1_Radioactivity and decays_fall07

1_Radioactivity and decays_fall07 - Nuclear physics and...

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1 Nuclear physics and radioactivity Part I: Nuclear structure The strong nuclear force The mass defect and nuclear binding energy Suggested reading: Cutnell’s Physics Knight’s Physics Nuclear structure Nuclear structure ± The nucleus of an atom consists of nucleons such as: ² Neutrons ² Protons ± The number of protons in the nucleus is given by the atomic number Z ; In a neutral atom, the number of protons equals the number of electrons in orbits around the nucleus: ± The total number of protons and neutrons in the nucleus is given by the atomic mass number or nucleon number A ; ± The number of neutrons in nucleus is N ; ± Thus, the symbol for atom X is X A Z and N Z A + = Nuclear structure ± Nuclei that have the same number of protons Z, but different A’s are isotopes ; ± Examples: m r 15 10 1 . 4 × The radius of potassium (A=39) is: 3 / 1 15 ) 10 2 . 1 ( A m r × U U U U 234 92 236 92 238 92 235 92 ± Protons and neutrons are clustered together in atom to forms a spherical region, whose radius depends on the atomic mass number A by: Cl Cl 35 17 37 17
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2 Problem : For lead (Pb) find (a) the net electrical charge of nucleus, (b) the number of neutrons, (c)the number of nucleons, (d) the approximate radius of nucleus, and (e) the nuclear density. a. The net electrical charge of the nucleus is equal to the total number of protons multiplied by the charge on a single proton: For lead, we have A = 208 and Z = 82. C C q net 19 19 10 31 . 1 ) 10 6 . 1 ( 82 × + = × + = b. The number of neutrons is N = A – Z = 208 – 82 = 126 c. By inspection, the number of nucleons is A = 208 d. The approximate radius of the nucleus rA (. / 12 10 10 13 –15 –15 1/3 –15 m) (1.2 10 m)(208) 7.1 m Problem (con’d) : For lead (Pb) find (a) the net electrical charge of nucleus, (b) the number of neutrons, (c)the number of nucleons, (d) the approximate radius of nucleus, and (e) the nuclear density. For lead, we have A = 208 and Z = 82. e. The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be found by multiplying the mass of a single nucleon by the total number A of nucleons in the nucleus. Treating the nucleus as a sphere of radius r , the nuclear density is ρ π == = × = × m V mA r A m total nucleon nucleon –15 nucleon –15 3 m) m) 4 3 3 4 3 3 4 3 / = × × 167 10 10 4 3 . –27 –15 3 17 3 kg m) 2.3 kg / m The atomic mass unit ± The atomic mass unit , u , is one-twelfth of the mass of 12 C atom of carbon. ± The atomic mass unit , u , is about the mass of one proton or neutron.The atomic masses of the elements in u are average masses, taking into account the different isotopes that exist. ± The energy equivalent of one atomic mass unit: u = 1.4924 x 10 -10 J = 931.5 MeV; ± One gram is about 600,000,000,000,000,000,000,000 u (that's 600 sextillion, or a 6 x 10 23 ). ± A pound is just 300 septillion u or 3 x 10 26 or 300,000,000,000,000,000,000,000,000. ± Now you can see why we use u's instead of grams!!
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3 The mass defect ± The standard is that one atom of carbon 12, the isotope of carbon with 6 protons and 6 neutrons, has a mass of exactly 12 u; ± The mass of a proton is 1.00728 u; a neutron is 1.00866 u (Table 31.1, Cutnell’s Physics).
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This note was uploaded on 09/24/2011 for the course PCS 229 taught by Professor Pejovic-milic during the Spring '11 term at Ryerson.

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1_Radioactivity and decays_fall07 - Nuclear physics and...

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