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NUCLEI LEARNING SHEET – 1 SOLUTIONS ANSWER KEY READ & EXPLAIN 1. Why Is the Atomic Mass of Chlorine Not an Integral Multi

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NUCLEI LEARNING SHEET – 1 SOLUTIONS ANSWER KEY READ & EXPLAIN 1. Why Is the Atomic Mass of Chlorine Not an Integral Multi NUCLEI LEARNING SHEET – 1 SOLUTIONS ANSWER KEY READ & EXPLAIN 1. Why is the atomic mass of chlorine not an integral multiple of hydrogen? A. Chlorine has two isotopes having masses 34.98u and 36.98u. These are nearly the integral multiples of the atomic mass of a hydrogen atom. Also, the relative abundance of these two isotopes is 75.4 % and 24.6% respectively. So, the average mass of chlorine is calculated by finding the weighted average of the masses of these two isotopes; which is, 35.47u. 2. 1 amu = ______ kg. A. 1.66 x 10-24 kg 3. What is the difference between isobars and isotones? A. Isobars are atoms that have the same mass number, i.e. the total number of protons and neutrons. Isotones have the same number of neutrons but different numbers of protons. 4. How many isotopes does Gold have? 197 195 A. Gold (79Au) has one stable isotope, Au, and 36 radioisotopes, with Au being the most stable with a half-life of 186 days. 5. Heavy stable nuclei have more neutrons than protons. This is because of the fact that (a) neutrons are heavier than protons. (b) electrostatic force between protons are repulsive. (c) neutrons decay into protons through beta decay. (d) nuclear forces between neutrons are weaker than that between protons. A. (b) electrostatic force between protons are repulsive. 6. The radius of a spherical nucleus as measured by electron scattering is 3.6 fm. What is the mass number of the nucleus? A. Given, r=3.6fm, r0=1.2fm 1/3 We know that the radius of spherical nucleus , r=r0A 푟 A=( )3 푟0 3.6 A=( )3 1.2 A=27 The correct answer is A. 7. What will be the ratio of the radii of the nuclei with mass numbers 8 to mass number 27? 1/3 A. Using r=r0A , we get answer 2/3 8. Given the mass of iron nucleus as 55.85u and A=56, find the nuclear density? A. mFe=55.85 u=9.27 x 10-26 kg −26 푚푎푠푠 9.27푋10 1 17 −3 nuclear density = = 4휋 푋 = 2.29푋10 푘푔푚 푣표푙푢푚푒 ( )(1.2푋10−15)3 56 3 The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus. 9. Calculate the energy equivalent of 1 g of substance. E=mc2 =1×10−3×9×1016 =9×1013J Answer the following questions in your notebook: 10. (a) Two stable isotopes of lithium 6 3 Li and 7 3 Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium. (b) Boron has two stable isotopes, 10 5B and 11 5B. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of 10 5B and 11 5 B. A. (a)The atomic mass is given by, (7.5푋6.01512) + (92.5 × 7.01600) 푚 = = 6.941 푢 100 10 (b) Mass of 5퐵, m1=10.01294u 11 Mass of 5퐵 ,m2=11.00931u 10 11 Let the abundance percentage of isotopes of boron, 5퐵= x% and 5퐵= (100−x)% The atomic mass of Boron, m=10.811u 푚1푥1+푚2푥2 푚 = 푚1+푚2 So, x=19.89%; 100−x=80.11% 20 21 22 11. The three stable isotopes of neon: 10푁푒, 10푁푒 푎푛푑 10푁푒 have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon. A. The atomic mass is given by, 푛 Average atomic mass= ∑푖=1(푚푎푠푠(푖))(푎푏푢푛푑푎푛푐푒(푖)) 90.51푋19.99 + 0.27푋20.99 + 9.22푋21.99 푚 = = 20.18 푢 100 12. In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on 24 25 their relative natural abundance on earth. The three isotopes and their masses are 12푀푔 (23.98504u), 12푀푔 26 24 (24.98584u) and 12푀푔 (25.98259u). The natural abundance of 12푀푔 is 78.99% by mass. Calculate the abundances of other two isotopes. A. Given that, Average atomic mass of magnesium, mavg = 24.312 u Mass of magnesium isotope (m1), Mass of magnesium isotope (m2), Mass of magnesium isotope (m3), Abundance of magnesium isotope (n1), We know that sum of abundances of all isotopes together is equal to total magnesium available on earth, 100% magnesium. Let, Abundance of magnesium isotope (n2) Thus, Abundance of isotope (n3), We have the relation for the average atomic mass is, ∴ ⇒ ⇒ -9.272526 u = -0.99675x u ∴ = Abundance of isotope (n2), Thus, Abundance of isotope (n3), READ & EXPLAIN 13. What is meant by mass defect? A. Mass defect is the difference between the predicted mass and the actual mass of an atom's nucleus. The binding energy of a system can appear as extra mass, which accounts for this difference. 14. What is the mass defect of helium atom? A. 0.03 u 15. Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass 16 2 defect of 8푂 in MeV/c . A. 1amu=1.66×1027kg C= speed of light =3×108m/s E=mc2 =1.66×1027(3×108)2=1.49×10−10 Joule 1meV=1.6×10−13J 1.49푋10−10 E= =931.6MeV 1.6푋10−13 16 8푂= mass no. =16 energy for 16amu=16×931.6=14905.6MeV 푒푛푒푟푔푦 14905.6푀푒푉 Mass defec t= = 푐2 푐2 Answer the following questions in your notebook: 14 14 16. Obtain the binding energy (in MeV) of a nitrogen nucleus 7푁, given m ( 7푁) = 14.00307 u. mH = 1.00783 amu mn = 1.0087 amu mN = 14.0030u amu 14 In 7푁, these are 7 protons and 7 neutrons. ∴ mass defect, Δ=(7mH+7mn)−mN =(7×1.00783)+(7×1.00807)-14.00307=0.11243 amu Binding energy of nitrogen nucleus = Δ×931 MeV =0.11243×931 =104.67 56 209 56 17. Obtain the binding energy of the nuclei 26퐹푒 and 83퐵푖 in units of MeV from the following data: m (26퐹푒) = 209 55.934939 u and m ( 83퐵푖) = 208.980388 u. 56 (i) 26Fe nucleus contains 26 protons . Number of neutrons =(56−26)=30 neutrons Now, Mass of 26 protons=26 X 1.007825=26.20345u Mass of 30 neutrons=30 X 1.008665=30.25995u Total mass of 56 nucleons =56.46340u 56 Mass of 26Fe nucleus =55.934939u Therefore, Mass defect, △m =56.46340−55.934939=0.528461u Total Binding Energy=0.528461 X 931.5Mev=492.26MeV 492.26 Average binding energy per nucleon= =8.790MeV 56 209 (ii) 83퐵푖 nucleus contains 83 protons Number of neutrons=209−83=126neutrons Now, Mass of 83 protons=83 X 1.007825=83.649475amu Mass of 126 neutrons=126X1.008665=127.091790amu Therefore, total mass of nucleons=83.649475+127.091790=210.741260amu Given, mass of nucleus=208.980388amu Now, mass defect △m =210.741260−208.980388=1.760872 Total binding energy=1.760872 X 931.5=1640.26MeV 1640.26 Therefore, average binding energy per nucleon = =7.848Mev 209.
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