Comprehensive CBSE Question Bank in Physics XII (Term-II)

258 Physics—XII (e) For a given energy, decrease in b implies decrease in cot (/2) and hence increase in scattering angle. 8. A hydrogen-like atom (described by the Bohr model) is observed to emit six wavelengths originating from all possible transitions between a group of levels. These levels have energies between – 0.85 eV and – 0.544 eV (including both these values). (a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in these transitions. (Take hc = 1240 eV – nm, ground state energy of hydrogen atom = – 13.6 eV) Ans. (a) Let the given transitions occur between nth and n+3 (n + x) th energy levels, where x is the difference in the quantum numbers of two energy levels. As number n of transitions occurring between these energy levels is six, hence x+ 1C2 = 6 or x = 3 i.e., if lowest energy Smallest wavelength has quantum number n, quantum number of highest energy level will be (n + 3). Given transitions are demonstrated in adjoining figure. Now, we have  Z2 (13.6 eV ) = 0.85 eV ...(1) and  Z2 (13.6 eV ) = 0.544 eV ...(2) n2 n2 Solving (1) and (2), we get n = 12 and Z = 3 (b) Smallest wavelength  is given by hc = (0.85 – 0.544) eV = 0.306 eV  or = hc or = 1240 nm = 4052.3 nm 0.306 eV 0.306 9. Choose the correct alternative from clues given at the end of each statement: (a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than.) (b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/Rutherford’s model.) (c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/Rutherford’s model.) (d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson’s model/Rutherford’s model.) (e) The positively charged part of the atom possesses most of the mass in .......... . (Rutherford’s model/both the models.) Ans. (a) no different from (b) Thomson’s model .......... Rutherford’s model (c) Rutherford’s model (d) Thomson’s model .......... Rutherford’s model (e) both the models.

Atoms 259 10. (a) Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. Ans. (a) vn = 1 c 137 n v1 = 1 × 3 × 108 m s–1 = 2.19 × 106 m s–1 137 v2 = 1  3  108 m s–1 = 1.09 × 106 m s– 1 137 2 v3 = 1  3  108 m s–1 = 7.29 × 105 m s–1 137 3 (b) T= 2r v Tn = 2 [n2 × 0.53 × 10–10]  1    1 3  108  137 n  Tn = 2n2  0.53  1010  137  n s 3  108 Tn = 1.52 × 10–16 n3 s T1 = 1.52 × 10–16 s T2 = 1.52 × 23 × 10–16 s = 1.22 × 10–15 s T3 = 1.52 × 33 × 10–16 s = 1.52 × 27 × 10–16 s = 4.10 × 10–15 s 11. A 12.75 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted? Ans. The energies of the electron in different states are: E1 = – 13.6 eV, E2 = – 3.4 eV, E3 = – 1.51 eV, E4 = – 0.85 eV Now, En = – 13.6 eV + 12.75 eV = – 0.85 eV = E4 Clearly, the electron is raised to the fourth orbit n=4 n=3 of principle quantum number n = 4. The quantum transitions to less excited state give six possible lines as follows: n = 4 (4  3) (4  2) (4  1) n=2 n = 3 (3  2) (3  1) n = 2 (2  1) (a) For n = 4 to n = 3 n=1 1 = R 1  1  7R 1  32 42  144 1 = 144  144 m = 18752.4 Å 7R 7  1.097  107 It belongs to Paschen series. It lies in infra-red region.

260 Physics—XII (b) For n = 4 to n = 2 1 = R 1  1 or 1 R 3 2  22 42  2 16  or 2 = 16 = 16 m = 4861.7 Å 3R 3  1.097  107 It belongs to Balmer series. It lies in the visible region. (c) For n = 4 to n = 1 1 = R 1  1  R 15  3 12 42  16  16 16 3 = 15 R = 15  1.097  107 m = 972.3 Å It belongs to Lyman series. It lies in the ultraviolet region. (d) For n = 3 to n = 2 1 = R 1  1  5R 4  22 33  36 4 = 36  36 m = 6563.3 Å 5R 5  1.097  107 This wavelength belongs to the Balmer series and lies in the visible region. (e) For n = 3 to n = 1 1 = R 1  1  R 8 5 12 32  9  or 5 = 9 = 9 m = 1025.5 Å 8R 8  1.097  107 This wavelength belongs to the Lyman series and lies in the ultraviolet region. (f) For n = 2 to n= 1 1 = R 1  1  3R 6 12 22  4 or 6 = 4 = 4 m = 1215.4 Å 3R 3  1.097 107 This wavelength belongs to Lyman series and lies in the ultraviolet region. 12. Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better. (a) Is the average angle of deflection of -particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model? (b) Is the probability of backward scattering (i.e., scattering of -particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

Atoms 261 (c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of -particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide? (d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of -particles by a thin foil? Ans. (a) About the same (b) Much less (c) It suggests that scattering is predominantly due to a single collision, because the chance of a single collision increases linearly with the number of target atoms, and hence linearly with the thickness of the foil. (d) In Thomson’s model, a single collision causes very little deflection. The observed average scattering angle can be explained only by considering multiple scattering. So, it is wrong to ignore multiple scattering in Thomson’s model. In Rutherford’s model, most of the scattering comes through a single collision and multiple scattering effects can be ignored as a first approximation. 13. The total energy of an electron in the first excited state of the hydrogen atom is about – 3.4 eV. (a) What is the kinetic energy of the electron in this state? (b) What is the potential energy of the electron in this state? (c) Which of the answers above would change if the choice of the zero of potential energy is changed? (d) Calculate the wavelength of light emitted if an electron makes a transistion to the ground state. [Ground state energy = –13.6eV] Ans. In Bohr’s model, mvr = nh and mv2 Ze2 which gives 2 r = 40r2 Ek = 1 mv2 = Ze2 r= 40h2 n2 2 80r ; Ze2m These relations have nothing to do with the choice of the zero of potential energy. Now, choosing the zero of potential energy at infinity, we have Ep = – Ze2 which gives Ep = – 2Ek and E = Ek + Ep = – Ek 40r (a) The quoted value of E = – 3.4 eV is based on the customary choice of zero of potential energy at infinity. Using E = – Ek, the kinetic energy of electron in this state is + 3.4 eV. (b) Using Ep = – 2 Ek, potential energy of the electron is – 2 × 3.4 eV = – 6.8 eV. (c) If the zero of potential energy is chosen differently, kinetic energy does not change. Its value is + 3.4 eV. This is independent of the choice of the zero of potential energy. The potential energy, and the total energy of the state, however, would alter if a different zero of the potential energy is chosen.

262 Physics—XII (d) h = (– 3.4) eV – (– 13.6) eV = 10.2 eV = 10.2 × 1.6 × 10–19 J hc = 10.2 × 1.6 × 10–19 J  = 6.62  1034  3  108 m = 1217 Å 10.2  1.6  1019 CASE STUDY BASED QUESTIONS Read the following passages and answer the questions that follow: CASE STUDY 1 The experiments carried out by Geiger and Marsden led to the Rutherford’s nuclear model of atom. They directed a beam of 5.5 MeV -particles emitted from a 21843Bi radioactive source at a thin metal foil made of gold as shown in figure below. Vacuum Lead bricks Gold foil target Most pass about 10–8 m thick through Beam of a-particles Source of q a-particles Zns Screen About 1 in 8000 Some are Detector is reflected back deviated through (Microscope) a large angle q Alpha particles emitted by a 21843Bi radioactive source were collimated into a narrow beam by their passage through lead bricks. The beam was allowed to fall on a thin foil of gold of thickness 2.1 × 10–7 m. The scattered alpha particle were observed through a rotatable detector consisting of zinc sulphide screen and a microscope. The scattered alpha particles on striking the screen produced brief light flashes or scintillations. These flashes may be viewed through a microscope and the distribution of the number of scattered particles may be studied as a function of scattering. 1. Rutherford’s experiment on scattering of -particles proved that (a) atoms are mostly empty (b) positive charge is uniformly distributed in the atom (c) atom contains electrons (d) none of these Ans. (a)

Atoms 263 2. Which of the following statements is incorrect? (a) Rutherford is credited with the discovery of the nucleus. (b) Rutherford in his experiment observed that most of the -particles pass through the foil. It means that they do not suffer any collisions. (c) Rutherford’s experiment suggested the size of the nucleus to be about 10–15 m to 10–14 m. (d) None of these. Ans. (d) 3. In a Rutherford scattering experiment when a projectile of charge z1 and mass M1 approaches a target nucleus of charge z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is (a) directly proportional to mass M1. (b) directly proportional to M1 × M2. (c) directly proportional to z1z2. (d) inversely proportional to z1. Ans. (c) Hint and Solution: Energy of projectile = Potential energy at closest approach = 1 z1z2 40 r  Energy  z1z2 4. In Rutherford’s -particle scattering experiment, kinetic energy of -particle striking the gold foil, is 7.9 × 10–13 J. What is the distance of minimum approach? (a) 4.6 × 10–12 m (b) 4.61 × 10–14 m (c) 3.86 × 10–14 m (d) 3.8 × 10–12 m Ans. (b) Hint and Solution: The K.E. of the -particles is equal to the potential energy of the -particles, at the distance of closest approach (r0).  K.E. = 1  (Ze)(2e) or r0 = 1  2Ze2 40 r0 40 (K.E.) Putting 1 = 9 × 109 in SI units, Z = 79 (for gold), e = 1.6 × 10–19 C and 40 K.E. = 7.9 × 10–13 J Then r0 = 9 × 109 × 2  79  (1.6  1019 )2 or r0 = 4.61 × 10–14 m. 7.9  1013 CASE STUDY 2 Hydrogen is the simplest of all atoms. In hydrogen atom, a single electron revolves around a nucleus of charge + e. For an electron moving with a uniform speed in a circular orbit of a given radius, the centripetal force is provided by the inverse square Coulomb force of attraction between the electron and the nucleus. It may be noted that the gravitational

264 Physics—XII attraction between the electron and the proton masses is weaker than the Coulomb attraction by a factor of 1040 and its contribution to the centripetal force can be neglected. 1. The angular momentum of electron in the second orbit of Bohr’s model of hydrogen atom is: (a) h (b) h (c) 2h (d) None of these.  2  Ans. (a) 2. In the Bohr’s model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electron. If a0 is the radius of the ground state orbit, m is the mass, e is the charge on electron and 0 is the permittivity of the free space, the speed of the electron is (a) e (b) zero e (d) 40a0m Ans. (c) 0a0m (c) e 40a0m Hint and Solution: mv2 = 1 e2 Here r = a0. r 40 r2 . 3. According to the Bohr theory of hydrogen atom, the speed of the electron, its energy and the radius of its orbit vary with the principal quantum number n respectively, as (a) 1, 1 , n2 (b) 1, n2 , 1 (c) n2 , 1 , n2 (d) n, 1 , 1 n n2 n n2 n2 n2 n2 Ans. (a) 4. Which of the following postulates of the Bohr model led to the quantisation of energy of the hydrogen atom? (a) The electron goes around the nucleus in circular orbits. (b) The angular momentum of the electron can only be an integral multiple of h/2. (c) The magnitude of the linear momentum of the electron is quantised. (d) Quantisation of energy is itself a postulate of the Bohr model. Ans. (b)

Chapter 6: Nuclei (NCERT Textbook Chapter 13) SUMMARY OF THE CHAPTER  A nuclide is a specific nucleus of an atom, which is characterized by its atomic number Z and mass number A. If X is chemical symbol of the element, we may represent the element as Z X A , where Z represents the number of protons and A represents the number of nucleons.  The radius of the nucleus is given by the relation, R = R0A1/3 where R0 is constant whose value is 1.1 × 10–15 m.  The density () of nuclear matter is same for all elements. = Atomic mass  3A = 3 Atomic volume 4R03 A 430 where R0 is constant, therefore,  is same for all elements. It is experimentally found to be 2.98 × 1017 kg m–3.  The unit in which atomic and nuclear masses are measured is called atomic mass unit. 1 amu = 1.66 × 10–27 kg Also, 1 amu  931 MeV  Isotopes of an element are the atoms of the same element that have the same atomic number but different atomic weights. 1H1, 1H2, 1H3 are the isotopes of hydrogen.  Isobars are the atoms of different elements that have the same atomic weight but different atomic numbers. 11Na22 and 10Ne22 are isobars.  Nuclear forces are the strongest attractive forces between nucleons. These forces are independent of charge, have short range and are non-central in nature. They do not obey the inverse square law.  Nuclear fission is the phenomenon of spliting of a heavy nucleus, generally A > 230, into two or more lighter nuclei, along with the release of energy. The energy released is of the order of about 200 MeV. 92U235 + 0n1  56Ba141 + 36Kr92 + 3(0n1) + Q It is important to note that the fission of 92U235 does not always yield Ba and Kr, but the energy evolved is always about 200 MeV.  The thermal neutrons are the slowly moving neutrons. Their velocity is about 2.2 km/s and energy is about 1 eV. 40  Reproduction factor or neutron reproduction factor is the ratio of the rate of production of neutrons to the rate of loss of neutrons due to leakage and absorption. It is denoted by k. If k = 1, the chain reaction will be steady or sustained. 265

266 Physics—XII  Nuclear fusion is the phenomenon of fusing two or more lighter nuclei to form a single heavy nucleus. It is always accompanied with the release of energy. 1H2 + 1H2  2He4 + 24 MeV  The stellar energy is the energy obtained continuously from the sun and the stars. The fusion reactions going inside the sun are responsible for the energy emitted by the sun. 4(1H1)  2He4 + 2(+1e0) + Q, where Q is of the order of 26.7 MeV.  Nuclear holocaust is the name given to large-scale destruction that would be caused by the use of nuclear weapons.  Radiation hazards are the harmful effects on an organism due to the nuclear radiations. These harmful effects are of two types: (i) pathological effects (ii) genetic effects The radiations are measured in terms of roentgen (R). Important Results 1. The radius of the nucleus is given by, R = R0A1/3 where R0 is constant and A is atomic mass. R0 = 1.2 × 10–15 m = 1.2 fm 2. The density () of the nucleus given by, = 3A = 3 4R03 A 4R03 The nuclear mass density is of the order of 1017 kg m–3, and is independent of mass number A. 3. 1 amu = 1.66 × 10–27 kg  931 MeV, where 1 MeV = 1.6 × 10–13 J. 4. Mass defect, m is given by, m = [Zmp + (A – Z)mn – mN] where mp is mass of proton, mn is mass of neutron and mN is mass of the atom. 5. 1 curie = 3.7 × 1010 disintegrations per second 6. 1 rutherford = 106 disintegrations per second 7. 1 bq (SI unit of activity) is 1 disintegration per second. VERY SHORT ANSWER TYPE QUESTIONS (1 Mark) 1. Which of the two is bigger—1 kWh or 1 MeV? Ans. 1 kWh. 2. Who discovered the neutron? Ans. Chadwick. 3. What is the mass of an electron in amu? Ans. 0.00055 amu. 4. What is the mass of a proton in amu? Ans. 1.007825 amu.

Nuclei 267 5. Why electron capture is more common in heavy atoms? Ans. This is due to small radius of K-shell. 6. How many joule are contained in 1 kWh? Ans. 1 kWh = 103 J s–1 × 3600 s = 3.6 × 106 J. 7. What exactly makes large nuclei unstable? Ans. The electrostatic force dominates the nuclear force. 8. What is one roentgen? Ans. It is that dose of radiation which produces 1.6 × 1012 pairs of ions in 1 g of air. 9. Name two elementary particles which have almost infinite life time. Ans. Electron and proton have almost infinite life time. 10. Is free neutron a stable particle? Ans. No. A free neutron decays spontaneously into a proton, an electron and antineutrino. 1 n  11H + 0 e +  0 1 11. Cadmium rods are provided in a reactor. Why? Ans. Cadmium rods are used as controller in nuclear reactor. Cadmium has large absorption affinity for neutrons. 12. How many electrons, protons and neutrons are there in a nucleus of atomic number 11 and mass number 24? Ans. Number of protons, Z = 11 = Number of electrons Number of neutrons, A – Z = 24 – 11 = 13 13. How many kg are there in one atomic mass unit? Ans. 1 amu = 1.66 × 10–27 kg. 14. You are given two nuclides 7 X and 4 Y . Are they the isotopes of the same element? 3 3 Why? Ans. Since an element is characterised by its atomic number therefore both are isotopes of the same element (Li). 15. In a particular fission reaction, a 235 U nucleus captures a slow neutron. The fission 92 products are three neutrons, a 142 La nucleus and a fission product ZX. What is the 57 value of Z? Ans. Applying conservation of charge, Z + 57 = 92 or Z = 35. 16. Write one equation representing nuclear fusion reaction. Ans. 2 H  12H  24He + 24 MeV. 1 17. How many joule are contained in 1 MeV? Ans. 1 MeV = 106 × 1.6 × 10–19 J = 1.6 × 10–13 J 18. What is Einstein’s mass-energy relation? Ans. E = mc2. 19. What is atomic mass unit? Ans. It is 1 th of the mass of one 12 C atom. 12 6

268 Physics—XII 20. What are isotopes? Ans. Isotopes are the atoms of an element which have the same atomic number but different mass numbers. 21. You are given two nuclides 7 X and 4 Y . Which one of the two is likely to be more 3 3 stable? Give reason. Ans. Due to larger number of neutrons in 7 X, the nuclear force dominates the Coulomb 3 force. 22. Select the pairs of isobars and isotones from the following nuclei: 14 C, 13 N, 14 N, 186O. 6 7 7 Ans. Isobars – 14 C, 147N ; Isotones – 14 C, 168O. 6 6 23. Select the pairs of isotopes and isotones from the following nuclei: 13 C, 14 N, 30 P, 31 P . 6 7 15 15 Ans. Isotopes – 30 P, 1351P; Isotones – 13 C, 147N. 15 6 24. What holds nucleons together in a nucleus? Ans. Nuclear force. 25. What is the source of solar energy? Ans. Nuclear fusion reactions. 26. What is the ratio of kWh to MeV? Ans. 1 kW h  3.6  106 J = 2.25 × 1019. 1 MeV 1.6  1013 J 27. What is the ratio of the radii of two nuclei of mass numbers A1 and A2? Ans.  A1 1/3  A2    28. Why is nuclear fusion not possible in laboratory? Ans. The nuclear fusion takes place at a very high temperature (= 107 K) and pressure. These conditions cannot be realised in laboratory. 29. What is the nuclear radius of 125Fe, if that of 27Al is 3.6 fermi? Ans. R  A1/3 RFe =  A Fe 1/3 or RFe = 3.6  125 1/3 fm = 6 fm RAl  A Al   27    30. Two nuclei have mass numbers in the ratio 8 : 125. What is the ratio of their nuclear radii? Ans. R1 : R2 = 81/3 : 1251/3 = 2 : 5. 31. Two nuclei have mass numbers in the ratio 1 : 3. What is the ratio of their nuclear densities? Ans. Since nuclear density is independent of the mass number, the ratio of nuclear densities will be 1 : 1.

Nuclei 269 32. Write the relationship between the size and the atomic mass number of a nucleus. [CBSE 2012 Comptt.] Ans. R = R0 A1/3 33. Calculate the energy equivalent of 1g of substance. Ans. E = mc2 = 1 × 10–3 × (3 × 108)2 J = 9 × 1013 J SHORT ANSWER TYPE–I QUESTIONS (2 Marks) 1. Very high temperatures as those obtained in the interior of the sun are required for fusion reaction to take place. Why? Ans. When two positively charged nuclei come near each other to fuse together, strong electrostatic repulsive force acts between them. To overcome this repulsive force, they require sufficient energy of the order of 0.1 MeV. This energy is avaliable at extremely high temperature ( 108 K) and extremely high pressure. This is not practicable. These conditions are available in the interior of the sun. 2. Which of the two is more stable— 7 Li or 4 Li? 3 3 Ans. 7 Li is more stable because the number of neutrons in 7 Li is more than in 4 Li. Due 3 3 3 to greater number of neutrons, the nuclear force dominates the electrostatic force of repulsion. 3. It is said that a very powerful crane is required to lift a nuclear mass of microscopic size. Comment on this. Ans. We know that the density of the nuclear matter is enormously high (1017 kg m–3). Thus, even a small nuclear matter will have a very large weight and may require a crane to lift it. As an illustration, 1 cm3 of nuclear matter weighs approximately 1011 kg. 4. Obtain approximately the ratio of the nuclear radii of the gold isotope 197 Au and the 79 silver isotope 107 Ag. What is the approximate ratio of their nuclear mass densities? 47 Ans. r (197Au)   197 1/3 ~ 1.23 r (107 Ag )  107  The ratio of their nuclear mass densities is roughly one. 5. In a nuclear reactor, explain clearly the role of delayed neutrons. Ans. Some (about 0.01 of the total) of the neutrons produced in fission are delayed by a few second because they are produced in subsequent decays of the initial fission fragments. This circumstance is crucial to mechanical control of the reactor. If all the fission neutrons were produced instantly in fission, there would be no time for the minute adjustment required in a reactor to keep it critical. 6. Is it possible that a nucleus has negative mass defect? Ans. This is not possible. This is because if a nucleus has negative mass defect, this would mean that the mass of the nucleus is greater than the combined mass of the nucleons taken individually. This is not possible.

270 Physics—XII 7. A velocity selector selects positive ions of certain velocity. Can the same selector be used to select negative ions of the same velocity? Ans. We know that the selector velocity is the ratio of electric field E and magnetic field B. Clearly, this is independent of the sign of charge. So, the same selector can be used to select negative ions of the same velocity. 8. Why heavy stable nucleus must contain more neutrons than protons? Ans. Nuclear forces are ordinarily attractive and Coulomb force between protons is repulsive. Clearly, the nuclear forces must dominate the Coulomb forces for nuclei to be stable. This is of course possible if the number of neutrons is more than the number of protons. 9. What are the number of protons and the number of neutrons in a nucleus of 238 U? 92 Ans. Number of protons, Z = 92 Number of neutrons, A – Z = 238 – 92 = 146. 10. In a nuclear reactor, explain clearly the role of control rods. (Why are they made of cadmium?) Ans. For a controlled chain reaction, the average number of available neutrons should never exceed one per fission. Any excess neutrons over this ‘critical limit’ should be absorbed. This is what the control rods do. They are made of cadmium because cadmium has a high cross-section for neutron absorption. 11. Why ionic crystals are poor conductors of electricity? Ans. Ionic crystals have no free electrons. So, they are poor conductors of electricity. [Note that the valence electrons in the ionic crystals are tightly bound to the ionic cores.] 12. Express 16 mg mass into equivalent energy in electron volt. Ans. E = mc2 = 16 × 10–6 × (3 × 108)2 = 14.4 × 1011 J = 14.4  1011 eV = 9 × 1030 eV. 1.6  1019 13. Why 29328U is not suitable for chain reaction? Ans. There are three isotopes 233U, 235U and 238U in natural uranium. Their relative abundances are 0.006%, 0.714% and 99.28% respectively. The fission of 235U can be caused by a slow moving neutron having energy of the order of 0.025 eV. The 238U nuclei are fissioned by neutrons having energy more than 1 MeV. When these fast neutrons are used to fission 238U nuclei, the secondary neutrons produced are slow. These secondary neutrons are not able to carry further the fission reaction. Thus, 238U cannot be used for chain reaction. 14. Compare the radii of two nuclei with mass numbers 1 and 27 respectively. Also compare their nuclear densities. Ans. R1   A1 1/3   1 1/3  1 R2  A2  27  3   As nuclear density is independent of mass number and is same for all nuclei, therefore 1 = 1. 2

Nuclei 271 15. Express one joule in eV. Taking 1 amu = 931 MeV, calculate the mass of C–12 atom. Ans. 1 eV = 1.6 × 10–19 J or 1J= 1 eV = 6.25 × 1018 eV 1.6  1019 Mass of C–12 atom = 12 amu = 12 × 931 MeV = 12 × 931 × 1.6 × 10–13 J = 12  931  1.6  1013 = 1.99 × 10–26 kg. (3  108 )2 16. What are thermal neutrons? Why are neutrons considered as ideal particles for nuclear fission? Ans. Thermal neutrons are low-energy neutrons having an approximate energy of 0.025 eV. Neutrons are considered as ideal particles for nuclear fission because they are uncharged. 17. Give the mass number and atomic number of elements on the right-hand side of the decay process 220 Rn  Po  He. 86 Ans. 220 Rn  28146Po  24He 86 Mass number of Po = 216; Atomic number of Po = 84 Mass number of He = 4; Atomic number of He = 2. 18. Define atomic mass unit. Find its energy equivalent in MeV. Ans. Atomic mass unit is defined as 1 th of the mass of one 12 C atom. 12 6 E = mc2 = 1.66 × 10–27 × 3 × 108 × 3 × 108 J = 1.66 × 9 × 10–11 J = 1.66  9  1011 MeV  931 MeV. 1.6  1013 19. Why is the mass of a nucleus always less than the sum of the masses of its constituents, neutrons and protons? Ans. During the formation of a nucleus, the protons and neutrons come closer to a distance of 10–14 m. The energy required for the purpose is spent by the nucleons at the expense of their masses. So mass of the nucleus formed is less than the sum of the masses of the individual nucleons. 20. Draw a graph showing the variation of potential Potential energy (MeV) 100 energy of a pair of nucleons as a function of their 0 separation. Indicate the regions in which nuclear force is (i) attractive (ii) repulsive. –100 2 3 ro 1 r ( fm) Ans. Figure shows potential energy of a pair of nucleons as a function of their separation. For a separation greater than r0, the force is attractive. For separation less than r0, the force is strongly repulsive. The attractive force is strongest when the separation is nearly 1 fm.

272 Physics—XII 21. State the reason, why heavy water is generally used as a moderator in a nuclear reactor. Ans. Neutrons produced during fission get slowed if they collide with a nucleus of the same mass. As ordinary water contains hydrogen atoms (of mass nearly that of neutrons), so it can be used as a moderator. But it absorbs neutrons at a fast rate via reaction: n + p  d + where d is deuteron. To overcome this difficulty, heavy water is used as a moderator which has negligible cross-section for neutron absorption. 22. The nuclear radius of 16 O is 3 ×10–15 m. What is the nuclear radius of 20852Pb? 8 Ans. Mass number of 16 O nucleus, A1 = 16 8 Nuclear radius of 16 O nucleus, R1 = 3 × 10–15 m or R1 = 3 fermi 8 Mass number of 205 Pb nucleus, A2 = 205 82 Nuclear radius of 205 Pb nucleus, R2 = ? 82 Nuclear radius is given by R = R0A1/3 For 16 O , R1 = R0A11/3 8 For 205 Pb , R2 = R0A21/3 82 Dividing, R2 = A21/3 or R2 = R1 × A 1/3 =3× (205)1/3 fermi = 7.02 fermi R1 A11/3 2 (16)1/3 A11/3 23. Find the density of nuclear mass in 29328U nucleus. Given: R0 = 1.5 fermi, mass of each nucleon = 1.67 × 10–27 kg. Ans. Mass number of 238 U = 238 92 R0 = 1.5 fm = 1.5 × 10–15 m Mass of each nucleon = 1.67 × 10–27 kg Density,  = Mass  1.67  1027  A = 1.67  1027  A  3  7 Volume 4  22 (R0A1/3 )3 4 R3 3 or = 1.67  1027  21  A kg m–3 = 1.18 × 1017 kg m–3 88  (1.5  1015 )3 A 24. If 200 MeV energy is released in the fission of a sample nucleus of 29325U, how many fissions must occur per second to produce a power of 1 kW? Ans. Energy released by the fission of one nucleus of 235 U 92 = 200 MeV = 200 × 1.6 × 10–13 J = 3.2 × 10–11 J 1 kW = 1000 W = 1000 J s–1 If x be the number of fissions per second required to generate a power of 1 kW, then x × 3.2 × 10–11 = 1000 or x= 1000 = 3.125 × 1013 3.2  1011

Nuclei 273 25. Alpha particle, electron and proton have equal kinetic energies. Which particle has more ionisation power in same gas medium? Explain it. Ans. Ionisation power  momentum p = 2mK All the given particles have equal kinetic energy.  Ionisation power  m Now, m > mp > me So, alpha particle has maximum ionisation power. 26. 29328U  237 Pa  11H 91 Given: MPa = 237.05115 u; MU = 238.0289 u; MH = 1.0073 u Why is this reaction not spontaneous? Ans. Mass defect = MU – (MPa + MH) = 238.0289 u – (237.05115 + 1.0073 ) u = – 0.02955 u The nuclear reaction is spontaneous if its mass defect is positive; otherwise the reaction is not spontaneous. Here the mass defect is negative ; hence the process is not spontaneous. 27. The atomic masses are generally not whole numbers. Why? Ans. The atomic mass of the element is the average mass of isotopes weighted over their natural abundance. The individual isotopes have a whole number. But the average is not a whole number. 28. Energy released per fission of a nucleus is of the order of 200 MeV whereas that per fusion is of the order of 10 MeV. But a fusion bomb (Hydrogen bomb) is said to be more powerful than a fission bomb. Explain why? Ans. Energy produced per fusion is less but the number of nuclei per unit mass is larger on account of smaller mass number. So, energy released per unit mass is greater. On the other hand, the fissionable materials have high atomic weight. The number of fission nuclei per unit mass is less. So, energy released is less. 29. The three stable isotopes of neon: 20 Ne, 21 Ne and 22 Ne have respective abundances 10 10 10 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. Ans. The average atomic mass of neon is m(Ne) = [90.51 × 19.99 + 0.27 × 20.99 + 9.22 × 21.99] × 10–2 = 20.18 u. 30. Obtain approximately the ratio of the nuclear radii of the gold isotope 17997Au and the silver isotope 107 Ag. 47 Ans. R1   A1 1/3   197 1/3 = 1.23 R2  A2   107   

274 Physics—XII SHORT ANSWER TYPE–II QUESTIONS (3 Marks) 1. Obtain the approximate value of the radius of (a) a nucleus of 4 He and (b) a nucleus 2 of 29328U. (c) What is the ratio of these radii? (Assume R0 = 1.2 × 10–15 m.) Ans. (a) R = R0A1/3 = 1.2 × 10–15(4)1/3 m = 1.9 × 10–15 m (b) R = 1.2 × 10–15(238)1/3 m = 7.4 × 10–15 m (c) R  7.4 = 3.89 R 1.9 2. Give reasons for the following: Lighter elements are better moderators for a nuclear reactor than heavier elements. Ans. The neutrons produced by fission are fast, with kinetic energies of about 2 MeV. However, fission is induced most effectively by thermal neutrons (0.0235 eV energy). The fast neutrons can be slowed down by mixing the uranium fuel with a substance called moderator. Moderator should have two properties: (i) It should be effective in slowing down neutrons via elastic collisions. We know that if a moving particle has a head-on elastic collision with a stationary particle, the moving particle loses all its kinetic energy if the two particles have the same mass. (ii) The moderator should not absorb neutrons. These properties are satisfied by lighter elements and not by heavier elements. 3. Answer the following: Are the equations of nuclear reactions ‘balanced’ in the sense a chemical equation (e.g., 2H2 + O2  2H2O) is? If not, in what sense are they balanced on both sides? Ans. A chemical equation is balanced in the sense that the number of atoms of each element is the same on both sides of the equation. A chemical reaction merely alters the original combination of atoms. In a nuclear reaction, elements may be transmuted. Thus, the number of atoms of each element is not necessarily conserved in a nuclear reaction. However, the number of protons and the number of neutrons are both separately conserved in a nuclear reaction. [Actually even this is not strictly true in the realm of very high energies–what is strictly conserved is the total charge and total ‘baryon number.’] In nuclear reactions, the number of protons and the number of neutrons are the same on the two sides of each equation. 4. In a nuclear reactor, explain clearly the role of moderator (why is heavy water used as a moderator?). Ans. The neutrons produced in fission have energies of the order of a few MeV. The cross- section (i.e., probability) for fission of 235U by these fast neutrons is negligible compared to that by slow (thermal) neutrons. The role of the moderator is to slow down neutrons for further fission of 235U and thus sustain a chain reaction. Now energy loss of a neutron in a collision is maximum if it hits a nucleus of the same mass. Thus, ordinary water (consisting of hydrogen atoms) can be used as an effective moderator. There is, however, one problem. In a neutron-proton collision, there is a considerable chance of another process, namely absorption of a neutron by a proton via the reaction: n + p  d + g (d : deuteron). To avoid this difficulty, one uses heavy water as a moderator, which has negligible cross-section for neutron absorption.

Nuclei 275 5. If ratio of mass numbers is 3 : 1, find the ratio of nuclear densities. Ans. Radius of nucleus, R = R0A1/3 Mass, M = Amp; Density,  = M  M = A mp  mp = constant V R3  R30 A 4 4 4  R30 3 3 3 So, the nuclear density is independent of nuclear mass number. So, the ratio of nuclear densities is 1. 6. A general impression is that mass-energy inter-conversion takes place only in nuclear reactions and never in chemical reactions. This is strictly speaking incorrect. Explain. Ans. From the point of view of mass-energy inter-conversion, a chemical reaction is similar to a nuclear reaction in principle. The energy released or absorbed in a chemical reaction can be traced to the difference in chemical (not nuclear) binding energies of atoms and molecules on the two sides of a reaction. Since, strictly speaking, chemical binding energy also gives a negative contribution (mass defect) to the total mass of an atom or molecule, we can equally well say that the difference in the total mass of atoms or molecules on the two sides of a chemical reaction gets converted into energy or vice versa. However, the mass defects involved in a chemical reaction are almost a million times smaller than those in a nuclear reaction. This is the reason for the general impression (which is incorrect) that mass-energy inter-conversion does not take place in a chemical reaction. 7. Give the order of magnitude of nuclear mass density and average atomic mass density. Compare these densities with the typical mass density of solids, liquids and gases (at ordinary temperatures and pressures). Ans. Nuclear mass density is of the order of 1017 kg m–3. This estimate is obtained by using the empirical relation r = r0 A1/3 (r0 = 1.2 × 10–15 m) for the radius of a nucleus of mass number A. It is typically 1013 to 1014 times the average mass density of an atom. That is, the average atomic mass density is of the order of 103 to 104 kg m–3. Typical mass densities of solids and liquids are of the same order since atoms are tightly packed in these phases. Typical densities of gases at STP are of the order of 10–1 to 1 kg m–3. 8. With respect to power generation, what are the relative advantages and disadvantages of Fusion type and Fission type reactors? Ans. Fusion requires high temperature. Controlled reaction is not yet obtained. Highly sophisticated technology will be required. However, the fuel is easily available, cheap and causes very less pollution. There is no problem of waste management. Fission controlled chain reaction is possible. The technology is well developed and established. There is high radioactive pollution. There also exists the problem of waste management. 9. Consider the case of bombardment of 235U nucleus with a thermal neutron. The fission products are 95Mo and 139La and two neutrons. Calculate the energy released. (Rest masses of the nuclides: 235U = 235.0439 u, 1 n = 1.0087 u, 95Mo = 94.9058 u, 0 139La = 138.9061 u, Take 1 u = 931 MeV. ) Ans. Total rest mass (initial) = 236.0526 u Total rest mass (final) = 235.8293 u

276 Physics—XII Decrease in rest mass due to fission = 0.2233 u Energy released = 0.2233 u × 931 MeV = 207.9 MeV. u 10. A power reactor develops energy at the rate of 30,000 kW. How many gram of 235U would be consumed daily? Assuming that on an average 200 MeV energy is released per fission. Ans. Number of atoms undergoing fission per second = 3  107 = 0.9375 × 1018 200  1.6  1013 Number of atoms undergoing fission in 24 hour = 0.9375 × 1018 × 24 × 3600 = 0.81 × 1023.  Mass of uranium undergoing fission = 235 × 0.81 × 1023 = 31.6 g 6.023  1023 11. The fission properties of 239 Pu are very similar to those of 29325U. The average energy 94 released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure 239 Pu undergo fission? 94 Ans. Energy released per fission of 239 Pu is 180 MeV. 94 Number of atoms of 239 Pu which undergo fission = 6.023  1023  1000 = 25.2 × 1023 94 239 Energy released in the fission of 1 kg of 239 Pu = 180 × 25.2 × 1023 MeV 94 = 4.536 × 1026 MeV 12. A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 29325U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 29325U and that this nuclide is consumed only by the fission process. Ans. Energy generated per gram of 29325U = 6  1023 × 200 × 1.6 × 10–13 J = 1.2  1.6 × 1013 J 235 235 Amount of 29325U consumed in 5 year operating for 80% of the time = 0.8  5  3.154  1016  235 g = 1544.1 kg 1.2  1.6  1013 Initial amount of 235 U = 3088.2 kg 92

Nuclei 277 LONG ANSWER TYPE QUESTIONS (5 Marks) 1. In a nuclear reactor, 235U undergoes fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 year, find the total mass of uranium required. Ans. The reactor produces 1000 MW power or 109 W power or 109 Js–1 of power. The reactor is to function for 10 year. Therefore, total energy which the reactor will supply in 10 year is E = (Power) (time) = (109 Js–1) (10 × 365 × 24 × 3600 s) = 3.1536 × 1017 J. But since the efficiency of the reactor is only 10%, therefore actual energy needed is 10 times of it or 3.1536 × 1018 J. One uranium atom liberates 200 MeV of energy or 200 × 1.6 × 10–13 J or 3.2 × 10–11 J of energy. So, number of uranium atoms needed are 3.1536  1018 = 0.9855 × 1029 3.2  1011 or number of kg-moles of uranium needed are n = 0.9855  1029 = 163.7 6.02  1026 Hence, total mass of uranium required is m = (n) M = (163.7) (235) kg or m  38470 kg or m = 3.847 × 104 kg 2. About 185 MeV of usable energy is released in the neutron induced fissioning of a 235 U nucleus. If the reactor using 29325U as fuel continuously generates 100 MW of 92 power, how long will it take for 1 kg of the uranium to be used up? Ans. Number of fissions occurring/s to give an energy output of 100 MW (= 108 Js–1) = 108 J s1 J = 3.378 × 1018 fission s–1 185  106  1.6  1019 Number of nuclei contained in 1 kg of 235U = 1 kg × 6.023 × 1026 nuclei = 2.563 × 1024 nuclei 235 kg/k mol k mol Time taken to exhaust 1 kg of 235U = 2.563  1024 second = 8.78 day 3.378  1018 3. In a fusion reactor, the reaction occurs in two stages.    (i) Two deuterium2 3 T nucleus with a proton as 1 D nuclei fuse to form a tritium 1 product.  (ii) A tritium nucleus fuses with another deuterium nucleus to form a helium 4 He 2 nucleus with neutron as another product. Find (a) the energy released in each stage. (b) the energy released in the combined reaction per deuterium and (c) what percentage of the mass energy of the initial deuterium is released? Given: 2 D = 2.014102 u, 3 T = 3.016049 u, 4 He = 4.002603 u, 1 1 2 1 H = 1.007825 u, 1 n = 1.008665 u, Take 1 u = 931 MeV 1 0

278 Physics—XII Ans. (a) 2 D + 12D  13T  11H + Q1 1  Q1 = 0.00433 × 931 MeV = 4.031 MeV 3 T + 12D  24He + 10n + Q2 1 Q2 = 0.01888 × 931 MeV = 17.577 MeV (b) Energy released per deuterium nucleus = 21.61 = 7.203 MeV 3 (c) Percentage of rest mass of deuterium released = 7.203 MeV = 0.384% (2.014102  931) MeV 4. For the D-T fusion reaction, find the rate at which deuterium and tritium are consumed to produce 1 MW. The Q-value of D-T reaction is 17.6 MeV. Assume all the energy from the fusion reaction is available. Ans. In the D-T fusion reaction, the energy released in each fusion is Q = 17.6 MeV. Number of reactions per second required to produce an energy of 106 J s–1 = 106 = 3.55 × 1017 reactions s–1 17.6  106  1.6 1019 In each reaction, one atom of deuterium and one atom of tritium are consumed.  Mass of deuterium consumed per second = 2 kg  1 k mol × 3.55 × 1017 atoms = 1.179 × 10–9 kg s–1 1k mol 6.023  1026 atoms s Mass of tritium consumed/second = 3 × 1.179 × 10–9 kg s–1 = 1.769 × 10–9 kg s–1 2 5. (a) Two stable isotopes of lithium 6 Li and 7 Li have respective abundances of 7.5% 3 3 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 B and 11 B. Their respective masses are 5 5 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of 10 B and 11 B. 5 5 Ans. (a) Atomic weight of lithium = 7.5  6.01512  92.5  7.01600 = 45.1134  648.98 = 6.94093 100 100 (b) Let the percentage of 10 B in natural boron be x. Then the percentage of 11 B in 5 5 natural boron will be (100 – x). Now, atomic mass of natural boron = weighted average masses of its isotopes 10.811 = 10.01294  x  11.00931(100  x) 100 1081.1 = 1100.931 – 0.99637x

Nuclei 279 or 0.99637x = 19.831 or x = 19.831 = 19.9 0.99637 So, percentage of 10 B is 19.9. Percentage of 11 B is (100 – 19.9) or 80.1. 5 5 6. How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as 2 H  2 H  3 He +n+ 3.27 MeV. 1 1 2 Ans. When two nuclei of deuterium fuse together, energy released = 3.27 MeV Number of deuterium atoms in 2 kg = 6.023  1023  2000 = 6.023 × 1026 2 When 6.023 × 1026 nuclei of deuterium fuse together, energy released = 3.27 × 6.023 × 1026 MeV 2 = 3.27 × 6.023 × 1026 × 1.6 × 10–13 J = 1.575 × 1014 J or Ws 2 Power of electric lamp = 100 W If the lamp glows for time t, then the electrical energy consumed by the lamp is 100 t.  100 t = 1.575 × 1014 or t = 1.575 × 1012 s = 1.575  1012 years = 4.99 × 104 years 3.154  107 7. Calculate the height of the potential barrier for a head on collision of two deuterons. Assume that they can be taken as hard spheres of radius 2.0 fm. Ans. The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Suppose the two particles are fired at each other with the same kinetic energy K so that they are brought to rest by their mutual Coulomb repulsion when they are just touching each other. We can take this value of K as a representative measure of the height of Coulomb barrier. 2K = potential energy 1 e2 4 0 (2R) e2 = (1.6  1019 )2 J K= 16  3.14  8.85  1012  2  1015 16 0 R = 2.8788 × 10–14 J = 2.8788  1014 keV = 179.9 keV 1.6  1019  103 8. Calculate and compare the energy released by (a) fusion of 1.0 kg of hydrogen deep within Sun and (b) the fission of 1.0 kg of 235U in a fission reactor. Ans. Fusion taking place deep within the sun is given by: 411H  24He + 2e+ + 2 + 26.7 MeV If Q1 is the energy released by fusion of 1.0 kg of hydrogen, then Q1 = 1000  6.023  1023  26.7 MeV = 40.2 × 1026 MeV 4

280 Physics—XII Again, during the fission of one nucleus of 235U, energy released is 200 MeV. If Q2 be the energy released by the fission of 103 g of 235U, then Q2 = 1000  6.023  1023  200 MeV = 5.1 × 1026 MeV 235 Q1 = 40.2  1026 = 7.88  8 Q2 5.1  1026 So, energy released per kg of fuel in fusion is nearly 8 times that in fission. 9. Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten per cent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilisation (i.e., conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200 MeV. Ans. Target of power, P = 200,000 MW = 2 × 1011 W Power required from nuclear power plants = 10% of P = 10 × 2 × 1011 W = 2 × 1010 W 100 Electrical energy required from nuclear power plants in 1 year = 2 × 1010 × 3.154 × 107 J = 6.308 × 1017 J Heat energy produced per fission of 235U = 200 MeV Available electrical energy per fission = 25 × 200 MeV = 50 × 1.6 × 10–13 J = 8 × 10–12 J 100 Number of fissions required in one year = 6.308  1017 = 7.885 × 1028 8  1012 Number of moles of 235U required per year = 7.885  1028 6.023  1023 Mass of 235U required per year = 235  7.885  1028 kg = 3.076 × 104 kg. 6.023  1023  1000 CASE STUDY BASED QUESTIONS Read the following passages and answer the questions that follow: CASE STUDY 1 An atom bomb is a fission bomb involving an uncontrolled chain reaction. It is a nuclear war weapon. Two pieces of fissionable material namely Uranium (U) or Plutonium (Pu), each of which is less than the critical size are taken and separated by a cadmium partition. These pieces are enclosed in a very thick envelope called tamper. To start the explosion, the two pieces are brought together by a mechanical device i.e., one may be the target and the other may be taken as a bullet. Both the pieces fit into each other and the total mass now becomes slightly more than critical, with

Nuclei 281 reproduction factor just above one. A source of neutrons is placed nearby. The neutrons get into the mass and finding it critical, start their uncontrolled chain reaction. The temperature and pressure both rise tremendously in a matter of a few micro-second (10–6 s). The intense energy released then breaks off the tamper and a huge, violent and devastating explosion results. As a by-product of this, high energy -rays are given out which are extremely injurious to human tissues, causing intense and permanent damage to the human system. One kg of uranium gives off energy which can cause devastation equivalent to about 20,000 tonnes of dynamite (T.N.T). 1. Energy generation in stars is mainly due to (a) chemical reactions (b) fision of heavy nuclei (c) fusion of light nuclei (d) fusion of heavy nuclei Ans. (c) 2. When a Uranium atom 92U235 undergoes fission, number of neutrons liberated is (a) 1 (b) 2 (c) 3 (d) about 2.6 Ans. (d) 3. The wavelength of -rays is of the order of (a) 10–8 m (b) 10–5 m (c) 10–12 m (d) 10–22 m Ans. (c) 4. What fissionable material was used in the bomb dropped at Nagasaki (Japan) in 1946? (a) Uranium (b) Neptunium (c) Plutonium (d) none of these Ans. (c) CASE STUDY 2 Nuclear reactors are designed in such a way that they cannot explode like atomic bombs. The course of a chain reaction is determined by the probability that a neutron released in fission will cause a subsequent fission. If the neutron population in a reactor decreases over a given period of time, the rate of fission will decrease and ultimately drop to zero. If over the course of time the neutron population is sustained at a constant rate, the fission rate will remain steady, and the reactor will be in what is called a critical state. Finally, if the neutron population increases over time, the fission rate and power will increase, and the reactor will be in a supercritical state. Material like Cadmium or Boron are used to control the chain reactions. Cadmium absorbs neutrons without suffering disintegration. Thus by adjusting the length of these Cd rods in or out of the reactors, the number of neutrons taking part in the nuclear reaction can be controlled. 1. Nuclear fuel is a material that can be fissioned by (a) fast neutrons (b) thermal neutrons (c) alpha particles (d) gamma rays Ans. (b) 2. When 92U235 undergoes fission, 0.1% of the original mass is converted into energy. The energy released by an atom which contain 10 kg of 92U235 is (a) 7.8 × 1010 J (b) 6.9 × 1011 J (c) 8.2 × 1011 J (d) 9.4 × 1012 J

282 Physics—XII Ans. (a) Hint and Solution: Mass of U-235 converted into energy = 10 × 0.1 kg = 10 g 100 Number of U235 atoms in 10 g = 6.023  1023 × 10 = 25.63 × 1021 235  Energy released = 25.63 × 1021 × 200 MeV = 51.26 × 1023 × 1.6 × 10–13 J = 8.2 × 1011 J 3. Which of the following statement is true? (a) Fusion is the process in which a heavy nucleus breaks up into two lighter nuclei. (b) Moderator is a material which is used to speed up the neutrons produced by nuclear fission. (c) The more the Cadmium rod are withdrawn, stronger is the intensity of chain reaction. (d) None of these. Ans. (c) Hint and Solution: Because Cadmium rod absorbs neutrons without suffering disintegration and slow down the intensity of chain reaction. 4. How much 235U is consumed in a day in an atomic power house operating at 400 MW, provided the whole of the mass of 235U is converted into energy? (a) 0.289 g (b) 0.34 g (c) 0.267 g (d) 0.384 g Ans. (d) Hint and Solution: Rate of production of energy, P = 400 MW = 400 × 106 Js–1 Energy produced in a day E = P × 86400 J = 400 × 86400 × 106 J = 3.456 × 1013 J If m kg is the required mass, then m= E or m= 3.456  1013 kg = 0.384 × 10–3 kg = 0.384 g c2 (3  108 )2

ISBN: 978-93-93268-65-5 789393 268655 T12-6739-349-COMP.CBSE QB PHY T-II XII