Physics-29.23-Atoms-Molecules & nuclei

1. RADIOACTIVITY It was observed by Henri Becquerel in i896 that some minerals like pitchblende emit radiation spontaneously, and this radiation can blacken photographic plates if photographic plate is wrapped in light proof paper. He called this phenomenon radioactivity. It was observed that radioactivity was not affected by temperature, pressure or chemical combination, it could therefore be a property that was intrinsic to the atom-more specifically to its nucleus. Detailed studies of radioactivity resulted in the discovery that radiation emerging from radioactive substances were of three types : -rays, -rays and -rays. -rays were found to be positively charged, -rays (mostly) negatively charged, and - rays uncharged. Some of the properties of and – rays, are explain below: (i) Alpha Particles: An -particle is a helium nucleus, i.e. a helium atom which has lost two electrons. It has a mass about four times that of a hydrogen atom and carries a charge +2e. The velocity of -particles ranges from 5 to 7 percent of the velocity of light. These have very little penetrating power into a material medium but have a very high ionising power. (ii) Beta Particles: -particles are electrons moving at high speeds. These have greater (compared to -particles) penetrating power but less ionising power. Their emission velocity is almost the velocity of light. Unlike -particles, they have a spectrum of energy, i.e., beta particles possess energy from a certain minimum to a certain maximum value. (iii) Gamma Rays: -rays are electromagnetic waves of wavelength of the order of ~ 10–12 m. These have the maximum penetrating power (even more than X–ray) and the least ionising power. These are emitted due to transition of excited nucleus from higher energy state to lower energy state. Property Nature Positive charged Negatively charged Uncharged particles, 2He4 particles (electrons). nucleus electromagnetic radiation Charge +2e –e 0 Mass 6.6466 × 10–27 kg 9.109×10–31kg 0 Range ~10 cm in air, can be Upto a few m in Several m in air stopped by 1mm of A can be stopped stopped by ~ cm of ~cm of A Pb Natural By natural radioiso- By radioisotopes Excited nuclei formed sources topes e.g. 92U236 e.g. 29Co68 as a result of , decay The other simple laws of radioactivity that were discovered by were the Rutherford-Soddy displacement laws: (a) During a decay, the daughter element was always two position below the parent in the periodic table; the mass number of the daughter nucleus was 4 units smaller than that of the parent nucleus. (b) During -decay, the daughter nucleus had the same mass number as the parent while its atomic number was greater than that of the parent by 1 (smaller by 1 unit in decay). (c) emission did not result in change of atomic number or mass number. Radioactivity is observed to be a random process, it cannot be predicted when a particular nucleus will decay. One can only discuss the probability of its decay. 2. RADIOACTIVE DECAY LAW The number of atoms disintegrating per second is directly proportional to the number of atoms (N) present at that instant. For most radioisotopes, is very small in the SI system it take a large number N(~ Avogadro number, 1023) to get any significant activity. ….(i) Where  is the radioactivity decay constant. If No is the number of radioactive atoms present at a time t = 0, and N is the number at the end of time t, then The term is called the activity of a radioactive substance and is denoted by 'A'. Units: 1 Becquerel (Bq) = 1 disintegration per second (dps) 1 curie (Ci) = 3.7 x 1010 dps 1 Rutherford =106 dps The changes in N, due to radioactivity, are very small in 1 s compared with the value of N 'its self. Therefore. N may be approximated by means of a continuous function N(t). For the case considered in equation (i), we can write, A = rate of decrease of N = or .....(ii) Note the equation (ii) is valid only for large value of N, typically –1018 or so. If the number of atoms at time t = 0 (initial) is N = N0, then one can solve equation (ii): or In or N(t) = N0 ......(iii) The activity A (t) = (t) and is given by A (t) = = A0 taking (A0 = N0) ....(iv) The variation of N is represented below: Graphs of A vs t or In A vs t are similar to the above graphs 2.1 HALF-LIFE A significant feature of the above graph is the time in which the number of active nuclei (N) is halved. This is independent of the starring value, N = N0. Let us compute this time: = or, = In 2 = 0.693 (approx) or, t' = = This half-life represents the time in which the number of radioactive nuclei falls to of its starting value. Activity, being proportional to the number of active nuclei, also has the same half-life. Illustration 1: A count rate-meter is used to measure the activity of a given sample. At one instant the meter shows 4750 counts per minute. Five minutes later it shows 2700 counts per minute. Find: (a) decay constant (b) the half life of the sample. Solution: Initial activity = A0 = dN/dt at t = 0 Final activity = At = dN/dt at t = t 2.2 MEAN-LIFE A very significant quantity that can be measured directly for small numbers of atoms is the mean lifetime. If there are n active nuclei, (atoms) (of the same type, of course), the mean life is Where , ,....... represent the observed lifetime of the individual nuclei and n is a very large number. It can also be calculated as a weighted average: Where N1 nuclei live for time , N2 nuclei live for time ....... and so on. This quantity may be related with . Using calculus (vi) may be rewritten as: Where |dN| is the number of nuclei decaying between t, t + dt; the modulus sign is required to ensure that it is positive. dN = and |dN| = = = (integrating the numerator by parts) or Illustration 2: Cascade Decays: Frequently, a nucleus A decays into another nucleus B, which again decays into C, and so on until the series ends in a stable nuclide. These are known as cascade decays. We will consider a simple example of a cascade decay where a nuclide a decays into B, and then, B decays into C: Suppose that there are N0 atoms of A to begin with the problem is to find the numbers 0f atoms of A, B, C respectively as a function of time t. Solution: Suppose that, at time t, there are NA atoms of A. Nb atoms of B. and We write the equations: –activity of A = .....(i) = + (activity of A) – (activity of B) = ....(ii) Since each decay of A increases the number of atoms of B ....(iii) Solving equation (i), we get NA (t) = N0 ....(iv) Substituting this in equation (ii), we get. the solution, NB (t) = ......(v) Putting the initial value (t = 0) of NB, we get C1, and substituting this value of C1 we get the expression for NB : NB (t) = .......(vi) Substitution the value of NB in equation (iii) and solving, we get NC = ......(vii) Using these values of NA, NB and NC. the activity can be calculated, if required. 3. THE NUCLEUS It exists at the centre of an atom, containing entire positive charge and almost whole of mass. The electron revolve around the nucleus to form an atom. The nucleus consists of protons (+ve charge) and neutrons. (i) A proton has positive charge equal in magnitude to that of an electron (+1.6 x 10–19 C) and a mass equal to 1840 times that of an electron. (ii) A neutron has no charge and mass is approximately equal to that of proton. (iii) The number of protons in a nucleus of an atom is called as the atomic number (Z) of that atom. The number of protons plus neutrons (called as Nucleons) in a nucleus of an atom is called as mass number (A) of that atom. (iv) A particular set of nucleons forming an atom is called as nuclide. It is represented as ZXA. (v) The nuclides having same number of protons (Z), but different number of nucleons (A) are called as isotopes. (vi) The nuclide having same number of nucleons (A), but different number of protons (Z) are called as isobars. (vii) The nuclide having same number of neutrons (A - Z) are called as isotones. 3.1 MASS DEFECT & BINDING ENERGY The nucleons are bound together in a nucleus and the energy has to be supplied in order to break apart the constituents into free nucleons. The energy with which nucleons are bounded together in a nucleus is called as Binding Energy (B.E.). In order to free nucleons from a bounded nucleus this much of energy (= B.E.) is to be supplied. It is observed that the mass of a nucleus is always less than the mass of constituent (free) nucleons. This difference in mass is called as mass defect and is denoted as m. If mn : mass of a neutron ; mp : mass of a proton M (Z, A) : mass of bounded nucleus Then, m = Z . mp + (A – Z). mn – M (Z, A) This mass-defect is in form of energy and is responsible for binding the nucleons together. From Einstein's law of inter-conversion of mass into energy: E = mc2 (c : speed of light; m : mass) binding energy = m . c2 Generally, m is measured in amu units. So let us calculate the energy equivalent to 1 amu. It is calculated in eV (electron volts; 1 eV = 1.6 x 10–19 J) There is another quantity which is very useful in predicting the stability of a nucleus called as Binding energy per nucleons. From the plot of B.E./nucleons Vs mass number (A), we observe that: (i) B.E./nucleons increases on an average and reaches a maximum of about 8.7 MeV for A  50 – 80. (ii) For more heavy nuclei, B.E./nucleons decreases slowly as A increases. For the heaviest natural element U238 it drops to about 7.5 MeV. (iii) From above observation, it follows that nuclei in the region of atomic masses 50-80 are most stable. Illustration 2: If mass of proton = 1.008 amu and mass of neutron = 1.009 amu, then the binding energy per nucleon for 4Be9 (mass = 9.012 amu) will be – (A) 0.0672 MeV (B) 0.672 MeV (C) 6.72 MeV (D) 67.2 MeV Solution: Mass defect Illustration 4: The energy released in the following -decay process will be - Given that (A) 0.931 MeV (B) 0.731 MeV (C) 0.511 MeV (D) 0.271 MeV Solution: Mass defect Illustration 5: The binding energy per nucleon for 3Li7 will be, if the mass of 3Li7 is 7.01653 amu. (A) 5.6 MeV (B) 39.25 MeV (C) 1 MeV (D) zero. Solution: Illustration 6: How much energy is released in the following reaction? If the B.E./Nucleon of 1H2 and 2He4 are 1.123 MeV and 7.2 MeV respectively. (A) 12 MeV (B) 24.3 MeV (C) 36 MeV (D) zero Solution: B. E. of 1H2 B.E. of an  -particle = 4  7.2 E = 28.8  energy released ER = E – Ed ER = 28.8 – 4.5 = 24.3 MeV 3.2 NUCLEAR FORCES The protons and neutrons are held together by the strong attractive forces inside the nucleus. These forces are called as nuclear forces. (i) Nuclear forces are short-ranged. They exist in small region (of diameter 10–15 m = 1 fm). The nuclear force between two nucleons decrease rapidly as the separation between them increases and becomes negligible at separation more than 10 fm. (ii) Nuclear force are much stronger than electromagnetic force or gravitational attractive forces. (iii) Nuclear force are independent of charge. The nuclear force between two proton is same as that between two neutrons or between a neutron and proton. This is known as charge independent character of nuclear forces. In a typical nuclear reaction (i) In nuclear reactions, sum of masses before reaction is greater than the sum of masses after the reaction. The difference in masses appears in form of energy following the Law of inter-conversion of mass & energy. The energy released in a nuclear reaction is called as Q Value of a reaction and is given as follows. If difference in mass before and after the reaction is m amu (m = mass of reactants minus mass of products) then Q value = m (931) MeV (ii) Law of conservation of momentum is also followed. (iii) Total number of protons and neutrons should also remain same on both sides of a nuclear reaction. Illustration 7: Calculate the Q-value of the nuclear reaction: 0C12 10Ne20 + 2He4 The following data are given: m(6C12) = 12.000000 u m(10Ne20) = 12.000000 u m(2Ne4) = 4.002603 u Solution: The Q-value of this reaction may be easily calculated may be easily calculated from the masses of the individual nuclei. Q = [2m (6C12) – { m (10Ne20) + m (2He4)}]e2 = 24.000000 – (19.992439 + 4.002603)} u × c2 = 4.618 MeV Illustration 8: Binding energy of the calculate the mass defect and the binding energy of an -particle given the following date: mn = 1.0086665 u mp = 1.007825 u = 4.0026 u (1u = 931.5 MeV/c2) Solution: The mass defect of an -particle is given by = (2 × 1.008665 + 2 × 1.007825) – 4.0026)u = 0.03038u The Binding energy is related to the mass defect by the reaction. B.E. = = 0.03038 × 931.5 MeV = 28.3 MeV (approximately) 3.3 NUCLEAR FISSION The breaking of a heavy nucleus into two or more fragments of comparable masses, with the release of tremendous energy is called as nuclear fission. The most typical fission reaction occurs when slow moving neutrons strike 92U235. The following nuclear reaction takes place. If more than one of the neutrons produced in the above fission reaction are capable of inducing a fission reaction (provided U235 is available), then the number of fissions taking place at successive stages goes increasing at a very brisk rate and this generates a series of fissions. This is known as chain reaction. The chain reaction takes place only if the size of the fissionable material (U235) is greater than a certain size called the critical size. If the number of fission in a given interval of time goes on increasing continuously, then a condition of explosion is created. In such cases, the chain reaction is known as uncontrolled chain reaction. This forms the basis of atomic bomb. In a chain reaction, the fast moving neutrons are absorbed by certain substances known as moderators (like heavy water), then the number of fissions can be controlled and the chain reaction is such cases is known as controlled chain reaction. This forms the basis of a nuclear reactor. Illustration 9: When a beta particle is emitted from a nucleus the effect on its neutron-proton ratio is (A) increased (B) decreased (C) remains same (D) first (1) then (2) Solution: Illustration 10: In which sequence, the radioactive radiations are emitted in the following nuclear reactions? (A) ,  and  (B) ,  and  (C) ,  and  (D) ,  and  Solution: We know that on emission of -particle from the nucleus, atomic number reduces by 2 and mass number reduces by 4. And on emission of -particle from the nucleus, the atomic number increases by 1. Similarly on emission of -particle from the nucleus, there is no change in atomic number of mass number. 3.4 NUCLEAR FUSION The process in which two or more light nuclei are combined into a single nucleus with the release of tremendous amount of energy is called as nuclear fusion. Like a fission reaction, the sum of masses before the fusion (i.e. of light nuclei) is more than the sum of masses after the fusion (i.e. of bigger nucleus) and this difference appears as the fusion energy. The most typical fusion reaction is the fusion of two deuterium nuclei into helium. For the fusion reaction to occur, the light nuclei are brought closer to each other (with a distance of 10–14 m). This is possible only at very high temperature to counter the repulsive force between nuclei. Due to this reason, the fusion reaction is very difficult to perform. The inner core of sun is at very high temperature, and is suitable for fusion, in fact the source of sun's and other star's energy is the nuclear fusion reaction 4. ASSIGNMENT 1. The mass number of a nucleus is (A) always less than atomic number (B) always more than atomic number (C) equal to atomic number (D) some times more than and some times equal to atomic number. Solution- (D) Mass number of a nucleus represents the number of nucleons (neutrons + protons). When no neutrons are present in the nucleus, mass number equals to the atomic number. 2. In which of the following decays, the element does not change? (A)  – decay (B) + – decay (C)  – decay (D) y – decay Solution- (D) -ray has no charge and no mass during -decay the element will not have any change in atomic number or mass number. 3. In the reaction the minimum energy of the particle is (A) 1.21 MeV (B) 1.62 MeV (C) 1.89 MeV (D) 1.96 MeV. (MN = 14.00307 amu, MHe = 4.00260 amu and MO = 16.99914 amu, MH = 1.00783 amu and 1 amu = 931 MeV) Solution- (A) 7N14 + 2He4 8O17 + 1H1 Total mass of reactants = 18.00567 amu Tota mass of products = 18.00697 amu Mass defect = 18.00697 – 18.00567 = 0.0013 amu Energy (E) = 931 (0.0013) = 1.2103 MeV 4. In the carbon cycle of fusion (A) Four 1H1 fuse to form 2He4 and two positrons (B) Four 1H1 fuse to form 2He4 and two electrons (C) Two 1H2 fuse to form 2He4 (D) Two 1H2 fuse to form 2He4 and two neutrons Solution- (A) 41H1 2He4 + 2. +1e0 + Energy. 5. In each fission of U235, 200 MeV of energy is released. If a reactor produces 100 MW power, the rate of fission in it will be (A) 3.125 × 1018 per minute > (B) 3.125 × 1017 per second (C) 3.125 × 1017 per minute> (D) 3.125 × 1018 per second Solution- (D) We know = 3.125 × 1018 /sec 6. To generate a power of 3.2 MW, the number of fissions of U235 per minute is (Energy released per fission = 200 MeV, 1 eV = 1.6  10–19 J) (A) 6×1018¡ (B) 6×1017 (C) 1017 (D) 6×1016 Solution- (A) Power of reactor Where 'n' is number of fissions 't' is time 'E' is energy released per fission 7. If in nuclear reactor using U235 as fuel, the power output is 4.8 MW, the number of fissions per second is (Energy released per fission of U235 = 200 MeV watts, e eV = 1.6  10–19 J) (A) 1.5×1017 (B) 3×1019 (C) 1.5×025 (D) 3×1025 Solution- (A) P = 4.8 MW = 4.8×106 W Power of a nuclear reactor = 1.5 × 1017 8. In the following nuclear reaction . What does X stand for ? (A) A neutron (B) A neutrino (C) an electron (D) A proton Solution- In -decay process a neutrino is accompanied with the emission of a positron. 9. If 200 MeV energy is released in the fission of a single U235 nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (given 1 eV = 1.6  10–19 J) (A) 3.125×1013 (B) 3.125×1014 (C) 3.125×1015 (D) 3.125×1016 Solution- (A) Energy released in the fission E = 200 MeV = 200×106×1.6×10–19 = 3.2×10–11 Joule = 3.125×1013 10. MP = 1.008 a.m.u., MN = 1.009 a.m.u. and 2He4 = 4.003 a.m.u. then the binding energy of -particle is (A) 21.4 MeV (B) 8.2 MeV (C) 34 MeV (D) 1.35 A/m. Solution- (D) BE = 2Mp + 2Mn – Ma = 2×1.008 + 2 × 1.009 – 4.003 = 0.031 amu = 0.031 × 931 = 28.8 MeV 11. Energy released in fusion of 1 kg of deuterium nuclei (A) 9×1013 J (B) 6×1027 J (C) 2×107 KwH (D) 8×1023 MeV Solution- (A) Fusion reaction of deuterium 1H2 + 1H2 2He3 + 0n1+3.27 MeV 12. The energy produced in the sun is due to (A) fission reaction (B) fusion reaction (C) chemical reaction (D) motion of electrons and ions Solution- (A) We know that source of the huge solar energy is the fusion of lighter nuclei. Fusion of hydrogen nuclei into helium nuclei is continuously taking place in the plasma, with the continuous liberation of energy. Therefore energy produced in the sun is due to fusion reaction. 13. Atomic number of a nucleus is Z, while its mass number is M. What will be the number of neutrons in its nucleus ? (A) M (B) Z (C) (M–Z) (D) (M+Z) Solution- (C) M = Z + N 14. In uncontrolled chain reaction, the quantity of energy released, is (A) very high (B) very low (C) normal (D) first (A) to (B) Solution- (A) We know that in the uncontrolled chain reaction, more than one of the neutrons produced in a particular fission cause further fissions so that the number of fissions increases very rapidly. Thus this is a very fast reaction and the whole substance is fissioned in a few moments. A huge quantity of energy is released. 15. The net force between two nucleons 1 fm apart is F1 if both are protons, F2 if both are neutrons, and F3 if one is a neutron and the other is a proton. (A) F1 < F2 < F3 (B) F2 < F1 < F3 (C) F1 < F2 = F3 (D) F1 = F2 < F3 Solution: (C) The nuclear force of interaction between any pair of nucleons is identical i.e. force between two neutrons (F2) equals the force between neutron and proton (F3). However, between two protons net force is equal to the resultant of nuclear force between them (attractive in nature) and electrostatic force between them (repulsive in nature). Hence F1 is a value lesser than F2 and F3. So F1 < F2 = F3. 16. Which of the following process represents a -decay? (A) (B) (C) (D) Solution: (C) In gamma decay neither the proton number (Z) nor the neutron number (A-Z) changes. Only the quantum states of the nucleons change. 17. The activity of a sample of radioactive material is R1 at time t1 and R2 at time t2 (t2 > t1). If mean life of the radioactive sample is T, then: (A) R1t1 = R2t2 (B) (C) (D) Solution: (C) If R0 be the initial activity of the sample, then and Where 18. When 15P30 decays to become 14Si30, the particle released is (A) electron (B) -particle (C) neutron (D) positron Sol: (D) 19. Among electron, proton, neutron and -particle the maximum penetration capacity is for (A) electron (B) proton (C) neutron (D) -particle Sol: (C) 20. Plutonium decays with a half-life of 24000 years. If plutonium is stored for 72000 years, the fraction of it that remains is (A) (B) (C) (D) Sol (D) Number of half-life periods  The fraction that remains 21. The decay constant () and the half-life (T) of a radioactive isotope are related as: (A) (B) (C) (D) Sol (D) Half-life 22. The decay constant of a radioactive sample is . The half-time and the average-life of the sample will be respectively (A) (B) (C) (D) Sol (B) Because ln 2 = 0.301  2.303 = 0.693 23. In the uranium radioactive series, the initial nucleus is 92U238 and the final nucleus is 82Pb206. When the uranium nucleus decays to lead, the number of -particle emitted will be (A) 1 (B) 2 (C) 4 (D) 8 Sol (D) 24. The activity of a radioactive sample is 1.6 curie, and its half-life is 2.5 days. Its activity after 10 days will be: (A) 0.8 curie (B) 0.4 curie (C) 0.1 curie (D) 0.16 curie Sol (C) A = 0.1 Curie. 25. In a mean life of a radioactive sample: (A) about 1/3 of substance disintegrates (B) about 2/3 of the substance disintegrates (C) about 90% of the substance disintegrates (D) almost all the substance disintegrates. Sol (B) Given 26. Concept of neutrino was given by (A) Sommerfield (B) Pauli (C) P-Fermy (D) J. Chadwick Sol (B) 27. Half-life of an element is 30 days. How much part will remain after 90 days (A) 1/4th part (B) 1/16 part (C) 1/8th part (D) 1/3rd part Sol (C) 28. In the radioactive decay number of emitted  and  particles emitted are (A) 3, 3 (B) 3, 1 (C) 5, 3 (D) 3, 5 Sol (B) 29. After five half lives what will be the fraction of initial substance remaining undecayed (A) (B) (C) (D) Sol (B) After 5 half-lives 30. If half-life of a substance is 3.8 days & its initial quantity is 10.38 gm, then quantity of substance reaming after 19 days will be: (A) 0.151 gm (B) 0.32 gm (C) 1.51 gm (D) 0.16 gm Sol (B) No. of half periods Quantity remaining after n half lives For Test 1. 1 atomic mass unit is equal to (A) 1/12 (mass of one C-atom) (B) 1/16 (mass of O2-molecule) (C) 1/14 (mass of N2-moleucle) (D) 1/25 (mass of F2-moleucle) Solution- (A) We know that the atomic mass unit is defined as th of the mass of one 6C12 atom. 2. For the construction of nuclear-bomb which of the following substances is taken ? (A) U-235 (B) Pu-239 (C) F-14 (D) both (A) and (B) Solution- (D) We know that uranium (U235) and plutonium (Pu) are used as fissionable substances in a nuclear bomb. They can be fissioned easily by neutrons. 3. Energy obtained when 1 mg mass is completely converted to energy is (A) 3×108 J (B) 3×1010 J (C) 9×1013 J (D) 9×1015 J Solution- (C) (E) = mc2 = (1 × 10–3) (3×108)2 = 9 × 1013 J 4. 1 MeV energy is (A) 1.6×10–19 J (B) 1.6×10–16 J (C) 1.6×10–13 J (D) 1.6×10–11 J Solution- (C) 1eV = 1.6 × 1019 J = 1.6 × 10–13 J 5. Nuclear fission and fusion can be explained on the basis of (A) conversion of energy principle (B) Einstein mass-energy equivalence relation (C) binding energy per nucleon variation with nucleon number (D) variation of mass with increasing atomic nucleus Solution- (B) We know that is nuclear fission and fusion, the mass is converted into the energy. Therefore nuclear fission and fusion can be explained on the basis of Einstein mass-energy equivalence relation. 6. If 1 gm hydrogen is converted into 0.993 gm of helium, in a thermonuclear reaction, the energy released in the reaction is (A) 63×107 J (B) 63×1010 J (C) 63×1014 J (D) 63×1020 J Solution- (B) = {(1×10–3) – (0.993×10–3)} × (3 × 108)2 = (7 × 10–6) × (9 × 1016) = 63 × 1010 J 7. In the given particles, which of the following is stable ? (A) electron (B) proton (C) positron (D) neutron Solution: (D) We know that neutrons do not have any charge. Therefore a neutron is more stable than proton, electron or position. 8. 1 a.m.u. is equal to (A) 1 gm (B) 4.8×10–10 esu (C) 6.023×1023 gm (D) 1.66×10–27 kg Solution: (D) We know that one atomic mass unit (a.m.u.) is defined as of the mass of the isotope of carbon atom (6C12) and is numerically equal to = 1.66 × 10–27 kg. 9. Hydrogen bomb is based on the phenomenon of (A) nuclear fission (B) nuclear fusion (C) radioactive decay (D) none of these Solution: (B) We know that in a hydrogen bomb, a uranium fission bomb is exploded, first, which produce high temperature of pressure. Due to this the heavy hydrogen nuclei come extremely close and fuse together, liberating huge energy. Thus the massive energy liberated is due to nuclear fusion. 10. When two deuterium nuclei fuse together in addition to tritium, we get a (A) neutron (B) proton (C) -particle (D) deuteron Solution: (B) 1H2 + 1H2 1H3 + 1H1 + Q 11. In nuclear reaction there is conservation of (A) energy only (B) mass, energy and momentum (C) mass only (D) momentum only Solution: (B) Mass, energy as well as momentum are conserved in nuclear reaction. 12. An atomic power reactor furnace can deliver 300 MW. The energy released due to fission of each of uranium atom U238 is 170 MeV. The number of uranium atoms fissioned per hour will be (A) 5×1015 (B) 10×1020 (C) 40×1021 (D) 30×1025 Solution: (C) No. of Atoms fissioned per sec No. of Atoms fissioned per hour 13. r1 and r2 are the radii of atomic nuclei of mass numbers 64 and 27 respectively. The ratio is (A) (B) (C) (D) 1 Solution: (A) 1H3 + 1P1 2He4 14. Two radioactive materials. X1 and X2 have decay constants 10 respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2. will be 1/e after a time (A) (B) (C) (D) Solution: (D) N1 = N0 N2 = N0 9 = 1, or t = 15. The wave length of the characteristic X-ray line emitted by a hydrogen like atom is 0.32 Å. The wave length of line emitted by the same element is: (A) 0.18Å (B) 0.48Å (C) 0.27Å (D) 0.38A Solution: (C) The wavelength of x-rays emitted for line and line are given by and 16. The ratio of the mean life of a radionuclide to its half-life is (A) 1 (B) 1.44 (C) 2 (D) 3.12 Sol (B) Mean-life (t)mean and half-life 17. The half-life of radioactive substance is 48 hr. How much time it will take to disintegrate to its 1/16th part. (A) 12 hr. (B) 16 hr. (C) 48 hr. (D) 192 hr. Sol (D) 18. Radioactive substance emits (A) -rays (B) -rays (C) -rays (D) all of these. Sol (D) We know that when one -particle is emitted, the mass (A) is reduced by 4 and atomic number (Z) is reduced by 2. Therefore for new nucleus, mass no. is (A – 4) and atomic number is (Z – 2). 19. A radioactive substance has a half-life of 1 year. The fraction of this material, that would remain after 5 years will be (A) (B) (C) (D) Sol (A) 20. Half-life of a substance is 20 minutes. What is the time between 33% decay and 67% decay? (A) 40 minutes (B) 20 minutes (C) 30 minutes (D) 25 minutes Sol (B) Let N0 be the number of nuclei at the beginning. Number of undecayed nuclei after 33% decay = 0.67N0 Number of undecayed nuclei after 67% decay = 0.33N0 Also And in one half-life the number of undecayed nuclei becomes half. 21. The half-life of a radioactive substance is 3.6 days. How much of 20 mg of that radioactive substance will remain after 40 days? (A) 2.68  103 mg (B) 4.31  10–2 mg (C) 6.20  10–3 mg (D) 9.76  10–3 mg Sol (D) 22. A substance reduces to 1/16th of its original mass in 2 hours. The half-life period of the substance will be: (A) 15 min (B) 30 min (C) 60 min (D) 120 min Sol (B) 23. Which of the following processes represents a gamma-decay? (A) (B) (C) (D) Sol (C) 24. The half-life of 215At is 100 s. The time taken for the radioactivity of a sample of 215At to decay to (1/16) th of its initial value is (A) 400 s (B) 6.3 s (C) 40 s (D) 300 s Sol (A) Number of half lives = 4 Total time = 25. A sample of a radioactive substance contains 2828 atoms. If its half-life is 2 days, how many atoms will be left intact in the sample after one day? (A) 1414 (B) 1000 (C) 2000 (D) 707 Sol (C) 26. If N0 is the original mass of the substance of half life period t1/2 = 5 years, then the amount of substance left after 15 years is (A) N0 / 8 (B) N0 / 16 (C) N0 / 2 (D) N0 / 4 Sol (A) 27. At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit (i) electrons (ii) protons (iii) He2+ (iv) neutrons The emission at the instant can be (A) i, ii, iii (B) i, ii, iii, iv (C) iv (D) ii, iii Sol (A) 28. A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is (A) 19 : 81 (B) 18 : 92 (C) 81 : 19 (D) 92 : 18 Sol (A) or or 29. A nucleus with Z = 92 emits the following in a sequence: , , –, –, , , ,  ; –, –, , +, +, . The Z of the resulting nucleus is (A) 76 (B) 78 (C) 82 (D) 74 Sol (B) 30. Which of the following cannot be emitted by radioactive substances during their decay? (A) protons (B) neutrinos (C) helium nuclei (D) electrons Sol (A)