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Tiêu đề 9. P2C9 (Atomic Model & Nuclear Physics_With Solve_Ridoy 09.12.24.pdf
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Nội dung text 9. P2C9 (Atomic Model & Nuclear Physics_With Solve_Ridoy 09.12.24.pdf
cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Varsity Practice Sheet 1 MCQ weMZ mv‡j DU-G Avmv cÖkœvejx 1. nvB‡Wav‡Rb cigvYyi Avw` kw3 ̄Íi ni †_‡K P‚ovšÍ kw3 ̄Íi nf-G ̄ vbvšÍ‡ii d‡j wb¤œwjwLZ †KvbwUi †ÿ‡Î me©vwaK kw3 wbM©Z nq? [DU 23-24] ni = 80, nf = 2 ni = 1, nf = 2 ni = 3, nf = 90 ni = 2, nf = 1 DËi: ni = 2, nf = 1 e ̈vL ̈v: En = RH     1 n 2 1 – 1 n 2 2 En      1 n 2 1 – 1 n 2 2 D”P kw3 ̄Íi †_‡K wb¤œ kw3 ͇̄i †M‡j kw3 wbM©Z n‡e| 2. 27 13Al + 4 2He  30 14Si + (?) wbDK¬xq wewμqvwU‡Z Abycw ̄ Z KYvwU n‡jvÑ [DU 20-21] Avjdv KYv †cÖvUb B‡jKUab wbDUab DËi: †cÖvUb e ̈vL ̈v: 27 13Al + 4 2He  30 14Si + (27 + 4 – 30) (13 + 2 – 14)X X = 1 1H + 3. A ̈vjywgwbqvg, wnwjqvg Ges wmwjK‡bi cvigvYweK msL ̈v h_vμ‡g 13, 2 Ges 14 n‡j, 27Al + 4He  27Si + (?) wbDwK¬qvi wewμqv‡Z Abycw ̄ Z KYv †KvbwU? [DU 19-20] an  particle an electron a positron a proton DËi: Blank e ̈vL ̈v: awi, KYvwU A ZX  A = 27 + 4 – 27 = 4 Z = 13 + 2 – 14 = 1  KYvwU 4 1X 4. wb‡Pi †KvbwU f‡ii GKK bq? [DU 19-20] amu Nm–1 s 2 MeV MeV c 2 DËi: MeV 5. nvB‡Wav‡Rb cigvYyi cÖ_g †evi K‡ÿ B‡jKUa‡bi †gvU kw3 –13.6 eV| Z...Zxq †evi K‡ÿ †gvU kw3 KZ? [DU 18-19; JnU 10-11] –1.5 eV –3.4 eV –4.5 eV –40.8 eV DËi: –1.5 eV 6. GKwU wbDwK¬qvm GKwU wbDUab MÖnY K‡i, GKwU weUv KYv ( – ) wbtmiY K‡i I `yBwU Avjdv KYvq cwiYZ nq| Avw` wbDwK¬qv‡mi A Ges Z h_vμ‡g wQjÑ [DU 18-19] 6, 3 7, 2 7, 3 8, 4 DËi: 7, 3 e ̈vL ̈v: A ZX + 1 0 n  2 4 2He2+ +  – A = 4 + 4 – 1 = 7 Z = 2 + 2 – 1 = 3 7. wb‡Pi mgxKi‡Y U-235 Gi wdkvb wewμqv †`Lv‡bv n‡q‡Q| Lvwj e·wU‡Z wb‡Pi †Kvb msL ̈vwU n‡e? [DU 16-17] 235 92U + 1 0n   56Ba + 92 36Kr + 3 1 0n 141 142 143 144 DËi: 141 e ̈vL ̈v: 235 92U + 1 0 n  141 56Ba + 92 36Kr + 3 1 0 n 8. †Kv‡bv †ZRw ̄...q †g.‡ji ÿq aaæe‡Ki gvb 0.01 s–1 | Gi Aa©vqyÑ [DU 16-17] 0.693 s 6.93 s 69.3 s 693 s DËi: 69.3 s e ̈vL ̈v: T1/2 = 0.693 0.01 = 69.3 s 9. a†cv‡jvwbqvg 214Po (Z = 84) Gi -wewKi‡Yi gva ̈‡g cÖvß †g.j n‡”QÑ [DU 15-16] 214Po (Z = 84) 210Pb (Z = 82) 214At (Z = 85) 210Bi (Z = 83) DËi: 210Pb (Z = 82) e ̈vL ̈v: 214 84 Po  4 2He2+ + 210 82 Pb
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