 10 5 nuclei. There is one decay every 4.2 hours. (a) What is the decay constant for this sample? (b) What is the half-life for the sample?