Title 13. NUCLEI(H).pdf ✅

NEET REVISION 7. () : Explana on For Hydrogen, 4 atoms involve in one fusion number of fusions in of Hydrogen Similarly no. of fissions involved in of . 8. () : Explana on The sun radiates energy in all direc ons in a sphere. At a distance , the energy received per unit area per second is (given). Therefore, the energy released in area per sec‐ ond is joule. The energy released per day , where . Thus, The equivalent mass is 9. () : Explana on Number of atoms that decay in me Probability that a nucleus decays (probability) But Here, 10. () : Explana on Energy released per fission Total energy released per sec Energy per second So, power output 11. () : Explana on From the given graph, we can write As slope 12. () : Explana on Frac on le , Hence, Now, If is half-life period, Again, 13. () : Explana on 14. () : Explana on Half life, (showing undecayed amount) Total me taken is Thus, a er three half life substance's part will disintegrate. ⇒ 1 kg = N1 ⇒ N1 = = 125 N 500×N 4 N2 → 1 kg U236 ⇒ N2 = × N 1000 236 ⇒ = E1 E2 125×N×28 ×N×200 1000 236 = 125×28×236 1000×200 = 4.13 R 1.4 kJ 4πR2 = 1400 × 4πR2 = 1400 × 4πR2 × 86400J R = 1.5 × 10 11 m ΔE = 1400 × 4 × 3.14 × (1.5 × 10 11) 2 × 86400 Δm = ΔE/c 2 Δm = [ 1400 × 4 × 3.14× (1.5 × 10 11) 2 × 86400 ] 9 × 10 16 Δm = 3.8 × 10 14 kg t = N0 − N0e −λt = N0 (1 − e −λt) ∴ = Number of nuclei decayed total number of nuclei = N0(1−e−λt) N0 ∴ P = 1 − e −λt λ = ln 2 T ∴ P = 1 − e − ln 2( ) 1 T P = 1 − e − ln 2(20/365) = 1 − e −0.693×4 73 = 1 − e −0.038 = 1 − 0.963 = 0.037 = 3.7% = 250MeV =250 × 10 6 × 10 14 × 40 100 = 100 × 10 20 = 10 22eV /sec = 10 22 × 1.6 × 10 −19 J/sec = 1.6 × 10 3 J/sec = 1600 W loge R0 = 2 ⇒ R0 = e 2 = 7.5 R0 = λN0 loge R = −λt + loge R λ = = = = 10 −4/sec 1 10 × 10 3 1 10 4 N0 = = = 75, 000 R0 λ 7.5 10 −4 f = 1 − = 20 100 8 10 N = N0,t = 10 days 8 10 N = N −λt 0 N0 = N0e 8 −λt 10 ⇒ e λt = 10 8 ∴ λ = = loge (10/8) 10 2.303 log(10/8) 10 λ = 0.022 T1/2 T1/2 = = 0.693 λ 0.693 0.022 T1/2 = 31.5 days N = N0( ) n = N0( ) 31.5/30 1 2 1 2 ≃ 51.2% n = = 3 × 10 17 9.6 × 10 6 200 × 10 6 × 1.6 × 10 −19 t1/2 = T 1 → → → T 1 2 T 1 4 T 1 8 t7/8 = 3T 7/8 th
NEET REVISION 15. () : Explana on Here, days, , Thus, As or 16. () : Explana on Energy released Energy released per deuterium, Number of deuterons in Energy released per of deuterium fusion 17. () : Explana on Energy release due to fission Number of atoms fissioned per second, Number of atoms fissioned per hour 18. () : Explana on As a result of emission of par cle, the mass number decreases by 4 units and atomic number de‐ creases by 2 units. And by the emmission of 1 positron the atomic number decreases by 1 unit but mass number remains constant. Mass number of final nucleus Atomic number of final nucleus Number of neutrons Required ra o 19. () : Explana on A er one half-life period, the ac vity of Tri um be‐ comes 50%. A er 2 half-life period 25% A er 5 half-life period 3.12% 3% It is years years i.e., 67 years only. 20. () : Explana on Given: Mass number of day Mass number of day Ini al mass days Number of ini al moles of and are given by moles. moles. As moles No. of atoms le 21. () : Explana on (momentum of alpha par cle) (momentum of daughter nucleus (ini al momentum) For -par cle For daughter nucleus, 22. () : Explana on From the ques on, and So, T1/2 = 24.1 N0 = 100 N = (100 − 90) = 10 = = 10 N0 N 100 10 = 2 t/T1/2 ⇒ 10 = 2 N t/Tl/2 0 N log 10 = log 2 t/T1/2 ⇒ t = × 24.1 d = 80 d 1 0.3010 Δm = 2(2.015) − (3.017 + 1.009) = 0.004 amu ∵ = (0.004 × 931.5)MeV = 3.726MeV = = 1.863MeV 3.726 2 1 kg = = 3.01 × 10 6.02×10 26 26 2 ∵ kg = (3.01 × 10 26 × 1.863) = 5.6 × 10 2MeV = 9 × 10 13 J Power = Energy Time = 300 × 10 6 W = 3 × 10 8 J/s = 170MeV = 170 × 10 6 × 1.6 × 10 −19 = 27.2 × 10 −12 J = 3×10 8 27.2×10−12 3×10 20 27.2 = 3×10 20×3600 27.2 = 4 × 10 22 m 1 α− ∴ = A − 12 = Z − 8 ∴ = (A − 12) − (Z − 8) = A − Z − 4 ∴ = A−Z−4 Z−8 ≈ 5 × 12 +7 A = 16, T1/2 = 1 B = 32, T1/2 = 1/2 = 320g,t = 2 A B N0(A) = = 20 320 16 N0(B) = = 10 320 32 n = t T1/2 N(A) = N0(A)( ) n = 20( ) 2/1 = 5 1 2 1 2 N(B) = N0(B)( ) n = 10( ) 2/1/2 = 10( ) 4 1 2 1 2 1 2 = 0.625 N = NA + NB = 5 + 0.625 = 5.625 = 5.625 × 6.023 × 10 23 = 3.38 × 10 24 p1 +p2 = 0 ⇒ p1 = −p2 α = 6.7MeV p 2 1 2m E = p2 2 2U ⇒ = E 6.7 m M ⇒ = E 6.7 4 214 ⇒ E = 0.125MeV N = N0 100 T1/2 = T yr λ = 0.693 T ∴ t = log[ ] 2.303 λ N0 N = log 100 2.303 T 0.693 = 6.65 T
NEET REVISION 23. () : Explana on Energy released in each fussion is No of atoms in is No of fusion is Total energy released is Energy released by is So, 24. () : Explana on , We know that there are two protons and two neu‐ trons in one nucleus of helium. 25. () : Explana on Let be the ini al ra o of . Let it be in the fossil. As remains constant, must have reduced by during the period. No. of half lives required Age of fossil years 26. () : Explana on Let is the ini al mass of the A er me the mass 27. () : Explana on Energy released in al final 28. () : Explana on 29. () : Explana on or 30. () : Explana on To become , it requires me of two half lives i.e., Centuries 31. () : Explana on 32. () : Explana on Let is ini al ac vity of whole blood. In al ac vity micro curie Ac vity in of blood at me 4.93 Ac vity of whole blood at me Total volume should be 33. () : Explana on half-lives We know, where, remaining of nuclei, and number of half-lives. 34. () : Explana on The person looses the radioac ve material both by excre on and by disintegra on.The effec ve rate constant is Where is the new half life. The concentra on falls from to in = 26.7 MeV 2 kg = NA × = NA × m M 2000 1 = × 2000 NA 4 EH = × NA × 26.7 MeV 2000 4 2 kgU 235 EU = × NA × 200 2000 235 = ≈ 7.62 EH EU 235 × 6.675 200 m(p) = 1.0073u m(n) = 1.0087u m(He) = 4.0015u Eb = Δm × 931.5 MeV ⇒ Eb = (2mp + 2mn − mHe) × 931.5 MeV ⇒ Eb = (2.0146 + 2.0174 − 4.0015) ×931.5 MeV ⇒ Eb = 28.4 MeV A14/A12 14C, 12C A14/A12 1 16 A12 A14 1 16 = 4 = 4 × 5730 = 22920 λ = = = 1.209 × 10 −4 y 0.693 −1 t t/2 0.693 5730 m0 14 C. t m = m0e −λt 0.21 m0 = m0e −λt 0.21 = e −λt ⇒ λ = 1.3 × 10 4y = BE − BE = 2x − y Q = 4 (x2 − x1) I = I0e −λt I = I0( ) n 1 2 ∴ = = ⇒ n = 3 I I0 1 2 n 1 8 ∴ = ⇒ n = 1 I ′ I0 1 2 t = nT ∴ 18 = 3 T ⇒ T = 6 mm 1 4 th t = 2 (T1/2) = 2 × 58 ZXA −−→ Z+1Y A −→ Z−1KA−4 → Z−1KA−4 −1β 0 (α) γ R0 R0 = = 1 = 3.7 × 10 4dps r = 1 cm3 t = 5 hrs, = 296 6 dps = dps R = t = 5 hr, V = = R r R0e−λt r = 3.7×10 4×0.7927 4.93 = 5.94 liters As, T1/2 = 2 min ⇒ 10 min = 5 = ( ) n N N0 1 2 N = N0 = initial number of nuclei n = ⇒ N = = = = 2.5 g N0 2 n 80 2 5 80 32 λ = λ1 + λ2 ⇒ = + 1 t 1 t1 1 t2 t = = = 4.8 h t1t2 t1+t2 24×6 24+6 t 6μci 3μci 4.8 h