NUCLEAR CHEMISTRY

UNIT 11 NUCLEAR CHEMISTRY Clearly state, what do you understand by the terms mass number nucleons and nuclides. Mass number is the sum of the number of protons and neutrons Neutrons and protons are called nucleons A particular nuclear species with a specific atomic number and mass number is called nuclide. Describe the properties of radiation which are emitted by radioactive nuclei (- rays ( rays consists of particles with 2 unit positive charge and 4 amu mass. They move with 1/10 th of the speed of light They have low penetrating power and high ionizing power. (- rays (- rays consists of negatively charged particles with mass of electron They travel with 90% of speed of light. They have higher penetrating power than (- rays but lower ionizing power. (-rays: ( –rays are electromagnetic radiations having high energy. They travel with speed of light. They have very high penetrating power and very low ionizing power. Give are example each of (i) (- emission (ii) ( -emission (iii) k-capture. Write the equation for these nuclear changes. i) 23892U ( 23490Th + 42He. ii) 23490Th ( 23491Pa + 0-1e. iii) 13356Ba + e- ( 13355Cs + X-rays. What is Group Displacement Law? An element belonging to group 1 decays by (- emission. To which group of the periodic table the daughter element will belongs. When an element emits an (- Particle the daughter element occupies two position left to the parent element and when an element emits B- particle, the daughter element occupies are position right to the parent element in the periodic table. If an element belonging to groupI undergoes (- decay the daughter element will belong to group II. 5. How many ( and ( particles will be emitted when 23290Th Changes into 20882Pb. 232 – 208 = 4x where x is the number of ( - particles. 24 = 4x x = 6. 90 = 82 + 2x – y where y is the number of ( - particles. = 82 + 12 – y y = 94 – 90 = 4. 6 ( particles and 4 ( particles are emitted. 6. Write the nuclear reactions for the following radioactive decay. a) 23892U Undergoes ( - decay b) 23491Pa Undergoes ( - decay c) 2211Na Undergoes ( + decay a) 23892U ( 42He + 23490Th b) 23491Pa ( 23492U + -10e c) 2211Na ( +10e + 2210Ne. 7. How are the radioactive decay series distinguished? Which one of the decay series is not natural but artificial. There are four decay series distinguished by whether the mass number are devided by 4 or whether when devided by four there are remainders 1, 2 or 3. 4n series Thorium series 4n + 1 series – Neptunium series 4n+2 series – Uranium series 4n+3 series – Actinium series Neptunium series is artificial. 8. What kinds of elementary particles are employed for the artificial transmutation of elements comment an their effectiveness. Protons neutrons, deutrons, alpha particles and (- particles (electrons) are used for artificial transmutation of elements. Neutrons are most effective because they are neutral and not repelled by positively charged nucleus. 9. What is meant by nuclear binding energy calculate binding energy per nucleon of Li isotope, which has isotopic mass of 7.016 mu . The individual masses neutron and proton are 1.008665mu and 1.007277mu respectively and mass of electron is 0.000548mu. The differences between the sum of the individual masses of various nuclear particles and nuclear mass is called mass defect. The energy corresponding to mass defect is called binding energy. Mass defect = [(4x1.008665) + (3 x 1.007277)] – 7.016 = 7.060327 – 7.016 = 0.044327 mu Energy released = 0.044327 x 931.48 Mev = 41.2 Mev Energy released per nucleon = 41.2 Mev 7 = 5.8 Mev ( 1 mu = 931.48 Mev) 10. The atomic mass of oxygen is 15.995 mu while the individual masses of proton and neutron are 1.0073 and 1.0087 mu. The mass of electron = 0.000548 mu. Calculate the binding energy of oxygen nucleus in joules? Mass defect= mass of 8 protons+ mass of 8 neutrons – mass of 016 . =[(8 x 1.0073)+(8 x 1.0087)] – 15.995 =16.136768-15.995=0.141768 u Energy released =931.48 x 0.141768 mu x1.6 10-13 x 6.023 x 1023 =1.236 x 1013 J mol-1 (1 Mev = 1.6 x 10 –13 J) 11. The isotopic composition of Rubidium is 85Rb 72% and 87Rb 28% 87Rb is weakly radioactive and decay by (-emission with a decay constant of 1.1 x 10 –11 per year. A sample of mineral pollucite was found to contain 450 mg Rb and 0.72 mg of 87Sr. Estimate the age of mineral pollucite, stating any assumption made (= 1.1 x 10 –11 year –1 Mass of 87 Rb in 450 mg of Rb =450x28/100=126mg Mass of Sr =mass of 87 Rb decayed=0.72mg Mass of 87Rb present initialy (No)=126+0.72 mg=126.72mg t = (2.303/() log (No/N) = (2.303 / 1.1 x 10-11 yr-1) x log (126.72 / 126) = (2.303 / 1.1 x 10-11) years x .0024 = 5.02 x 108 years. 12. The isotopic masses of 12H and 24He are 2.0141 and 4.0026 mu respectively and the velocity of light in vacuum is 2.998 x 108 m/s. Calculate the quantity of energy in Joule liberated when two moles of 12H undergo fusion to form 1 mole of 24He. Mass defect = Mass of 2 moles of 12 H- mass of 1 mole of 24 He = (2 x 2.0141- 4.0026) x 6.023x 1023 mu =0.0256 x 6.023 x 1023 x 1.66 x 10-27 kg (1 mu=1.66 x 10-27 kg) Energy= mc2=0.0256 x 6.023 x 1023 x 1.66 x 10-27kg x (2.998 x 108 ms -1) 2 = 2.30 x 1012 J 13. The radioactive isotope 6027Co which has now replace radium in the treatment of cancer can be made by a (n, p) or (n, () reaction. For each reaction, indicate the appropriate target nucleus. If the half-life of 6027Co is 7 years evaluate the decay constant in s-1 2860Ni +01n( 6027Co+ 11H 5927 Co+01n (6027Co+( t1/2 =7 years ( = (0.693 / t ½ ) = (0.693/ 7 x 365 x 24 x 60 x 60) = 3.14 x 10-9 s-1. (1 years =1x365x24x60x60 second) 14. A sample of wood from an archeological source shows a 14C activity which is 60% of the activity found in fresh wood today. Calculate the age of archeological sample (t1/2 of 14c =5770 years) t ½ = 5770 years ( = (0.693 / t ½ ) = (.693 / 5770 years) t = (2.303 / () log (No / N) =(2.303 x 5770 years /0 .693) x log100/60 = (2.303 x 5770 years /0 .693) x 0.2214 = 4252 years. 15. What is a nuclear fission reaction? Explain the principle of atomic bomb and working of a nuclear reactor to produce electricity. When a heavy nucleus is bombarded with neutron, it breaks into small nucleus along with release of a lot of energy. It is called nuclear fission reaction eg 23592u +10n-(14656Ba +9636Kr +310 n When three neutron bombard with three uranium nuclei, 9 neutrons will be ejected, 9 neutrons bombard with 9 uranium nuclei and 27 neutrons will be ejected and thus an uncontrolled chain reaction will be set up if the mass of uranium is more than critical mass and tremendous amount of energy will be released. This is the principle of atomic bomb. Working of nuclear reactor Nuclear reactor is a device in which controlled nuclear fission is carried act. A nuclear reactor consists of; a fissile material (uranium enriched in23592U) a moderator (graphite or heavy water) to slow down the neutrons control rods of boron steel or cadmium which are capable of absorbing neutrons and are used to ensure that neutron flux is under control. The control rods are inserted into unclear reactor and can be raised or lowered to control the chain reaction. The large amount of energy released is used for steam generation through heat exchangers which run turbine and electricity is produced For fig. See p. 22 of NCERT Text Book 16. What is meant by fissionable isotope or a fissile isotope? How are such isotopes produced artificially? Give an example. Fissionable isotopes are those that can be broken down into smaller elements by bombarding with a neutron. They are produced in breeder reaction. Eg:23892 U+10n ( 23992 U( 23993 Np +0-1e 23993 Np ( 23994 Pu + 0-1e 23994 Pu is a fissionable material . 17. In the neutron induced fission reaction of 23592 U, one of the products is 9537 Rb In this mode, another nuclide and three neutrons are also produced. Identify other nuclide. 23592 U + 10n ( 9537 Rb + 13855Ba + 3 10n Other nuclide is 13855Ba 18. Explain the principle of (a) activation analysis and (b) Breeder reactor. A breeder reactor is one that produces more fissionable nuclei than it consumes. For example when 23892 u is bombarded with neutrons it produces plutonium- 239, a fissionable nucleus by the following sequence of reactions. 23892 u + 10n ( 23992 u ( 23993 Np + 0-1 e 23993 Np ( 23994 Pu + 0-1 e When one atom of 23592U undergoes nuclear fission, three neutrons are produced out of which are is used to continue the chain reaction and other two are used to convert 2 nuclei of non fissionable 23892U into fissionable 23994Pu. Neutron activation analysis The sample to be analyzed is bombarded with neutrons in a nuclear reactor and the stable isotope of the element in the sample to be investigated is converted into its radioactive isotope, which can be detected, and from the measurement of rate of decay the concentration of element can be determined. -dN/dt = (Nt where ( ( = 0.693/ t ½) It is a nondestructive technique and the method is very useful for determination of elements in trace amounts. 19. Describe chief application of radioisotopes in (a) the study of reaction mechanism (b) Medicines. Radioisotopes help to determine the mechanism of reaction for eg : consider the esterification reaction. C6H5 – C = O + CH3 – O*H ( C6H5 – C = O + H2O | | OH O*CH3 The origin of starred oxygen can be determined by labeling oxygen atom of methanol with 18O and then using in the esterification, it can be proved that starred oxygen is from methanol. (b). Co – 60 is used in the treatment of cancer 20.Complete the following nuclear reaction a. 9642Mo ( _______n)9743 Tc _______ ((, 2n ) 21185 At 5525Mn (n,() ______ 96246Cm + 126 C( ________ +4 10 n 1327Al ((, n) _______ 23892U ((, (-)________ 9642 Mo (D, n) 9743Tc 20983 Bi ((, 2n) 1185At 5525Mn (n,() 5625Mn 24696 Cm + 126 C_____254102 No + 4 10n 2713 Al ((, n) 3015 P 23892 U((, (-) 24295Am 21. Complete the equations for the following nuclear processes 3517Cl + 01n ( ________+ 24 He 23592 U + 10 n( _______+ 13754 Xe + 2 10 n 2713 Al + 42 He ( _______ + 10 n _______ (n, p) 3516 S 23994 Pu ((, (- ) _______ 3517 Cl + 10 n ( 3215 P + 42 He 23592 U + 10 n ( 9738Sr + 13754 Xe + 201 n 2713 Al + 24 He ( 3015 P + 10n 3517 Cl (n, p )3516 S 23994 Pu ((, (-) 24397 Bk 22.Calculate the mass of 140La in a sample whose activity is3.7x10 10 Bq (1Bq=1disintegration per second) given that its t 1/2 is 40 hours [Hint; Mass=3.7x10 10 x 40 x 60 x 60 x1 40/(NAx ln2)] -dN/dt = (Nt = 0.693 Nt /t ½ 3.7 x 1010 s-1= 0.693 Nt / 40 x 60 x 60 s Nt = 7.688 x 1015nuclei Mass of 140La = nuclei x 140 gmol-1/ 6.022 x 10-23 mol-1 = 1.78 x 10-6 g 23.Calculate the binding energy per nucleon 12C,14N,16O and comment on magnitudes. Masses of neutron and proton are1.0078 and1.0087mu respectively.(1mu=931Mev,atomic masses of C,N and O are 12.01,14.01and15.995 respectively) (m = [(6x1.0078)+(6x1.0087)]-12.01 = 0.087mu Binding energy = (mx931Mev = 0.089x931Mev = 82.86Mev Binding energy per nucleon of C = 82.86/12 = 6.905Mev (m = [(7x1.0078) + (7x1.0087)] -14.01 = 0.1055amu Binding energy = (mx931Mev = 0.1055x931Mev = 98.22Mev Binding energy per nucleon of N = 98.22/14 = 7.015Mev (m = [(8x1.0078)+(8x1.0087)]-15.995 = 0.137mu Binding energy = (mx931Mev = 0.137x931Mev = 127.5Mev Binding energy per nucleon of O = 127.5/16 = 7.9717Mev Binding energy per nucleon increases with increase in atomic number. 24. The (- activity of a sample of CO2 prepared from a contemporary wood gave a count rate of 25.5 counts per minute. The same mass of CO2 from an ancient wooden statue gave a count rate of 20.5 c.p.m. In the same counter condition. Calculate the age to the nearest 50 years taking t ½ of 14C as 5770 years. What would be the expected count rate of an identical mass of CO2 from a sample that is 4000 years old? ( = (0.693 / t ½) = (0.693/5770) yr-1 t = (2.303 / () x log (N0 /Nt) = (2.303 x 5770 year/0 .693) x log (25.5 / 20.5) = (2.303 x 5770 year/0 .693) x 0.0947 = 1816 years. = 1800 years ( in nearest fifty years) Log (No / Nt) = ( x t / 2.303 = (0.693 x 4000 yr ) / (5770 yr x 2.303) = 0.2085 (No/ Nt ) = 1.616 ( anti log of 0.2085) Nt = (No / 1.161) = (25.5 / 1.161) = 15.78 c.p.m. 25. How is 14C produced in nature and what happen to it subsequently? Give equations for these processes. 14C is formed in the upper atmosphere by the action of cosmic radiation on 14N 147N + 01n ( 146C + 11H 14 C decays with (- emission 146C ( 147N + 0-1e 26. What do you understand by tracers? Give an example of a tracer that can be used in determining the mechanism of a reaction. A sample of radioisotope used to study the course (mechanism) of reaction is called tracer. 18O is used to study the mechanism of esterification reaction. 27. What are synthetic elements? Mention two synthetic elements and write the nuclear questions leading to their synthesis. Synthetic elements are the elements produced by bombardment technique ie, by bombardment of a nucleus with different particles. eg: 9642 Mo + 21H ( 9743Tc + 01n 9743 Tc is a synthetic element. 23994 Pu + 42 He ( 24296 Cm + 01n 24296 Cm is synthetic element. 28. What is meant by thermonuclear reactions and why are they so called? Why are these reactions not useful for peaceful purposes? The reactions in which two or more lighter nuclei combine to form heavier nuclide are called nuclear fusion [thermo-nuclear] reactions. The activation energies for fusion reactions are very high ie, they require high temperature to overcome electrostatic repulsion between the nuclei. So fusion reactions are known as thermonuclear reaction. These reactions are not useful for peaceful purpose because of technical difficulties in conducting fusion in a controlled way. 29. Describe the principle of an atom bomb. What is meant by critical mass? What is the critical mass of 23592U? The basic principle of atom bomb is uncontrolled nuclear chain fission reaction. A neutron form Ra-Be source initiates the fission process which starts the chain reaction. finally leading to explosion and also releasing large amount of heat energy. The rapid release of heat raises the temperature enormously and generates a very high pressure – front in the surrounding atmosphere. The minimum amount of fissionable material required to continue the chain reaction process under the given set of conditions is called the critical mass. The critical mass of 23592U is between 1 kg and 100kg. ****************