UNIT-VIII (ATOMS AND NUCLEI)

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1 UNIT-VIII (ATOMS AND NUCLEI) GIST OF LESSON/FORMULAE ANDSHORTCUT FORMULAE 1. Rutherford s -Particle scattering experiment (Geiger Marsden experiment) IMPOTANT OBSERVATION Scattering of -particles by heavy nuclei is in accordance with coulomb s law. Rutherford observed that number of -particles scattered is given by N 2. Distance of closest approach : Estimation of size of nucleus = 3. Impact Parameter (b) b = 4. Bohr s atomic model constant Radius of orbit = Frequency v = v = X = Where = = is called fine structure 5. Energy of electron E n = ( ) E n = R = = X 10 7 m -1 and is called Rydberg constant. E n = ev = [ ] where is called wave number. Short Cut Formula K.E. = - ( Total Energy ) P.E. = - 2 K.E. 6. Spectral Series of Hydrogen Atom 280A

2 7. Energy level diagram for hydrogen atom We know that for hydrogen atom, energy of an electron in n th orbit is given by E n = ev 8. NOTE- Bohr s quantisation condition of angular momentum Let us consider the motion of an electron in a circular orbit of radius r around the nucleus of the atom. According to de-broglie hypothesis, this electron is also associated with wave character. Hence a circular orbit can be taken to be a stationary energy state only if it contains an integral number of de-broglie wavelengths i,e, we must have 2 = nλ 281A

3 But λ = 2 = n = n, This is famous Bohr s quantisation condition for angular momentum. 9. Atomic Mass Unit (u) One atomic mass unit is defined as th of the actual mass of c-12 atom. 1 u = X mass of C-12 atom = X X kg = 1.66 X kg. 10. Electron Volt (ev) It is the energy acquired by an electron when it is accelerated through a potential difference of 1 volt. 1 ev = 1.6 X J & 1 MeV = 1.6 X J 11. Relation Between amu & MeV We know, 1 u = 1.66 X kg 931 MeV 12. Nuclear Density ( ) = 2.3 X Kg/m 3 obviously, nuclear density is independent of mass number A. 13. Isotopes The atoms of an element, which have the same atomic number but different mass numbers are called isotopes. For examples Hydrogen has three isotopes, & 14. Isobars The atoms having the same mass number but different atomic numbers are called isobars. For examples &, &, & 15. Isotones The nuclids having the same number of neutrons are called isotones. For examples &, & 16. Isomers These are the nuclei having the same atomic number & same mass number but existing in different energy states. For example A nucleus in its ground state and the identical nucleus in metastable excited state, are isomers. 17. Properties of nuclear Forces (i) Nuclear forces are very short range attractive forces. (ii) Nuclear forces are charge independent. (iii) Nuclear forces are non-central forces. (iv) Nuclear forces do not obey inverse square law. 18. Nuclear force as a separation between two nucleons 19. Potential energy of a pair of nucleons as a separation between two nucleons 282A

4 20. Mass Defect ( = [Z m p + (A Z) M n ] M N 21. Packing fraction (P.F.) It is defined as the mass defect per nucleon. i,e, P.F. = Nucleus is stable if P.F.>1 & unstable if P.F.< Binding Energy (B.E.) The binding energy of a nucleus may be defined as the energy required to break up a nucleus in to its constituent protons and neutrons and to separate them to such a large distance that they may not interact with each other. It is equivalent energy of mass defect. i,e, B.E. = X c 2 B.E. = [{Z m p + (A Z) M n } M N ] x c Binding Energy per nucleon B.E. per nucleon = 24. Binding Energy Curve 283A

5 25. Imortance of binding energy curve This curve can be used to explain the phenomenon of nuclear fission & fusion. (i) Nuclear Fission There is overall gain in the binding energy per nucleon, when we move from heavy to medium range nuclei, hence release of energy. This indicates that energy can be released when a heavy nucleus is broken in to small fragments. This is called nuclear fission. (ii) Nuclear Fusion Similarly there is overall gain in the binding energy per nucleon, when we move from lighter to medium range nuclei, hence release of energy. This indicates that energy can be released when two or lighter nuclei fuse together to form a heavy nucleus. This is called nuclear fusion. 26. Radioactivity The phenomenon of spontaneous and continuous disintegration of the nucleus of an atom of a heavy element on its own with the emission of certain type of radiations is called radioactivity. 27. particles, -particles & rays When radiations emitted by a radioactive element is placed under electric field or magnetic field, radiations split in to three rays, classified as particles, -particles & rays. 284A

6 particles An 28. particle is the nucleus of helium or it is a doubly ionised He- atom. It is denoted by. Charge on particle = + 2e =3.2 X C. Mass of particle = X Kg. particles particles are fast moving electrons. It is denoted by or. Charge on particle = X C. Mass of particle = 9.1 X Kg. rays rays are electromagnetic waves of wavelength 0.01 A 0. Obviously these are not having any charge. 29. Explanation of process of -decay -decay is a process in which an unstable nucleus transforms itself in to a new nucleus by emitting an particle. For example- 92U Th He 4 + Q In general, ZX A Z-2Y A He 4 + Q Where Q is the energy released in -decay and is shared by daughter nucleus and particle. During the decay ratio increases. 30. Explanation of process of -decay The process of spontaneous emission of an electron (e - ) or a positron (e + ) from a nucleus is called decay. For example- 15P 32 16S e 0 + In general, ZX A Z+1Y A + -1 e 0 + or ZX A Z-1 Y A + +1 e 0 + During the decay ratio decreases. 31. Explanation of process of -decay The process of spontaneous emission of a -ray photon during the radioactive disintegration of a nucleus is called -decay. In general, ZX A Z X A Radioactive decay 285A

7 The spontaneous emission of radiation from a radioactive element is called radioactive decay. 33. Decay Law The number of nuclei disintegrating per second of a radioactive sample at any instant is directly proportional to the number of undecayed nuclei present in the sample at that instant. i,e, N = N Half Life Period (T) The time interval in which one half of the radioactive nuclei originally present in a radioactive sample disintegrate, is called half-life of that radioactive substance. S.I. unit of half life is second. 35. Relation between decay constant ( )& Half Life Period (T) = 36. Mean Life( ) = 37. Activity or Decay rate (R) R = R = N R = R Units of activity (1) Becquerel (Bq) One Becquerel is defined as the decay rate of one disintegration per second. i,e, 1 Bq = 1 decay/sec S.I. unit of activity is Becquerel. (2) Curie (Ci) One curie is the decay rate of 3.7 X disintegration per second. i,e, 1 Curie = 3.7 X decays/sec (3) Rutherford (Rd) One Rutherford is the decay rate of 10 6 disintegration per second. i,e, 1 Rutherford = 10 6 decays/sec CONCEPT MAP OR MIND MAP DIAGRAM OF NUCLEI N 286A

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9 IMPORTANT TOPICS/CONCEPTS FOR SLOW LEARNERS 1. E n = ev = [ ] where is called wave number. Short Cut Formula K.E. = - ( Total Energy ) P.E. = - 2 K.E. 2. Atomic Mass Unit (u) One atomic mass unit is defined as th of the actual mass of c-12 atom. 1 u = X mass of C-12 atom = X X kg = 1.66 X kg. 3. Electron Volt (ev) It is the energy acquired by an electron when it is accelerated through a potential difference of 1 volt. 1 ev = 1.6 X J & 1 MeV = 1.6 X J 4. Relation Between amu & MeV We know, 1 u = 1.66 X kg 931 MeV 5. Nuclear Density ( ) = 2.3 X Kg/m 3 obviously, nuclear density is independent of mass number A. 6. Isotopes The atoms of an element, which have the same atomic number but different mass numbers are called isotopes. For examples Hydrogen has three isotopes, & 7. Isobars The atoms having the same mass number but different atomic numbers are called isobars. For examples &, &, & 8. Isotones The nuclids having the same number of neutrons are called isotones. For examples &, & 9. Isomers These are the nuclei having the same atomic number & same mass number but existing in different energy states. For example A nucleus in its ground state and the identical nucleus in metastable excited state, are isomers. 10. Properties of nuclear Forces (i) Nuclear forces are very short range attractive forces. (ii) Nuclear forces are charge independent. (iii) Nuclear forces are non-central forces. (iv) Nuclear forces do not obey inverse square law. 11. Mass Defect ( = [Z m p + (A Z) M n ] M N 12. Packing fraction (P.F.) It is defined as the mass defect per nucleon. i,e, P.F. = Nucleus is stable if P.F.>1 & unstable if P.F.< 1 288A

10 13. Binding Energy (B.E.) The binding energy of a nucleus may be defined as the energy required to break up a nucleus in to its constituent protons and neutrons and to separate them to such a large distance that they may not interact with each other. It is equivalent energy of mass defect. i,e, B.E. = X c 2 B.E. = [{Z m p + (A Z) M n } M N ] x c Binding Energy per nucleon B.E. per nucleon = 15. Binding Energy Curve Importance of binding energy curve This curve can be used to explain the phenomenon of nuclear fission & fusion. (i) Nuclear Fission There is overall gain in the binding energy per nucleon, when we move from heavy to medium range nuclei, hence release of energy. This indicates that energy can be released when a heavy nucleus is broken in to small fragments. This is called nuclear fission. (ii) Nuclear Fusion Similarly there is overall gain in the binding energy per nucleon, when we move from lighter to medium range nuclei, hence release of energy. This indicates that energy 289A

11 can be released when two or lighter nuclei fuse together to form a heavy nucleus. This is called nuclear fusion. 16. Decay Law The number of nuclei disintegrating per second of a radioactive sample at any instant is directly proportional to the number of undecayed nuclei present in the sample at that instant. i,e, N = N Half Life Period (T) The time interval in which one half of the radioactive nuclei originally present in a radioactive sample disintegrate, is called half-life of that radioactive substance. S.I. unit of half life is second. 18. Relation between decay constant ( )& Half Life Period (T) = 19. Mean Life( ) = 20. Activity or Decay rate (R) R = R = N R = R 0 N QUESTIONS FROM IMPORTANT TOPICS AND DERIVATION (3 AND 5 MARKS) Qn1. Define Nuclear forces and gives their important characteristics/properties. 3M Ans.The nucleus of an atom has a number of protons and neutrons (nucleons) which are held together by the forces known as Nuclear forces in the tiny nucleus, inspite of strong force of repulsion between protons. Characteristics/Properties of nuclear forces: 1. Nuclear forces are strongest forces in nature. 2. Nuclear forces are short range forces. 3. Nuclear forces are basically strong attractive forces but contain a small component of repulsive forces. 4. Nuclear forces are saturated forces. 5. Nuclear forces are charge independent 6. Nuclear forces are spin- dependent 7. Nuclear forces are exchange forces Qn2.Define atomic mass unit (a.m.u.) and calculate its value in SI unit of mass. Also find energy equivalent in MeV corresponding to it. 3M Ans.Atomic mass unit is defined as th of mass of one atom. According to Avogadro s hypothesis number of atoms in 12 g of is equal to Avogadro number. i.e 6.023x A

12 Therefore the mass of one carbon atom = = 1.99 x10-23 g = 1.99 x10-26 kg Or, 1a.m.u. = x1.99 x10-26 =1.665x10-27 kg Energy equivalent of 1 a.m.u, m = 1 a.m.u = 1.665x10-27 kg E = m C 2 J = 1.665x10-27 x ( 3x10 8 ) 2 / 1.6 x MeV = MeV Qn3. Define binding energy per nucleon and packing fraction? Draw the curve showing the variation of binding energy per nucleon with mass number (A). Discuss its conclusions and explain how nuclear fission and fusion processes are explained with its help. 5M Ans.The binding energy per nucleon is the average energy required to extract one nucleon from the nucleus. Binding energy per nucleon = Greater is the binding energy per nucleon greater is the stability of nucleus. PACKING FRACTION The packing fraction of a nucleus is defined as the mass defect per nucleon of the nucleus. Packing fraction = = BINDING ENERGY CURVE It is found that binding energy of 3 Li 7 is greater than that of 2 He 4,but the value of its binding energy per nucleon is lesser. However, 2 He 4 is found to be more stable than 3 Li 7. Therefore, it may be concluded that the stability of nucleus depends upon binding energy per nucleon rather than the total binding energy of nucleus. Fig. shows the graph between the binding energy per nucleon and mass number of different nuclei. From the binding energy curves the following conclusions can be drawn: 1. The binding energy per nucleon for light nuclei, such as 1 H 2, is very small. 2. The binding energy per nucleon increases rapidly for nuclei upto mass number 20 and the curve possesses peaks corresponding to nuclei 2 He 4, 6 C 12 and 8 O 16. The peaks indicate that these nuclei are more stable than those in their neighbourhood. 3. After mass number 20, binding energy per nucleon increases gradually and for mass number between 40 and 120, the curve becomes more or less flat 291A

13 4. After mass number 120, binding energy nucleon starts decreasing and drops to 7.6MeV for uranium such nuclei are unstable and are found to disintegrate. 5. The binding energy per nucleon has a low value for both very high and very heavy nuclei in order to attain very higher value of binding energy per nucleon, the lighter nuclei may unite together to form a heavier nucleus (process of nuclear fusion ) or a heavier nucleus may split into lighter nuclei (process of nuclear fusion). In both the nuclear processes, the resulting nucleus acquires greater value of binding energy per nucleon along with the liberation of energy. Qn4. Define decay constant (λ) & half-life period ( t h ) and hence derive the relationship between decay constant and half-life period. 3M Ans. According to radioactive decay law N = N 0 If t = 1/λ N = N 0 = N 0 = =0.368N 0 Hence radioactive decay constant of a substance may also be defined as the reciprocal of time during which no. of radioactive atoms remains 36.8% ( or 1/e times) of its initial value. HALF LIFE TIME Time during which radioactive atoms remains half of its initial value called half life time i.e. N= at t h = N 0 = or = 2 = log e 2 = log 10 2 = x = = Qn5. Define average or mean life of the radioactive element and derive its relationship with decay constant and half life time. 3M Ans. Average or mean life of the radioactive element is the ratio of the total life time of all the atoms of the element to the total number of atoms initially present in the sample. Let at t= 0, no. of atoms are = No At time t no. of atoms = N Let dn atoms disintegrate in small time dt so life of dn atoms lies between (t+dt) As dt is very small then age of each dn atom can be taken as t Therefore total age of dn atoms = t.dn Total life time of all elements = Average life time = = 292A

14 As dn = -λndt = - λn 0 dt when N=N 0, t=0 and when N=0, t= and the limits = λ = = - 0 dt = dt] = Thus average life time is reciprocal of decay constant or, = t h / = 1.44 t h Qn6. Using Bohr s postulates derive the expression for the total energy of electron in stationary orbits of hydrogen atom. Hence show that the totalenergy in the stationery orbits are in the ratio 1: 1/4 : 1/9 What is the significance of negative value of energy? 5M Ans. From the first postulate of Bohr's atom model, (Ze.e/4πε 0 )/r 2 = mv 2 /r (1/2)mv 2 = (Ze.e/4πε 0 )/2r i.e. K.E of electron = (1/2)mv 2 = (Ze 2 /44πε 0 )/ 2 r...(i) Potential due to the nucleus, in the orbit in which electron is revolving V = (Z e/4πε 0 )/r Potential energy of electron = Potential x charge = (Ze/4πε 0 )/r x (-e) = -(Ze 2 /4πε 0 )/ r...(ii) Total energy of electron in the orbit, E = K.E. + P.E. E = (Ze 2 /4πε 0 )/2r + (-Ze 2 /4πε 0 )/r = -Ze 2 /4πε 0 )/2 r Using, r = ε 0 h 2 n 2 / π mze 2 in above equation, we get E n = -me 4.Z 2 /8 ε 0 2 h 2 n 2 Substituting all the standard values and converting in ev, we get En = -13.6Z 2 /n 2 (in ev) For hydrogen atom, Z = 1 En= -[13.6/n 2 ]ev Putting n= 1,2,3. E1:E2 ;E3: = 1:1/4:1/ The negative value of total energy shows that the electron is bound to the nucleus and is not free to leave it. 293A

15 Q. 7 A nucleus undergoes β decay and becomes. Calculate the maximum kinetic energy of electrons emitted assuming that the daughter nucleus and anti neutrino carry negligible kinetic energy. 3M Solution:β decay of is given as: Where,Q = Kinetic energy of the daughter nucleus Ignoring the rest mass of the anti-neutrino and the mass of the electron, the mass defect involved in the nuclear reaction is given as: Hence, when the energy carried by is zero, the maximum kinetic energy of β particle is MeV. Q.8 The energy level diagram of an element is given below. Identify, by doing necessary calculations for each, which transition corresponds to the emission of a spectral line of wavelength nm. 3M Solution: For A: Energy State (E 1 ) = 1.5 ev Energy State (E 2 ) = 0.85 ev Energy of the emitted photon is given as: E = E 2 E 1 = 0.85 ( 1.5) = = 0.65 Hence, the wavelength of the emitted photon is given by: 294A

16 For B: Ground state energy, E 1 = 3.4 ev Excited state energy, E 2 = 0.85 ev Energy of the emitted photon is given as: E = E 2 E 1 = 0.85 ( 3.4) = = 2.55 ev For C: Ground state energy, E 1 = 3.4 ev Excited state energy, E 2 = 1.5 ev Energy of the emitted photon is given as: E= E 2 E 1 = 1.5 ( 3.4) = = 1.9 ev For D: Ground state energy, E 1 = 13.6 ev Excited state energy, E 2 = 1.5 ev Energy of the emitted photon is given as: E = E 2 E 1 = 1.5 ( 13.6) = 12.1 ev Hence, element D corresponds to a spectral line of wavelength nm. Q.9 (i) Define activity of a radioactive material and write its S.I. units. (ii) Plot a graph showing variation of activity of a given radioactive sample with time. (iii) The sequence of stepwise decay of a radioactive nucleus is If the atomic number and mass number of D 2 are 71 and 176 respectively, what are their corresponding values of D? 5M Solution: 295A

17 (i) The activity of a radioactive material is defined as the decay rate of a sample containing one or more radio nuclides.the SI unit of radioactivity is Becquerel (B). (ii) (iii) So, the corresponding values of atomic number 10. What is radioactivity? State the law of radioactive decay. Show that the radioactive decay is exponential in nature. Ans. Radioactive decay The spontaneous emission of radiation from a radioactive element is called radioactive decay. Decay Law The number of nuclei disintegrating per second of a radioactive sample at any instant is directly proportional to the number of undecayed nuclei present in the sample at that instant. i,e, = λ N (1) Where λ is constant of proportionality & is called decay constant. From equation (1) we have = λ N = λ dt Integrating on both sides we get, N 296A

18 = λ (2) = λ t + C But, when t= 0, N= N 0, therefore from equation (2) we get = λ X 0 + C N 0 = C On putting this value of C in equation (2) we get = λ t + = λ t = λ t = N = N (3) This equation is known as decay equation. From eqn (3) we have N = N 0 Substituting t = in the above equation we get N = N 0 N = N 0 ( ) Thus decay constant of a radioactive element may be defined as the reciprocal of the time in which number of UN decayed nuclei of that radioactive element falls to times of its initial value. S.I. unit of decay constant is sec A

19 Previous 8 year AISSCE questions (Nuclei) 1 Mark Questions 1. Define the activity of a given radioactive substance. Write its SI unit. (All India 2013) 2. Write any two characteristics properties of nuclear force. (All India 2011) 3. Two nuclei have mass numbers in the ratio 1:8.What is the ratio of their nuclear radii? (All India 2009) 4. Two nuclei have mass numbers in the ratio 8:125.What is the ratio of their nuclear radii?(all India 2009) 5. Two nuclei have mass numbers in the ratio 27:125.What is the ratio of their nuclear radii?(all India 2009) 6. Assuming the nuclei to be spherical in shape, how does the surface area of a nucleus of mass no. A 1 compare with that of a nucleus of mass no. A 2? (All India 2008C) 7. Define the term activity of a radio nuclide. Write its SI unit. (All India 2007) 2 Mark Questions 8. In a given sample,2 radio isotopes, A and B are initially present in the ratio of 1:4.The halflives of A and B are 100 yr and 50 yr respectively. Find the time after which the amounts of A and B become equal. (Foreign 2012) 9. How the size of a nucleus is experimentally determined? Write the relation between the radius and mass no. of the nucleus. Show that the density of nucleus is independent of its mass no. (Delhi 2011C) 10.Why is it necessary to slow down the neutrons, produced through the fission of U nuclei(by neutrons) to sustain a chain reaction? What type of nuclei is (preferably) needed for slowing down fast neutrons? (All India 2008C) 3 Mark Questions 11. State the law of radioactive decay. Plot a graph showing the no. (N) of undecayed nuclei as a function of time (t) for a given radioactive sample having half life T 1/2.Depict in the plot, the no. of undecayed nuclei at: a) t = 3 T 1/2 b) t = 5 T 1/2 (Delhi 2011) 298A

20 12.. State the law of radioactive decay. Plot a graph showing the no. (N) of undecayed nuclei as a function of time (t) for a given radioactive sample having half life T 1/2.Depict in the plot, the no. of undecayed nuclei at: a) t = 2 T 1/2 b) t = 4 T 1/2 (Delhi 2011) 13. State the law of radioactive decay. Plot a graph showing the no. (N) of undecayed nuclei as a function of time (t) for a given radioactive sample having half life T 1/2.Depict in the plot, the no. of undecayed nuclei at: a) t = 3 T 1/2 b) t = 5 T 1/2 (Delhi 2011) 14. What is the basic mechanism for the emission of β - and β + particles in a nuclide? Give an example by writing explicitly a decay process for β - emission. Is a) the energy of the emitted β-particles continuous or discrete? b) the daughter nucleus obtained through β-decay, an isotope or an isobar of the parent nucleus?(delhi 2010C) 15. a) What is meant by half-life of a radioactive element? b)the half-life of a radioactive substance is 30 s.calculate i)the decay constant and ii) time taken for the sample to decay by 3/4 th of the initial value. (Foreign 2009) 16. a) What is meant by half-life of a radioactive element? b)the half-life of a radioactive substance is 20 s.calculate i)the decay constant and ii) time taken for the sample to decay by 7/8 th of the initial value. (Foreign 2009) 5 Mark Questions 17. Using Bohr s postulates, derive the expression for the frequency of radiation emitted when electron in hydrogen atom undergoes transition from higher energy state quantum no.(n i ) to the lower state(n f ). When electron in hydrogen atom jumps from energy state n i = 4 to n f = 3,2,1, identify the spectral series to which the emission lines belong. (All India 2007) 18. i) Derive the law of radioactive decay N = N 0 e -λt viz.,. ii)explain, giving necessary reactions, how energy is released during a) fission b) fusion. (All India 2011C) 299A

21 19.i) Draw the plot of binding energy per nucleon (BE/A) as a function of mass no. A. Write two important conclusions that can be drawn regarding the nature of nuclear force. ii) Use this graph to explain the release of energy in both the processes of nuclear fusion and fission. iii) Write the basic nuclear process of neutron undergoing β-decay. Why is the detection of neutrinos found very difficult? (All India 2013) 20. Define the Q-value of a nuclear process. When can a nuclear process not proceed spontaneously? If both the no. of protons and the no. of neutrons are conserved in a nuclear reaction, in what way is mass converted into energy (vice versa)in the nuclear reaction? (All India 2010C) 21. Draw the plot of binding energy per nucleon (BE/A)vs mass no.(a),for a large no. of nuclei lying between 2<A<240.Using this graph, explain clearly how the energy is released in both the processes of nuclear fusion and fission? (All India 2009C) 22. Why is the mass of a nucleus always less than the sum of the masses of its constituents, neutrons and protons? If the total no. of neutrons and protons in a nuclear reaction is conserved, then how is the energy absorbed or evolved in the reaction? Explain. (Delhi 2006) 23. Draw a plot of binding energy per nucleon (BE/A) mass no. for different nuclei. Explain with the help of this graph, the release of energy by the process of nuclear fusion. (Delhi 2006) 24. Sketch a graph showing the variation of potential energy of a pair of nucleons as a function of their separation. Write three characteristic properties of nuclear force which distinguish it from the electrostatic force. (All India 2006C) SOLUTIONS 1. The total decay rate of a sample is called the activity of the sample. The unit of activity is Becquerel (Bq). 1Bq = 1disintegration/s 2. i) These are short range force. ii) These are strong force of attractive nature. 300A

22 3. Nuclear radius, R=R o A 1/3 Where, R o =constant, A= mass no.... R 1 /R 2 = (A 1 /A 2 ) 1/3 = (1/8) 1/3 = ½ R1 : R2 = 1 : 2 4.Refer to ans. 3. (2 : 5) 5. Refer to ans. 3. (3 : 5) 6. Nuclear radius, R = R o A 1/3...surface area, S = 4πR 2 = 4π(R o A 1/3 ) 2 = 4πR o. A 2/3...Ratio of surface area, S 1 /S 2 = (A 1 /A 2 ) 2/3 7. The rate or activity of a sample is defined as the rate of disintegration taking place in the sample of radioactive substance. SI unit = Becquerel (Bq) 8. Let Na be the concentration of A after time ta and Nb be the concentration after time tb. From radioactive disintegration eqn, Na = N o e -λt A t A Nb = 4N o e -λt B t B Now,half-life of A is 10 yr and B is 50 yr. So, λ A = ln2/100 and λ B = ln2/50 Dividing, we get λ A / λ B = 1/2 or λ B = 2 λ A let after 10 yrs, Na = Nb so, Na/Nb = e -λat / e -λbt Na = Nb 4 e -λbt = e -λat 4 = e -(λa-λb)t ln 4 = -(λa-2λa)t ln 4 = λa t t = (ln 4/ln 2) *100 = 200 yr 9.The size of the nucleus is exp. Determined using Rutherford s α-scattering experiment and the distance of closed approach and impact parameter. The relation between radius and mass no. of nucleus is, R = R o A 1/3, where R o = 1.2 fm Nuclear density, ρ = Mass of nucleus / Volume of nucleus = ma / (4/3)π(R o A 1/3 ) 3 M= mass of each nucleon ρ = ma / (4/3)πR o 3 A 301A

23 = m / (4/3)πR o 3 it is clear that ρ does not depend on mass no. 10. Average kinetic energy of neutrons produced in nuclear fission of 92 U 235 is nearly 2MeV whereas the chances of absorption of neutrons average kinetic energy of nearly MeV is high by U nuclei. So. there is need to slow down the fast neutrons using appropriate substance namely moderator, into slow thermal neutrons. Nuclei have comparable mass to that of neutrons should be preferable be used to slow down fast neutrons.it is due to the fact that the elastic collision between fast neutrons and slow moving protons in paraffin lead to interchange the velocities. 11. Law of Radioactive Decay - The rate of disintegration of radioactive sample at any instant is directly proportional to the no. of undisintigrated nuclei present in the sample at that instant i.e. dn/dt N dn/dt = -λn N = no. of undisintigrated nuclei present in the sample at any instant t and dn/dt is rate of disintegration. 12. Refer to ans Refer to ans During β - decay from the nucleus,nuclei undergoes a change in such a way that atomic no. increases by 1 and mass no. remains same. In β + decay,the mass no. of present radioactive nuclei remains same whereas atomic no. decreases by 1. i)the energy of emitted β particle is continuous. ii)as there is no change mass no. during β-decay.so,the daughter nucleus is isobar of the parent nucleus. 302A

24 15.i) HALF LIFE Half life of radioactive element is the time taken by the sample to disintegrate upto half of its original amount. Half-life period, T 1/2 = 0.693/λ where, λ is decay constant. ii) T 1/2 = 30s (a) λ =?... T 1/2 = 0.693/ λ λ = 0.693/T 1/2 = 0.693/30 = s -1 (b)... N = N o [1/2] n where, n = no. of half lives N = no. of undisintegrated nuclei present in the sample. N o = original no. of undisintegrated atom. Here, N = N o 3/4 N o = 1/4N o N = N o [1/2] n N o /4 = N o [1/2] [1/2] 2 = [1/2] n n=2 but no. orf half-lives= total time taken / half life 2 = total time taken / (30s) Total time taken= 60s = 1 min. 16. refer to ans In the hydrogen atom, Radius of electron orbit, r = n 2 h 2 /4π 2 kme 2...(i) Kinetic energy of electron, E k = 1/2 mv 2 = ke 2 /2r E k = ke 2 4πkme 2 / n 2 h 2 = 2πk 2 me 4 / n 2 h 2...[From eq. (i)] potential energy, E p = -k(e) * (e) / r = -ke 2 / r E p = -ke 2 * 4π 2 kme 2 / n 2 h 2 = -4k 2 π me 4 /n 2 h 2 Total energy of electron. = E k + Ep E = 2π 2 k 2 me 4 / n h - 4π 2 k 2 me 4 /n 2 h 2 = -2π 2 k 2 me 4 /h 2 * [1/n 2 ] Now according t6o bohr s frequency condition when electron in hydrogen atom undergoes transition from higher energy state to lower energy state(n f ) is, hv = E nj - E nf or hv =[ -2π 2 k 2 me 4 /h 2 * 1/n 2 j ] [ -2π 2 k 2 me 4 /h 2 * 1/n 2 f ] or hv = -2π 2 k 2 me 4 /h 2 2 [1/n f - 1/n 2 j ] or v = 2π 2 k 2 me 4 2 /h3 [1/n f - 1/n 2 j ] or v = c2π 2 k 2 me 4 /ch 3 2 [1/n f - 1/n 2 j ] 2 k 2 me 4 /ch 3 = R = Rydberg contant = * 10 7 m -1 2 Thus, v Rc * [1/n f - 1/n 2 j ] Now, higher state n i = 4, lower state, n f = 3,2,1 For the transition, n j = 4 to n f = 3: --- Paschen series n j = 4 to n f = 2:--- Balmer series n j = 4 to n f = 1:--- Lyman series 303A

25 18. i) For radioactive decay law,... dn/dt = -λn where,λ = decay constant N = no. of undisintigrated nuclei in the sample of radioactive substance. or dn/n = -λdt integrating both sides, we get ʃ dn/n = -λ ʃ dt ln N = -λt + C At t=0,n = N o = no. of undisintigrated nuclei in the sample, initially ln N o = - λ * 0 + C C = ln N o ln N = -λt + ln N o ln N ln N o = -λt ln N/N o = -λt log e N/N o = -λt N/N o = e -λt or N = N o e -λt ii) a) Nuclear fission The phenomena of splitting of heavy nuclei(mass no. > 120) into smaller nuclei of nearly equal masses is lnown as nuclear fission. In nuclear fission,the sum of the masses of the product is less than the sum of masses of the reactants.this difference of mass gets converted into energy as per E = mc 2 and hence ample amount of energy is released in a nuclear fission.e.g. 92U o n 1 36Kr o n 1 + Q Masses of reactant = amu amu = amu Masses of products = = amu Mass defect = = amu... 1 amu = 931 MeV Energy release = * 931 = 200 MeV nearly Thus, energy is liberated in nuclear fission. b) Nuclear fusion The phenomena of conversion of two lighter nuclei into a single heavy nucleus are called nuclear fusion. Since, the mass of the heavier product nucleus is less than the sum of masses of reactant nuclear and therefore certain mass defect occurs which converts into energy as per Einstein s mass energy relation. Thus energy is released during nuclear fusion. e.g., 1H H 1 1H 2 + e + + v MeV also, 1H H 2 1H H MeV 304A

26 19.i) a) The force is attractive and strong enough to produce a binding energy of few MeV per nucleon. b) The constancy of the binding energy in the range 30<A<170 is a consequence of the fact that the nuclear force is short range force. ii) Nuclear fission A very heavy nucleus (A=240) has lower binding energy per nucleon as compared to the nucleus with A=120.Thus if the heavier nucleus breaks to the lighter nucleus with high binding energy per nucleon, nucleons are tightly bound. This implies that energy will be released in the process which justifies the energy release in fission reaction. Nuclear fusion When 2 light nuclei are combined to form a heavier nuclei, the binding energy of the fused heavier nuclei is more that the binding energy per nucleon of the lighter nuclei. Thus the final system is more tightly bound than the initial system. Again the energy will be released in fusion reaction. iii) In β-decay n p + e- + v- Neutrinos interact very weakly with matter so, they have a very high penetrating power. That s why the detection of neutrinos is found very difficult. 20. The Q- value of a nuclear process refers the energy release in the nuclear process which can be determined using Einstein s mass energy relation, E = mc 2.The Q- value is equal to the difference of mass of products multiplied by square of velocity of light. The nuclear process does not proceed spontaneously when Q- value of a process is ve or sum of masses of products in greater than sum of masses of reactant. Mass defect occurs in nucleus which converts into energy=mc 2 and produces binding energy.this energy binds nucleons together. 21.The curve reveals that BE/A is smaller for heavier nuclei than middle level nuclei. This shows that heavier nuclei are less stable.in nuclear fission, BE/A of reactants changes from nearly 7.6MeV to 8.4MeV.Higher value of binding energy of the nuclear product results in the liberation of energy during the phenomena of nuclear fission. In nuclear fission, BE/A of lighter nuclei into heavier one changes from low value of binding energy per nucleon to high value and release of energy takes place in fusion. 305A

27 22.The difference in mass of sum of masses of nucleons and mass of nucleus is known as mass defect which converts into energy as per E=mc 2 This energy is known as binding energy which is used to hold the nucleons together in spite of repulsive Colombian force between +ve charged protons. Otherwise nucleons will no longer be stable. The no. of neutrons and protons is conserved, but it does not mean that the parent nuclei and product nuclei are same.the difference in the mass of parent and product nuclei, converts into energy. 23.Refer to answer Nuclear force i) Strongest short range force which operate upto distance of 2-3 fm ii) it does not obey inverse square law iii) it exhibit charge independent character. Electrostatic force i) It is not very short range force necessarily. ii)it obey inverse square law. iii)it depend on the nature of charge,like charge repel whereas opposite charge attract each other Previous Eight Year s Examination Questions (Atoms) 1 Marks Questions 1. Why is the classical (Rutherford) model for an atom of electron orbiting around the nucleus-not able to explain the atomic structure? [Delhi 2012] 2. Find the ratio of photons produced due to transition of an electron of hydrogen atom by its [All India 2010] (i) Second permitted energy level to the first level and (ii) The highest permitted energy level to the first permitted level. 3. The radius of innermost electron orbit of a hydrogen atom is m. what is the radius of orbit in the second excited state? [Delhi 2010] 4. Two nuclei have mass number in the ratio 1:3. What is the ratio of their nuclear densities? [Delhi 2009] 5. State Bohr s postulate of quantization of angular momentum of the orbiting electron in hydrogen atom. [Delhi 2009C] 306A

28 6. Out of the two characteristics, the mass number (A) and the atomic number (Z) of a nucleus, which one does not change during nuclear β-decay? [All India 2008C] 2 Marks Questions 7. (i) In hydrogen atom, an electron undergoes transition from 2nd state of the first excited state and then the ground state. (ii) Find out the ratio of the wavelength of the emitted radiations in the two cases. [Delhi 2012] 8. Explain, in brief, why Rutherford s model cannot account for the stability of an atom. [Delhi 2010C] 9. Using the relevant Bohr s postulates derive the expression for the radius of the electron in nth orbit of the electron in hydrogen atom? [Delhi 2010C] 10. The ground state energy of hydrogen atom is ev. The photon emitted during the transition of electron from n=2 to n=1 state, is incident on the photosensitive material of unknown work function. The photoelectron are emitted from materials with a maximum kinetic energy of 8 ev. Calculate the threshold wavelength of the material used. [Foreign 2008] 3 Marks Questions 11. Using Bohr s postulates, obtain the expression for the total energy of the electron in the stationary state of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition between energy levels. [Delhi 2013] 12. Using Bohr s postulates for hydrogen atom,, show that total energy (E) of the electron in the stationary states can be expressed as the sum of kinetic energy (K) and potential energy (U), where K=-2U. Hence, deduce the expression for the total energy in the nth energy level of hydrogen atom. [Foreign 2012] 13. The ground state energy of hydrogen atom is ev. If an electron makes a transition from an energy level ev to ev, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong? [All India 2012] 14. In a Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of Z=80, when an α-particle of 8 MeV energy impinges on it before it comes to 307A

29 momentarily rest and reverses its direction. How will the distance of closest approach be affected when the kinetic energy of the α-particle is doubled? [All India 2012] 15. The ground state energy of hydrogen is ev. If an electron makes a transition from an energy level ev to -3.4 ev, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong? [All India 2012] 16. The electron in a given Bohr orbit has a total energy of -1.5 ev. Calculate (i) Its kinetic energy (ii) Potential energy and (iii) Wavelength of light emitted, when the electron makes a transition to the ground state. Ground state energy is ev. [Delhi 2000, 2011C] 17. Using the postulate of Bohr s theory of hydrogen atom. Show (i) Radii of orbits increase as n 2, and (ii) The total energy of electron increase as 1/n 2, where n is the principal quantum number of the atom. [All India 2011C] 18. Using the relevant Bohr s postulate, derive the expressions for the (i) Speed of the electron in the nth orbit, (ii) Radius of the nth orbit of the electron, in hydrogen atom. [Delhi 2010C] 19. Draw a schematic arrangement of the Gieger-Marsden experiment. How did the scattering of α-particle by a thin foil of gold provide an important way to determine an upper limit on the nucleus? Explain briefly. [All India 2009] 20. The energy of the electron, in the ground state of hydrogen, is ev. Calculate the energy of proton that would be emitted if the electron were to make a transition corresponding to the emission of the first line of the (i) Lyman series (ii) Balmer series of the hydrogen spectrum. [All India 2009C] 21. The ground stets energy of hydrogen atom is ev. (i) What is the kinetic energy of an electron in the 2nd excited state? (ii) If the electron jumps to the ground state from 2nd excited state, calculate the wavelength of the spectral line emitted. [All India 2008] 22. The ground state energy of hydrogen atom is ev. The photon emitted during the transition of electron from n=3 to n=1 state, is incident on a photo sensitive material of unknown work function. 308A

30 The photoelectrons are emitted from the materials with a maximum kinetic energy of 9 ev. Calculate the threshold wavelength of the material used. [Foreign 2008] 5 Marks Question 23.Using postulate of Bohr s theory of hydrogen atom, show that (a) the radius of orbit increase as n 2 and (b) the total energy of an electron increases as 1/n 2, where n is the principal quantum number of the atom. (ii) Calculate the wavelength of H α line inbalmer seriesof hydrogen atom. Given, Rydberg constant, R= m -1. [All India 2011C] Step-by Step Solution 1 Marks Questions Answers 1. The classical method could not explain the atomic structure as the electron revolving around the nucleus are accelerated and emits energy as the result, the radius of circular path goes on decreasing. Ultimately electrons fall into the nucleus, which is not in practical. (1) 2. (i) Since, the second permitted energy level to the first level = E 2 E 1 = Energy of photon released = (-3.4 ev) (-13.6 ev) = 10.2 ev (ii) The highest permitted energy level to the first permitted level Ratio of energies of photon =E - E 1 = - (13.6) = 13.6 ev = 10.2/13.6 = ¾ = 3:4 3. The radius of atom whose principal quantum number is n,is given by r = n 2 r 0 (1) where, r0 = radius of innermost electron orbit for hydrogen atom and r0 = m 309A

31 For second excited state, n=3 r = 3 2 r 0 = R = Nuclear density is independent of mass number. 5. For Bohr s postulate of quantization of angular momentum of the orbiting electron in hydrogen atom is electrons are permitted to revolve in only those orbits in which the angular momentum of electron is integral multiple of h/2π i.e. mvr = nh/2π, where, n=1,2,3. 6. The mass number, A of a nucleus does not change during nuclear β-decay. (1) (1) (1) (1) 2 Marks Questions Answer 7. (i) An electron undergoes transitional from 2 nd excited state to the first excited state is Balmer series and then to the ground state is Lyman series. (ii) The wave length of emitted radiation in two cases n= ev Balmer Series (1) Lyman Series n= ev λ For n 2 n 1 E= ( ) = ev λ = m = 1218 Å 8. The following are the drawbacks of Rutherford s model (i) Ruther suggested that on revolving in the orbits, electrons radiates energy and shrinks consequently toward the nucleus 2-e, the radius followed by the electrons, gradually decreases based on it, electron should fall into nucleus and atom should be destroyed. (1) 310A

32 (1) (ii) According to it we should obtained radiation of all possible wavelength but in actual practice atomic spectrum in line de destroyed. (1) 9. The electron revolves in a stable orbit, the centripetal force is provided by electrostatic force of attraction acting on, it due to positive charge in the nucleus. Hence, mv 2 n/r n =1/4π 0 e 2 /r 2 n and from Bohr s quantum condition, we have....(ii) =... (i) Squaring Eq. (ii) and then equating it with (i), we get. (1) 10. Energy of electron in nth orbit of hydrogen atom For n=1 For n=2 Energy of photon released = = -3.4 (-13.6) =10.2 ev Also, K.E. max = 8 ev According to Einstein Equation K.E. max = hv - Ǿ 8 ev = Ǿ Wave function (Ǿ) = 2.2 ev threshold wavelength λ = nm 3 Marks Questions Answer 11. According to Bohr s postulate, in a 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 311A

33 given radius, the centripetal force provide by coulomb force of attraction between the electron and the nucleus. The gravitational attraction may be neglected as the mass of electron and proton is very small. So, i Where, m= mass of electron r= radius of electronic orbit v= velocity of electron Again, By Bohrs 2 nd postulate Where, n=1,2,3.. Putting the value of v in Eq. i ( ) ii K.E. of electron, Using Eq. ii, we get, Potential energy Using Eq. ii Hence total energy of electron in nth orbit E = PE KE E = = 13.6/ ev 312A

34 When the electron in hydrogen atom jumps from higher energy level to low energy level, the difference of energies of the two energy levels is emitted as a radiation of particular wavelength. It is called spectrum line. In H-atom, when an electron jumps from the orbit ni to nf, the wavelength of the radiation given by, 1/π =R Where. R- Rydberg s constant = m -1 For, Balmer serious 1/π =R Where ni = 1,2,3,. 12. According to Bohr s Postulate for hydrogen atom, electron revolves in a circular orbit around the positive charge nucleus. These are the stationary (orbit) state of the atom. For a particular orbit, electron moves there, so it has K.E.Also, there potential energy due to charge on electron and heavy positive charge nucleus. Hence, total energy E of an atom is sum of K.E. (K) and P.E. (U). i.e. E = K + U Let us assume that the nucleus has positive charge, Ze. An electron moving with a constant speed v along a circle of radius, e with center at the nucleus. Force acting on nuclies is given by F=Ze 2 /4π r 2 Theacceleration of electron = v 2 /r (towards the center) If m =mass of an electron Then, from Newton s 2 nd law F = m (v 2 /r) Ze 2 /4π r 2 = m (v 2 /r) r = Ze 2 /4π (mv 2 ) (i) From Bohrs quantisation rule mvr = nh / 2π...(ii) Where, n is positive integer. Plug-in the value of r from E.q (i) V= Ze 2 / 2 hn.(iii) So, K.E. = K = ½ mv 2 = mz 2 e 4 / 8 2 h 2 n 2..(iv) Potential energy of the atom U = Z 2 e 2 / 4π r..(v) Using equation (iii) and (i) 313A

35 r = h 2 n 2 / πmze 2 Using the value of r, in equation (v) U = -Z 2 e 4 m / 4π 2 h 2 n 2 So, the total energy, E = K + U E = mz 2 e 4 m / 8 2 h 2 n Proton is emitted when electron transits from higher energy to lower energy state, the difference of energy of the state appear in form of energy of photon. According to bohr s theory of hydrogen atom, energy of photon released, E 2 E 1 = hv Given, E 1 = ev E 2 = ev E 2 E 1 = (-1.51) E = 0.66 ev So, the wave length of emitted spectral line, λ = 1242 ev nm / E (in ev) = m = m The wavelength belongs to Pascher series of hydrogen spectrum. 14. Z= 80, KE = K= 8MeV = J The energy conservation law, K= (Ze)(2e) / 4π r Where, r = distance of closest approach. r = 2Ze 2 / 4π ϵ r r = ( ) / r = m r = 1 /K where KE gets double, distance of closest approach reduces to half. 15. Basic assumptions of Rutherford atomic model are given below i) Atom consists of small central core, called atomic nucleus in which whole mass and +ve charge is assumed to be concentrated. ii)the size of the nucleus is much smaller than size of the atom. iii) The nucleus is surrounded by electrons. Atoms are electrically neutral as total ve charge of electrons surrounding the nucleus is equal to total +ve charge on the nucleus. 314A

36 iv) Electrons revolves around the nucleus in various circular orbits and necessary centripetal force of attraction between +vely nucleus and vely charged electrons. Stability of Atom When an electron revolves around the nucleus,then it radiates electromagnetic energy and hence,radius of orbit of electron decreases gradually.thus, electron revolve on spiral path of decreasing radius and finally, it should fall into nucleus,but this not happen. Thus, Rutherford atomic model cannot account for stability of atom. 16. (i) As total energy in given bohr orbit E n = -1.5 ev Kinetic energy of electron K n = -E n = ee n = -1.5 ev (ii) The potential energy of electron Un = 2En = -3.0 ev (iii) If electron makes a transition from -1.5 ev state to ground state having energy ev, the energy of radiated photon E = -1.5 (-13.6 ) ev = 12.1 ev = J As E = hv = hc / λ, hence wavelength of light emitted is equal to λ = hc / E = = mm 17.(i) As, radius of electron s nth orbit in hydrogen atom r n = r n (ii) Also, the total energy of an electron belonging to nth orbits, E n 1 / n 2 i.e., the total energy of electron increasing as 1 / n 2 18.(i) Speed of electron in nth orbit Centripetal force of revolution is provided by electrostatic force of attraction. mv 2 / r = ke 2 /r 2 where, m = mass of electron r = radius of orbit o electron = ke 2 / mv 2..1 Also from Bohr s postulates, mvr= nh / 2π r= nh / 2πmv A

37 Comparing equation 1 and 2, we get V = (2πke 2 / ch ) c/n Where, c is the velocity of light Or v = α c /n Also α = 1 / 137 v = 1 / 137 (c/n) for n=1, v = 1 / 137 c In k shell of hydrogen atom, electron revolves with 1/137 times of speed of light (ii) For radius of nth orbit of electron, 19.Let an electron revolves around the nucleus of hydrogen atom. The necessary centripetal force is provided by electrostatic force of atteraction. + Ze e -, m mv 2 / r = ke 2 / r 2 r= ke 2 / mv 2 where, m mass of electron and v is its speed of a circular path of radius, r By Bohr s 2 nd postulates, mvr= nh / 2π were, n = 1,2,3,. r= nh / 2πmv comparing both r ke 2 / mv 2 = nh / 2πmv v = 2πke 2 / nh substuting value r= n 2 h 2 / 4π 2 mke 2 r n 2 20.Given,the energyof the electron,in the ground state hydrogen id ev.... E 1 = ev For n = 2, E 2 = -3.4 ev 316A

38 For n = 3, E 3 = -1.5 ev [... En = -13.6/n 2 ev]... Energy of photon corresponding to the first line of the i) Lyman series, E = E 2 E 1 = (-3.4) (-13.6) = 10.2 ev ii) Balmer series, E = E 3 E 2 = (-1.5 ev) (-3.4) = 1.9 ev 21. Ground state energy energy, E i =-13.6 ev As, E n = E i /n 2 = 13.6 ev / n 2 (i) KE of an electron = -Total energy of electron In each second exicited state, n = 3 Total energy, E n = -1.5 ev KE of the electron in 2 nd exicited state = E 3 = 1.5 ev (ii) Transition of electron from n 2 to n 1 We have wavelength of emitted radiation 1/ λ = R 1 / λ= 8R / 9 λ = m λ = 1030 Å 22.Given, ground state energy E 1 = ev Energy of electron in nth orbit E n = ev / n 2 For n= 1 E 1 = ev For n = 3 E 3 = ev / 3 2 = -1.5 ev The energy of photon released during the transition of electron from n =3 to n=1 is 317A

39 E = E 3 E 1 = -(1.5) (-13.6) = 12.1 ev Now, Einstein s photoelectric equation Energy of photon (E) = KE max + Ǿ Where Ǿ is the work function of metal 12.1 ev = 9 ev + Ø Ø = = 3.1 ev So the wave length λ = 1242 ev nm / Ø = 1241 ev nm / 3.1 ev = 401 nm 5 Marks Questions 23.i) a) Refer ans. 19 b) Refer ans. 19 ii) For Balmer series, α-line,wavelength is given by 1/λ = R {1/2 2 1/n 2 } where, n = 3 [for α-line] 1/λ = R {1/2 2 1/3 2 } 1/λ = R {1/4 1/9} 1/λ = R {(9 4)/36} 1/λ = 5R/36 λ = 36/5R = 36/5 * * 10 7 λ = * 10-7 m λ = 6563 Ȧ HOT QUESTIONS WITH ANSWERS - 2& 3 MARKS ATOMS &NUCLEI Qn1. Show that the instantaneous rate of change activity of a radioactive substance is inversely proportional to the square of its time period.3m Ans. A = A 0 e -λt da/dt = A 0 e -λt (-λ) = -λa = -λ(λn) = - λ 2 N 318A