Tiêu đề 9. P2C9 Atomic Model & Nuclear Physics_With Solve.pdf ✅

cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Mastery Practice Sheet 1 cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb Atomic Model & Nuclear Physics beg Aa ̈vq weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv wjwLZ cÖkœmg~n 1| GK LÐ cÖvPxb Kv‡V 14C I 12C Gi AbycvZ eZ©gvb Kv‡ji RxweZ Mv‡Qi Kv‡V IB Abycv‡Zi 1 12 Ask| 14C Gi Aa©vqy 5570 eQi| cÖvPxb MvQwUi eqm wbY©q Ki? [BUET 22-23] mgvavb: t = T1/2 ln2 ln     N0 N  t = 5570 ln2 ln (12)  t = 1.996 × 104 years (Ans.) 2| 235 92 U Gi GKwU wdkb wewμqvq 200 MeV kw3 Drcbœ nq| Giƒc 2 gm 235 92 U n‡Z wbM©Z kw3i cwigvY kWh- Gi wbY©q Ki| [BUET 20-21] mgvavb: w 235 = N 6.023 × 1023  2 235 = N 6.023 × 1023  N = 2 × 6.023 × 1023 235 GKwU wdk‡b kw3 Drcbœ nq 200 × 1.6 × 10–19 × 106 J  N wU U-235 n‡Z wbM©Z kw3i cwigvY = 200 × 1.6 × 10–19 × 106 × 2 × 6.023 × 1023 235 = 1.64 × 1011 J = 1.64 × 1011 3.6 × 106 kWh = 45564.06619 kWh (Ans.) 3| †ZRw ̄Œq AvB‡mv‡Uv‡ci 8.60 Ci cwigvY GKwU †WvR GKRb †ivMx‡K Bb‡RKk‡bi gva ̈‡g cÖ`vb Kiv nj| AvB‡mv‡UvcwUi Aa© Rxeb 3h| AvB‡mv‡UvcwUi KZ ̧‡jv Avw` wbDwK¬qvm Bb‡RKk‡bi gva ̈‡g cÖ`vb Kiv n‡qwQj? [BUET 19-20] mgvavb: dN dt = 8.60 × 10–6 × 3.7 × 1010 Bq = 318200 Bq dN dt  N = ln2 T1 2 N = 318200  N = 318200 × 3 × 60 × 60 ln2 = 4.9579 × 109 wU (Ans.) 4| 9.1 × 10–31 kg fi wewkó GKwU B‡jKUab hw` wbDwK¬qvm‡K †K›`a K‡i 0.53 × 10–10 m e ̈vmv‡a©i Kÿc‡_ Nyi‡Z _v‡K, Z‡e Zvi †K.wYK †eM †ei Ki| [cøv‡1⁄4i aaæeK = 6.63 × 10–34 Js] [BUET 18-19] mgvavb: †evi gZev` Abyhvqx, B‡jKUa‡bi †KŠwYK fi‡eM, L = nh 2  mr2 = nh 2   = 6.63 × 10–34 × 1 2 × 9.1 × 10–31 × (0.53 × 10–10) 2 = 4.16 × 1016 rads–1 (Ans.) [r = 0.53 × 10–10 m nj †evi e ̈vmva© Z_v nvB‡Wav‡Rb Gi 1g kw3 ͇̄ii e ̈vmva©] 5| GKwU cvigvYweK Pzwjø‡Z 235U wbDwK¬qvi wdkb cÖwμqvq 200 MeV kw3 Db¥y3 K‡i| H PzwjøwUi `ÿZv 10% Ges GwUi ÿgZv 1000 MW| PzwjøwU 10 eQi Pvjv‡Z KZUzKz BD‡iwbqvg jvM‡e? [1 eV = 1.602 × 10–19 J, Avogadro’s Constant = 6.023 × 1023 mol–1 ] [BUET 18-19] mgvavb: awi, w kg U-235 jvM‡e|  w 235 = N 6.023 × 1023  N = w × 6.023 × 1023 235 w = Pt = Pin × t = Pout  × t = 1000 × 106 0.10 × 10 × 365 × 24 × 60 × 60 = 3.1536 × 1018 J  w × 6.023 × 1023 235 × 200 × 1.6 × 10–19 × 106 = 3.1536 × 1018  w = 38.45 × 106 g = 38.45 × 103 kg  38.45 × 103 kg U-235 jvM‡e (Ans.) 6| †Kvb †ZRw ̄Œq c`v‡_©i Aa©vqy 1000 eQi| KZ eQi ci Dnvi †ZRw ̄ŒqZv ÿqcÖvß n‡q 1 10 th n‡e? H †ZRw ̄Œq c`v‡_©i Mo Avqy KZ n‡e? [BUET 17-18] mgvavb: N = N0e –t = N0e – ln2 T1 2 t  1 10 = e – ln2 1000 × t  t = 3321.928 years (Ans.) 1  = T1 2 ln2 = 1000 ln2 = 1442.69 years (Ans.) Atternative way:

cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Mastery Practice Sheet 3 dN dt = N  dN dt = ln2 T1 2 N  3.7 × 1010 = ln2 26.8 × 60 m × 6.023 × 1023 214  m = 3.0497 × 10–8 gram [†hûZz me SI GK‡K †m‡nZz mgq‡K I SI GK‡K wb‡Z n‡e] (Ans.) 16| GKLÛ †iW‡bi 60% ÿq n‡Z KZw`b jvM‡e? †iW‡bi Aa©vqy 4 w`b| [RUET 08-09] mgvavb: N = N0e –t = N0e – ln2 T1 2 t      1 – 60 400 = e – ln2 4 × t  t = 5.2877 days (Ans.) 17| Aa©vqy I Aeÿq aaæeK Gi cvi ̄úwiK m¤úK© wbY©q Ki| [CUET 13-14] mgvavb: N = N0e –t  N N0 = e–t  1 2 = e –  T1/2 – ln2 = –  T1 2  T1 2 = ln2  (Ans.) 18| †Kvb GKwU †ZRw ̄Œq e ̄‘i Aa©vqy 6.93 w`b| KZw`b c‡i wKQz cwigvY GB †ZRw ̄Œ‡qi gvÎv 1 10 th Aewkó _vK‡e? [CUET 09-10, 05-06, BUTex 06-07, 04-05] mgvavb: N = N0e –t = N0e – ln2 T1 2 t  1 10 = e – ln2 6.93 × t  t = 23.026 days (Ans.) 19| hw` nvB‡Wav‡Rb Gi GKwU B‡jKUab Z...Zxq kw3 ͇̄i hvq, Zvn‡j wewKwiZ kw3i K¤úv1⁄4 KZ? [CUET 05-06] mgvavb: 1  = RH     1 n1 2 – 1 n2 2 × Z2  1  = 109678     1 2 2 – 1 3 2 × 1   = 6.56 × 10–5 cm = 6.56 × 10–7 m  f = c  = 3 × 108 6.57 × 10–7 = 4.57 × 1014 Hz (Ans.) 20| 1 gm GKwU †ZRw ̄Œq e ̄‘i cÖwZ †m‡K‡Û 3.7 × 1010 cigvYy ÿq nq| e ̄‘i cvigvYweK IRb 226| Bnvi Mo Avqy wbY©q Ki| [CUET 04-05] mgvavb: dN dt = N = 3.7 × 1010  1  × 1 × 6.023 × 1023 226 = 3.7 × 1010   = 7.2028 × 1010 s (Ans.) w M = N 6.023 × 1023  N = 1 × 6.023 × 1023 226 21| GKwU e ̄‘‡Z hw` cÖviw¤¢K Ae ̄’vq 109 msL ̈K Au198 Gi cigvYy _v‡K Z‡e KZ mg‡q Zvi 3 × 108 msL ̈K cigvYy †f‡O hv‡e? [Au198 Gi Aa©vqy 2.70 d] [CUET 03-04] mgvavb: N = N0e –t  109 – 3 × 108 = 109 × e – ln2 2.7 t  t = 1.3893 days (Ans.)  = ln2 T1 2 22| GKRb †ivMxi †ivM wbY©‡qi Rb ̈ Zvi kix‡i 12 gm cwigvY I-131 AvB‡mv‡Uvc cÖ‡ek Kiv‡bv nq hvi Aa©vqy 8d| 24 N›Uv ci Zvi kix‡i Kx cwigvY Av‡qvwWb Dcw ̄’Z _vK‡e? [BUTex 22-23] mgvavb: W = W0e –t  W = W0e – ln2 T1/2 t = 12 × e– ln2 8 × 1  W = 11 gm (Ans.) 23| GKwU wnwjqvg (2He4 ) wbDwK¬qv‡mi KYv cÖwZ eÜbkw3 wbY©q Ki| [GKwU †cÖvU‡bi fi = 1.00728 amu, GKwU wbDUa‡bi fi = 1.00876 amu, wnwjqvg wbDwK¬qv‡mi cÖK...Z fi = 4.00276 amu Ges 1 amu = 931 MeV]| [BUTex 18-19] mgvavb: m = (2 × 1.00728 + 2 × 1.00876 – 4.00276) amu = 0.02932 amu ⸪amu  931 MeV  0.02932 amu = 27.297 MeV wbDwK¬qv‡mi KYv cÖwZ eÜb kw3 = 27.297 4 = 6.824 MeV/nucleon (Ans.) 24| (K) †iwWqv‡gi Mo Avqy 2294 eQi| Aa©vqy KZ? (L) †h mg ̄Í †g.wjK Dcv`v‡bi fimsL ̈v mgvb Zv‡`i wK ejv nq? (M) †Kvb †dvU‡bi Zi1⁄2 •`N© ̈ 4 × 10–7 m| Gi •iwLK fi‡eM KZ? (N) cvigvYweK †evgv •Zwi nq †Kvb c×wZ‡Z? [BUTex 09-10] mgvavb: (K) T1 2 = ln2  = ln2 ×   T1 2 = (ln2 × 2294) years = 1589.74 years (Ans.) (L) AvB‡mvevi (Ans.) (M) P = h  = 6.626 × 10–34 4 × 10–7 kgms–1 = 1.6565 × 10–27 kg ms–1 (Ans.)