Tiêu đề 13. NUCLEI.pdf ✅

13. NUCLEI NEET PREPARATION (MEDIUM PHYSICS PAPER) Date: March 12, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on Nuclear forces are charge independent and short ranged ( few fermi). So correct op on is (2) 2. () : Explana on The radius of a nucleus is where is a constant and be its mass number. As (given) or The atomic number of nucleus is 3. () : Explana on The energy released during the given reac on will be where mass defect 4. () : Explana on Given, half- me We know that So, me A er 5 half- me period ac vity reduces to 3.125% of ini al ac vity. Now, given We know that Taking on both sides, we get Also, we know that 5. () : Explana on Electrons are significantly lighter than par cles. Therefore, electrons cannot sca er par cles at large angles, in accordance with the law of conser‐ va on of momentum. In contrast, the mass of the atomic nucleus is comparable to that of the -par‐ cle, which is why the nucleus alone is responsible for the sca ering of -par cles. Reason is separate fact. So correct op on is (2). 6. () : Explana on Nuclear radius , ≈ R = R0A1/3 R0 A ∴ = ( ) 1/3 R RHe A 4 = (14) R 1/3 RHe ∴ (14) 1/3 = ( ) 1/3 ; 14 = A 4 A 4 A = 56 Z = A − N = 56 − 30 = 26 ΔQ = Δmc 2 Δm = = (mA − mB − mD) × 931.5 MeV = (238.05079 − 234.04363 − 4.00260) × 931.5 = 4.25 MeV T1/2 = T A = 3.125% of A0 ∴ = = A A0 3.125 100 1 32 = ( ) n ⇒ = ( ) n A A0 1 2 1 32 1 2 ⇒ ( ) 5 = ( ) n ⇒ n = 5 1 2 1 2 t = n × T1/2 = 5 T A = 1% of A0 ∴ = A A0 1 100 = e −λt ⇒ = e A −λt A0 1 100 loge1 − loge100 = −λtlogee ⇒ t = 4.6006 λ λ = = 0.693 T1/2 0.693 T t = = 6.66 T 4.606 0.693 α α α α r = r0(A) 1/3 ∴ = ( ) 1/3 = ( ) 1/3 = ( ) 1/3 = r1 r2 A1 A2 64 125 4 3 5 3 4 5
7. () : Explana on The mass defect of The binding energy per nucleon 8. () : Explana on 9. () : Explana on The neutron has no net charge and mass is slightly larger than proton. Protons and neutrons cons ‐ tute the nuclei of atoms. Outside the nucleus the neutron is unstable. The life of neutron outside the nucleus is about 932 seconds. ( minutes). A er this period, 10. () : Explana on Electron emission during decay is always ac‐ companied by an neutrino emission. Hence , statement (a) is incorrect While statements (b) and (c) are correct 11. () : Explana on In radioac vity, electrons are emi ed in the form of par cles. It is also called nega ve -de‐ cay, where an unstable nucleus emits an energe c electron and an an -neutrino and a neutron in the nucleus becomes a proton that remains in the product nucleus.In photoelectric effect, when light rays of frequency greater than or equal to thresh‐ old frequency incident on metal surface, then elec‐ trons are emi ed from metal surface. This phenom‐ enon is called photoelectric effect. For photoelec‐ tric emission it is not necessary that element is un‐ stable but for radioac vity, it is necessary that ele‐ ments are unstable. 12. () : Explana on Applying charge and mass conserva on, Hence, is neutron. 13. () : Explana on 14. () : Explana on 15. () : Explana on 16. () : Explana on Number of atoms in fuel Number of atoms fissioned in one second Each fission gives . Hence, energy ob‐ tained in one second, Megawa 17. () : Explana on Nucleus does not contain electron. So the only pos‐ sible op on is (4) 18. () : Explana on As, 19. () : Explana on At At is the sample le a er one half-life 20. () : Explana on Number of atoms undecayed number of atoms decayed Frac on of atoms that got decayed in me is i.e., frac on will rise up to 1, following exponen al path as shown for curve ( ). 4 2He, Δm = 0.03u = MeV Δm×931 4 = MeV = 6.9825MeV 0.03×931 4 Δmc 2 = J(MSΔt) Δm = ms(Δt) C2 = 1×120×4.2×1000 9×10 16 = 56 × 10 −13 kg = 5.6 × 10 −12 kg. ≃ 15.5 n → p + e− + v ̄ β − β (−1e 0) β 4 + A = A + 3 + A′ ⇒ A′ = 1 2 + Z = Z + 2 + Z ′ ⇒ Z ′ = 0 M E = mc 2 = 1 × 10 −3 × (3 × 10 8) 2J = eV 1×10 −3×3×10 8×3×10 8 1.6×10 −19 (∵ 1 J = eV ) 1 1.6×10 −19 E = 56.169 × 10 31eV E = 5.6 × 10 26MeV E = Δm × c 2 ΔE = eV 9.1 × 10 −31 × 9 × 10 16 1.6 × 10 −19 E = 0.511MeV = c 2 = 3.6 × 10 dE 26 dt dm dt ⇒ = = 4 × 10 9 kg s dm −1 dt 3.6×10 26 9×10 16 2kg × 6.02 × 10 26 = 5.12 × 10 2 2 235 = 5.12×10 24 30×24×60×60 = 1.975 × 10 18s −1 185 MeV P = 185 × 1.975 × 10 18MeV s−1 = 185 × 1.975 × 10 18 × 1.6 × 10 −19js−1 = 58.46 Q = ΣBr − ΣBp Here , ΣBr = 3Eb , ΣBp = Ea, Q = e ∴ e = 3Eb − Ea ⇒ Ea + e = 3Eb t = 0, N = No t = 3.47, N = No/2 No/2 ∴ t1/2 = 3.47 min ⇒ = 3.47 min ⇒ λ = 0.2 min 0.693 −1 λ ⇒ At t = 60 min ∴ N = Noe−λt = Noe−0.2×60 = No e 12 N = Noe−λt = No − N = No(1 − e−λt) ⇒ t f = = 1 − e No−N −λt No B
21. () : Explana on Pair produc on refers to the crea on of an elemen‐ tary par cle and its an -par cle usually from a photon (or another neutral boson). This is allowed, provided there is enough energy available to create the pair. 22. () : Explana on Slow neutron bombardment of ini ates a chain reac on, yielding products like , and addi onal neutrons. The process generates more neutrons, causing subsequent uranium nuclei to undergo fission, crea ng a self - sustaining chain reac on. So correct op on is (1). 23. () : Explana on 24. () : Explana on 25. () : Explana on Momentum of a photon Hence, recoil energy of the nucleus is ( is the mass of the nucleus) 26. () : Explana on Let is the ini al concentra on of . The concentra on of at me is 27. () : Explana on Binding energy per nucleon increases with atomic number and is maximum for iron. A er that it de‐ creases. 28. () : Explana on Change in atomic mass Number of - par cles emi ed Atomic number will decrease by Atomic number will be increase by 4 due to emis‐ sion par cles. So, net decrease in atomic number 29. () : Explana on By adding the given two reac ons and have same atomic numbers but different neutron numbers. The given and are isotopes 30. () : Explana on reac on is fusion and fission. 31. () : Explana on Within heavy nuclei, the repulsive forces among the protons exert a greater influence compared to the nuclear a rac ve forces. To adjust , radioac‐ vity takes place. So correct op on is (2). 32. () : Explana on is the ini al number of radioac ve nuclei and is the number of nuclei that remains undecayed. Thus, where is the number of half lives As 33. () : Explana on The mother and daughter elements a er the emis‐ sion of -rays are called isomers having same atomic number and mass number. Isotopes have same atomic number but different atomic mass and isobars have same mass number but different atomic number. Isodiaphers have same difference in proton and neutron numbers. 34. () : Explana on The binding energy per nucleon of a deuterium . Total binding energy . The binding energy per nucleon of a helium nuclei . Total binding energy . Hence, energy released. 235U 141Ba, 92Kr 92U 238→90Th 234+2He 4− p = hv c E = p 2 2M M ∴ E = ⇒ E = ( ) 2 hv c 2M h 2v 2 2Mc 2 N0 X X t NX = N0e−λt NY = N0 − N0e−λt = = NX NY N0e−λt N0−N0e−λt 1 15 ⇒ 1 = 16e−λt ⇒ t = = 200yr ln 16 λ 90Th 232→82Pb 208 = 232 − 208 = 24 α = 24/4 = 6 6 × 2 = 12 4β = 12 − 4 = 8 A → C+2He 4 + 2 0 e −1 A C A C I st IV th n p N0 N = ( ) n N N0 1 2 n n = = = t T1/2 T1/2 1 2 T1/2 1 2 ∴ = ( ) 1/2 = N N0 1 2 1 √2 γ 1H2 + 1H2 → 2He 4 + ΔE = 1.1 MeV ∴ = 4 × 1.1 = 4.4 MeV = 7 MeV ∴ = 4 × 7 = 28 MeV ΔE = (28 − 4.4) MeV = 23.6 MeV .
NEET PREPARATION (MEDIUM PHYSICS PAPER) NEET PREPARATION 35. () : Explana on Binding energy, 36. () : Explana on In -decay, the atomic number (the number of pro‐ tons) of the daughter nucleus decreases by 2 units, not by 1 unit. So, the asser on is not correct. An -par cle consists of two protons and two neu‐ trons, which gives it a mass of four atomic mass units (amu). This is a valid and accurate statement. So op on (4) is correct. 37. () : Explana on This is decay (In this, remains same while decreases by 1) 38. () : Explana on Radioac ve emissions involve alpha-rays , which are helium nuclei with a posi ve charge, beta-rays , comprising high-speed electrons with a nega ve charge, and gammarays , which are neutral electromagne c radia ons. So correct op on is (3). 39. () : Explana on Since nuclear force is a short range force , for a separa on of , nuclear force between protons will be negligible 40. () : Explana on Energy liberated As four nucleons are involved in the reac on the energy liberated per each nucleon (1 ) is 41. () : Explana on As 42. () : Explana on Fission rate 43. () : Explana on Given reac on Energy released of products of reac‐ tants. 44. () : Explana on Cobalt-60 is indeed used in radiotherapy for cancer treatment. Cobalt-60 is a radioac ve isotope that emits -radia ons, which are used to target and kill cancer cells during radiotherapy. Both Asser on and Reason are correct, and Reason is the correct explana on of Asser on. So op on (1) is correct. 45. () : Explana on The total mass of the ini al par cles and the total mass of the ini al and final mass of par ‐ cles. The - value is given by . B. E. = (8mp + 9mn − M0)c 2 α α 22 11Na → X + e+ + v β +− A Z 22 11Na →22 10 Ne + e+ + v (α) (β) (γ) (= 10 −15m) 10 −8m ∴ Fe >> Fn E = Δmc 2 = 0.028664 u × 931 = 26.7MeV MeV u u = 6.675MeV 26.7MeV 4 P = n ( ) E t ⇒ 10kWh = n [ ] 200MeV t ⇒ 10 × 10 3 × 60 × 60 = n × 200 × 10 6 × 1.6 × 1 (∵ t = 1s) ∴ n = = 11.25 × 10 10×10 17 3×60×60 200×10 6×1.6×10 −19 = 200MeV Energy Fission = 200 × 10 6 × 1.6 × 10 −19 J = 5 200×10 6×106×10 −19 = 1.56 × 10 11 fission/sec 3B → A = BE −BE ⇒ Q = Ea − 3Eb γ mi = 1.007825 + 7.016004 = 8.023829u mf = 2 × 4.002603 = 8.005206 Δm = mi − mf = 8.023829 − 8.005206 = 0.018623u Q Q = (Δm)c 2 = 0.018623 × 931.5 = 17.35 MeV