Tiêu đề 13.NUCLEI - Explanations.pdf ✅

1 (b) ∆m = 1 − 0.993 = 0.007 gm ∴ E = (∆m)c 2 = (0.007 × 10−3)(3 × 108) 2 = 63 × 1010J 2 (d) Applying principle of energy conservation. Energy of proton=total BE of 2 α −energy of Li 7 = 8 × 7.06 × 7 × 5.6 = 56.48 − 39.2 = 17.28 MeV 3 (c) Energy released from 1 kg of uranium = 200 ×106×1.6×10−19× 6.023 ×1026 235 = 8.2 × 1013 J 4 (c) 1 λ = R ( 1 2 2 − 1 4 2 ) = 3R 16 ⇒ λ = 16 3R = 16 3 × 10−5 cm Frequency n = c λ = 3×1010 16 3 ×10−5 = 9 16 × 1015Hz 5 (d) n = 24 24×138.6 = 1 138.6 ; Now N N0 = ( 1 2 ) n = ( 1 2 ) 1/138.6 ⇒ N = 10,00000 ( 1 2 ) 1/138.6 = 995011 So number of disintegration = 1000000 − 995011 = 4989 = 5000 7 (a) The activity or decay rate R of radioactive substance is the number of decays per second. ∴ R = λN or R = λN0 ( 1 2 ) t/T1/2 or R = R0 ( 1 2 ) t/T1/2 where R0 = λN0 is the activity of radioactive substance at time t = 0. According to question, R R0 = 1 − 75 100 = 25% ∴ 25 100 = ( 1 2 ) t/T1/2 or ( 1 2 ) 2 = ( 1 2 ) t/T1/2 or t T1/2 = 2 ∴ t = 2T1/2 = 2 × 3.20 = 6.40 h or t ≈ 6.38 h 8 (d) In time t = T , N = N0 2 In another half-life,(ie, after 2 half-lives) N = 1 2 N0 2 = N0 4 = N0 ( 1 2 ) 2 After yet another half-life ,(ie, after 3 half-lives) N = 1 2 ( N0 4 ) = N0 8 = N0 ( 1 2 ) 3 and so on. Hence, after n half-lives N = N0 ( 1 2 ) n = N0 ( 1 2 ) t/T where t = n × T = total time of n half-lives. Here, n = t T = 19 3.8 = 5 ∴ The fraction left N N0 = ( 1 2 ) n = ( 1 2 ) 5 = 1 32 = 0.031 9 (b) A = A0 ( 1 2 ) t/T1/2 ⇒ 5 = A0 ( 1 2 ) 2×60 30 = A0 16 ⇒ A0 = 80s −1 10 (b) (Z=92)U (A=238) (8α,6β) → Z ′X A ′ So A ′ = A − 4nα = 238 − 4 × 8 = 206 and Z ′ = nβ − 2nσ + z = 6 − 2 × 8 + 92 = 82 11 (a) Excitation energy ∆E = E2 − E1 = 13.6 Z 2 [ 1 1 2 − 1 2 2 ] ⇒ 40.8 = 13.6 × 3 4 × Z 2 ⇒ Z = 2 Now required energy to remove the electron from ground state = +13.6Z 2 (1)2 = 13.6(Z) 2 = 54.4 eV 12 (d) Number of α −particles emitted = 238−222 4 = 4 This decreases atomic number to 90 − 4 × 2 = 82 Since atomic number of 83Y 222 is 83, this is possible of one β −particle is emitted 14 (a) (4n + 2) series starts from U 238 and it’s stable end product is Pb 206 16 (a) Rest energy of an electron = mec 2 Here me = 9.1 × 10−31kg and c = velocity of light ∴ Rest energy = 9.1 × 10−31 × (3 × 108) 2 joule = 9.1 × 10−31 × (3 × 108) 2 1.6 × 10−19 eV = 510 keV 17 (b) Recoil momentum = momentum of photon = h λ
= hR ( 1 n1 2 − 1 n2 2 ) = hR × 15 16 = 6.8 × 10−27N × s 18 (c) In nuclear fission, neutrons are released 21 (d) For first line in Lyman series λL1 = 4 3R ...(i) For first line in Balmer series λB1 = 36 5R ...(ii) From equations (i) and (ii) λB1 λL1 = 27 5 ⇒ λB1 = 27 5 λL1 ⇒ λB1 = 27 5 λ 23 (c) In hydrogen atom En = − Rhc n2 Also En ∝ m; where m is the mass of the electron. Here the electron has been replaced by a particle whose mass is double of an electron. Therefore, for this hypothetical atom energy in n th orbit will be given by En = − 2Rhc n2 The longest wavelength λmax (or minimum energy) photon will correspond to the transition of particle from n = 3 to n = 2 ⇒ hc λmax = E3 − E2 = 2Rhc ( 1 2 2 − 1 3 2 ) This gives λmax = 18 5R 24 (d) N N0 = ( 1 2 ) t/T ⇒ 1 16 = ( 1 2 ) t/48 ⇒ ( 1 2 ) 4 = ( 1 2 ) t/48 ⇒ t = 192 hours 25 (d) In practise, nuclear fusion is very difficult process. This is so when positively charged nuclei come very close for fusion, the force of electrical repulsion between them becomes very strong. For fusion against this force, they require very high energy. To impart, so much energy to them, very high temperature and very high pressure is required. 26 (b) N = N0 ( 1 2 ) n = 4 × 1 2 = 2 27 (b) Ionizing property depends upon the charge and mass 30 (c) 1 λ = RZ 2 ( 1 n1 2 − 1 n2 2 ) ⇒ λ ∝ 1 Z 2 λLi ++: λHe +: λH = 4: 9: 36 31 (a) Speed of electron in n th orbit (in CGS) vn = 2πZe 2 nh (k = 1) For first orbit of H1; n = 1 and Z = 1 So v = 2πe 2 h ⇒ v c = 2πe 2 hc 32 (a) Half-life of a radioactive element T = 0.693 λ or T ∝ 1 λ ∴ λA λB = TB TA 33 (b) E = −Rch ⇒ R = − E ch = 13.6 × 1.6 × 10−19 3 × 108 × 6.6 × 10−34 = 1.098 × 107per m 34 (a) ∵ E1 > E2 ∴ v1 > v2 i. e., photons of higher frequency will be emitted if transition takes place from n = 2 to 1 36 (d) Because radioactivity is a spontaneous phenomenon 37 (a) The proton is the most stable in the Baryon group 38 (a) Mass number decreases by 8 × 4 = 32 Atomic number decreases by 8 × 2 − 5 = 11 39 (d) Rate R = − dN dt = λN0e −λt = λN ⇒ R N = λ (constant) i. e., graph between R N and t, is a straight line parallel to the time axis 41 (c) A = A0e −λt = A0e −t/τ ; where τ = mean life So A1 = A0e −t1/T ⇒ A0 = A1 e −t1/T = A1e t1/T ∴ A2 = A0e −t/T = (A1e t1/T )e −t2/T ⇒ A2 = A1e (t1−t2)/T 42 (d) Half-life T/2 = T 1.44 = 100 1.44 s = 69.44 s = 69.44 60 ≈ 1.155 min 44 (b) The electron in a hydrogen atom, moves with constant acceleration, called centripetal acceleration, round the nucleus. Acceleration of electron a = v 2 r Given, v = 2.18 × 106m/s r = 0.528 Å = 0.528 × 10−10m n=6 n=5 n=4 n=3 n=2 n=1 E2 E1
∴ a = (2.18 × 106) 2 0.528 × 10−10 = 9 × 1022 m/s 2 45 (c) The number of counts left after time t N = N0 ( 1 2 ) t T1/2 ∴ 30 = 240 ( 1 2 ) 60 T1/2 or (∵ t = 1h = 60 min) 30 240 = ( 1 2 ) 60 T1/2 or ( 1 2 ) 3 = ( 1 3 ) 60 T1/2 Comparing the powers, we get ∴ 60 T1/2 = 3 T1/2 = 60 3 T1/2= 20 min 46 (a) N = N0 ( 1 2 ) n ⇒ N N0 = ( 1 2 ) n ⇒ 1 100 − ( 1 2 ) n ⇒ 2 n − 100 n comes out in between 6 and 7 47 (a) Moderator slows down neutrons 48 (b) v ∝ Z 2 ⇒ vH2 vHe = ( 1 2 ) 2 = 1 4 ⇒ vHe = 4vH2 = 4v0 49 (d) 3 2 kT = 7.7 × 10−14 J T = 2 ×7.7 ×10−14 3 × 1.38 ×10−23 = 3.7 × 109 K 50 (b) From conservation of momentum, two identical photons must travel in opposite directions with equal magnitude of momentum and energy hc λ from conservation of energy hc λ + hc λ = m0c 2 + m0c 2 ⇒ λ = h m0c 51 (b) Nuclear fusion takes place in stars which results in joining of nuclei accompanied by release of tremendous amount of energy 52 (d) In vector form of Coulomb’s law proves that the forces F12 and F21 are equal and opposite. or F21 = F12 Fpe = Fep F’pe = F’ep And Fpe + Fep = − F’ep + F’pe So option (d) is incorrect. 53 (c) ‘Rad’ is used to measure biological effect of radiation. 54 (a) (32 4He + 1−1e 0 ) result in decrease in mass number = 3 × 4 and Decrease in charge number = 3 × 2 + 1(−1) = 5 ∴ Isotope (X) has mass number = 236 − 12 = 224 and charge number = 88 − 5 = 83 55 (b) In 10 s, number of nuclei has been reduced to one fourth (25% to 6.25%) Therefore it’s half life is T1/2 = 5 s ∴ Mean life T = T1/2 0.693 = 5 0.693 = 7.21 s 56 (c) mv 2 a0 = 1 4πε0 e 2 a0 2 ⇒ v = e √4πε0a0m 58 (b) Because atom is hollow and whole mass of atom is concentrated in a small centre called nucleus 59 (a) From radioactive decay law. − dN dt ∝ N or − dN dt = λN Thus, R = − dN dt Or R = λN or R = λN0e −λt ... (i) Where R0 = λN0 is the activity of the radioactive materialat time t = 0. At time t1, R1 = R0e −λt ... (ii) At time t2, R2 = R0e −λt ... (iii) Dividing Eq. (ii) by (iii), we have R1 R2 = e −λt1 e −λt2 = e −λ(t1−t2) or R1 = R2e −λ(t1−t2) 60 (b) In positive beta decay a proton is transformed into a neutron and a positron is emitted. p + → n 0 + e + Number of neutrons initially was A − Z. Number of neutrons after decay (A − Z) − 3 × 2(due to alpha particles) + 2 × 1(due to positive beta decay). The number of protons will reduce by 8 [as 3×2 (due to alpha particles) + 2(due to positive beta decay)]. Hence, atomic number reduces by 8. So, the ratio number of neutrons to that of protons = A−Z−4 Z−8
62 (d) Because sound waves require medium to travel through and there is no medium (air) on moon’s surface 63 (b) Nuclear radius is proportional to A 1/3 R = R0A 1/3 ∴ R1 R2 = [ 7 56] 1/3 = 1 2 64 (c) Radiocarbon dating relies on a simple natural phenomenon. As the earth’s upper is bombarded by cosmic radiation, atmospheric nitrogen is broken down into an unstable isotope of carbon- carbon (C-14). The unstable isotope is brought to earth by atmospheric activity, such as storms, and becomes fixed in the biosphere. Because it reacts identically to C-12 and C-13, C-14 attached to complex organic molecules through photosynthesis in plants and becomes their molecular makeup. Animals eating those plants in turn absorb carbon−14 as well as stable isotopes. This process of ingesting C − 14 continues as long as the plant or animal remains alive. The C − 14 within an organism is continually decaying into stable carbon isotopes, but organism is absorbing more C − 14 during its life, the ratio of C − 14 to C − 12 remains about same as the ratio in the atmosphere. Where the organism dies, the ratio of C-14 within its carcass begins to gradually decrease. 65 (d) 3 – 1 transition has higher energy so it has higher frequency (v − E h ) 66 (c) A0 3 = A0 ( 1 2 ) 9/T1/2 A’ = A0 3 ( 1 2 ) 9/T1/2 ∴ A′ A0/3 = 1 3 or A’ = A0 9 67 (c) N = N0 ( 1 2 ) t T1/2 ⇒ N N0 = ( 1 2 ) 30 10 = 1 8 = 0.125 68 (b) Power = energy time = 300 × 106 watt = 3 × 108 J/s 170 MeV = 170 × 106 × 1.6 × 10−19 = 27.2 × 10−12 J Number of atoms fissioned per second = 3 ×108 27.2 ×10−12 = 3×1020 27.2 Number of atoms fissioned per hour = 3 ×1020× 3600 27.2 = 3×36 27.2 × 1022 = 4 × 1022 m 69 (c) Average life 1 λ = 1600 0.693 = 2308 ≈ 2319 years 70 (a) N0 32 = N0 ( 1 2 ) 60/T ⇒ 5 = 60 T ⇒ T = 12days 71 (c) By using v = Rc [ 1 n1 2 − 1 n2 2 ] ⇒ v = 107 × (3 × 108) [ 1 4 2 − 1 5 2 ] = 6.75 × 1013Hz 72 (b) Mass of H2 nucleus = mass of proton = 1 amu energy equivalent to 1 amu is 931 MeV so correct option is (b) 74 (a) As N N0 = ( 1 2 ) n ; where, Number of half lives, n = t T T is the half life period For X sample, 1 16 = ( 1 2 ) 8/TX or ( 1 2 ) 4 = ( 1 2 ) 8/TX ⇒ 4 = 8 TX ...(i) For Y sample, ( 1 256) = ( 1 2 ) 8/TY or ( 1 2 ) 8 = ( 1 2 ) 8/TY ⇒ 8 = 8 TY ...(ii) Divide (i) by (ii) we get 4 8 = 8 TX × TY 8 ⇒ 1 2 = TY TX or TX TY = 2 1 76 (a) Energy required to ionize helium atom = 24.6 eV 77 (a) v ∝ 1 λ ∝ Z 2 ⇒ λZ 2 = constant ⇒ λ = λ 4 Z 2 ⇒ Z = 2 79 (b) T = 2πr v ; r = radius of n th orbit = n 2h 2 πMZe 2 v = speed of e − in n th orbit = ze 2 2ε0nh ∴ T = 4ε0 2n 3h 3 mZ 2e 4 ⇒ T ∝ n 3 Z 2 81 (a) Remaining amount = 16 × ( 1 2 ) 32/2 = 16 × ( 1 2 ) 16 = ( 1 2 ) 12 < 1mg 82 (d) Law of conservation of momentum gives m1v1 = m2v2 ⇒ m1 m2 = v2 v1 But m = 4 3 πr 3ρ