Tiêu đề 9. P2C9. পরমানুর মডেল ও নিউক্লিয়ার পদার্থবিজ্ঞান (With Solve).pdf ✅

cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Engineering Practice Sheet ...................................................................................... 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 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 92U Gi GKwU wdkb wewμqvq 200 MeV kw3 Drcbœ nq| Giƒc 2 gm 235 92U n‡Z wbM©Z kw3i cwigvY kWh 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 wU cigvYy GKwU wdk‡b kw3 Drcbœ nq 200 × 1.6 × 10–19 × 106 J  N wU wdkb 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 3 h| 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  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 †K.wYK fi‡eM, L = nh 2  mr2 = nh 2   = 6.63 × 10–34 × 1 2 × 9.1 × 10–31 × (0.53 × 10–10) 2 = 4.13 × 1016 rads–1 (Ans.) [r = 0.53 × 10–10 m n‡jv †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 235U jvM‡e|  w 235 = N 6.023 × 1023  N = w × 6.023 × 1023 235 E = Pt = Pin × t = Pout  × t      = Pout Pin = 1000 × 106 0.10 × 10 × 365 × 24 × 60 × 60 = 3.1536 × 1018 J  w × 6.023 × 1023 235 × 200 × 1.602 × 10–19 × 106 = 3.1536 × 1018  w = 38.4033 × 106 g = 38.4033 × 103 kg  38.4033 × 103 kg 235U 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: t = T1/2 ln2 ln     R0 R = 1000 ln2 ln (10)  t = 3321.92 years (Ans.)  = T1/2 ln2 = 1000 ln2 = 1442.69 years (Ans.)
2 ........................................................................................................................................  Physics 2nd Paper Chapter-9 7. †iwWqv‡gi Aa©vqy 1620 eQi| 1 g f‡ii †iwWqvg n‡Z cÖwZ †m‡K‡Û KZ ̧‡jv †iwWqvg cigvYy wbM©Z n‡e? [†iwWqv‡gi cvigvYweK fi = 226 kg/kmol, A ̈v‡fvMv‡Wav msL ̈v = 6.02 × 1026 atoms/kmol] [BUET 16-17] mgvavb: dN dt = N = ln2 T1/2 × N = ln2 1620 × 365 × 86400 × 1 × 6.02 × 1026 × 10–3 226  dN dt = 3.614 × 1010 Bq (Ans.) 8. U 238 Gi Aa©vqy1.42 × 1017 s| 1 g U238 †_‡K cÖwZ †m‡K‡Û KZ ̧‡jv cigvYy †f‡1⁄2 hv‡e? [A ̈v‡fv‡M‡Wav msL ̈v, NA = 6.02 × 1023 mol–1 ] [BUET 14-15] mgvavb: dN dt = N = ln2 T1/2 N  dN dt = ln2 1.42 × 1017 × 1 × 6.023 × 1023 238  dN dt = 1.2353 × 104 Bq (Ans.) 9. nvB‡Wav‡Rb cigvYy‡Z GKwU B‡jKUab GKwU †cÖvUb‡K †K›`a K‡i 0.53 A  e ̈vmva© wewkó Aiwe‡U AveZ©b K‡i| B‡jKUabwUi †K.wYK Z¡iY wbY©q Ki| [me = 9.1 × 10–31 kg, e = 1.6 × 10–19 C and 0 = 8.85 × 10–12 C 2N –1m –2 ] [BUET 07-08] mgvavb: cÖ‡Z ̈K kw3 ͇̄i _vKv B‡jKUab Gi Rb ̈ •iwLK Ges †K.wYK †eM aaæe _v‡K| ZvB †K.wYK Z¡iY 0 nq| (Ans.) 10. †Kvb GKwU †ZRw ̄Œq c`v‡_©i Aa©vqy3.8 w`b| AvU w`b ci GB c`v‡_©i kZKiv KZ Ask Aewkó _vK‡e? [BUET 07-08] mgvavb: N = N0e –t = N0e – ln2 T1/2 t  N N0 = e – ln2 3.8 × 8  N N0  100% = 0.2324 × 100% = 23.24% (Ans.) 11. KvD›U †iU wgUv‡ii mvnv‡h ̈ †Kvb †ZRw ̄Œq e ̄‘i mwμqZv gvcv hvq| †Kvb gyû‡Z© KvD›U wgUv‡i 4750 KvD›U cÖwZ wgwbU cvV †`q| cvPu wgwbU ci GwU 2700 KvD›U cÖwZ wgwbU cvV †`Lvq| †ZRw ̄Œq e ̄‘wUi Aa©vqy Ges ÿq aaæeK wbY©q Ki| [BUET 05-06] mgvavb:  = 1 t ln     R0 R   = 1 5 ln     4750 2700   = 0.113 min–1 (Ans.) T1/2 = ln2  = 6.135 min (Ans.) 12. †iW‡bi Aa©vqy4 w`b| Gi Mo Avqy KZ? [BUET 04-05] mgvavb:  = 1  = T1/2 ln2 = 4 ln2   = 5.77 days (Ans.) 13. GK MÖvg †iwWqvg 5 ermi -KYv wewKi‡Yi d‡j 2.1 mg K‡g hvq| †iwWqv‡gi Aa©vqy †ei Ki| [BUET 03-04] mgvavb: m = m0e –t  (1 – 2.1 × 10–3 ) = 1 × e – ln2 T1/2  5  T1/2 = 1648.61 years (Ans.) 14. GKwU †ZRw ̄Œq e ̄‘‡Z 1018 wU cigvYy Av‡Q| e ̄‘wUi Aa©vqy n‡”Q 2000 w`b| 5000 w`b ci KZ fMœvsk Aewkó _vK‡e? [BUET 01-02] mgvavb: N N0 = e – ln2 T1/2  t = e – ln2 2000 × 5000  N N0 = 0.17677 (Ans.) 15. 214Pb82 Gi Aa©vqy 26.8 wgwbU| Kx cwigvY fi †_‡K GK Kzix †ZRw ̄ŒqZv cvIqv hv‡e Zv †ei Ki| [BUET 00-01] mgvavb: 1 Curie = 3.7 × 1010 Bq m cwigvY f‡i N wU atom Av‡Q,  m 214 = N 6.023 × 1023  N = m × 6.023 × 1023 214 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 g (Ans.) weMZ mv‡j RUET-G Avmv cÖkœvejx 1. GKLÛ †iW‡bi 60% ÿq n‡Z KZw`b jvM‡e? †iW‡bi Aa©vqy 4 w`b| [RUET 08-09] mgvavb: t = T1/2 ln2 ln     N0 N  t = 4 ln2 ln     100 40 = 5.287 days (Ans.) weMZ mv‡j CUET-G Avmv cÖkœvejx 1. 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.)
cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Engineering Practice Sheet ..................................................................................... 3 2. †Kvb GKwU †ZRw ̄Œq e ̄‘i Aa©vqy 6.93 w`b| KZw`b c‡i GB †ZRw ̄ŒqZvi 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.) 3. hw` nvB‡Wav‡Rb Gi GKwU B‡jKUab Z...Zxq kw3 ̄Íi †_‡K wØZxq 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 6.56 × 10–7 = 4.57 × 1014 Hz (Ans.) 4. 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   = 7.2028 × 1010 s (Ans.) 5. 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: t = T1/2 ln2 ln     N0 N  t = 2.7 ln2 ln     109 109 – 3 × 108 = 1.389 days (Ans.) weMZ mv‡j BUTex-G Avmv cÖkœvejx 1. 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 8 d| 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.) 2. 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 wbDwK¬qv‡mi KYv cÖwZ eÜb kw3 = m  931 4 = 0.02932  931 4 = 6.824 MeV/nucleon (Ans.) 3. (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 = 1590.08 years (Ans.) (L) AvB‡mvevi (Ans.) (M) P = h  = 6.626 × 10–34 4 × 10–7 kgms–1 = 1.6565 × 10–27 kgms–1 (Ans.) (N) Nuclear Fission (Ans.) MCQ weMZ mv‡j BUET-G Avmv cÖkœvejx 1. †Kvb c`v‡_©i Aa©vqy 10 w`b n‡j 75% ÿq n‡Z KZ mgq jv‡M? [BUET Preli 22-23] 10 d 15 d 20 d 30 d DËi: 20 d e ̈vL ̈v: t = T1/2 ln2 ln     N0 N = 10 ln2 ln     100 25  t = 20 days 2. †Kvb iwk¥‡Z PvR© _v‡K bv? [BUET Preli 21-22]    None DËi:  3. wb‡¤œi †KvbwU‡K Zvcxq wewKiY e‡j? [BUET Preli 21-22] IR-radiation X-radiation -radiation UV-radiation DËi: IR-radiation 4. cÖ_g wZbwU †evi Kÿc‡_i e ̈vmv‡a©i AbycvZ n‡”QÑ [BUET 13-14] 1 : 1 4 : 1 9 1 : 2 : 3 1 : 4 : 9 1 : 8 : 27 DËi: 1 : 4 : 9 e ̈vL ̈v: r  n 2  r1 : r2 : r3 = 12 : 22 : 32 = 1 : 4 : 9
4 ........................................................................................................................................  Physics 2nd Paper Chapter-9 5. wdkb wewμqvq cÖwZwU wbDwK¬qvm †_‡K wbtm„Z kw3i cwigvYÑ [BUET 13-14] 200 MeV 931 MeV 200 eV 200 GeV DËi: 200 MeV 6. me‡P‡q kw3kvjx bbAv‡qvbvBwRs †iwW‡qkb nj- [BUET 12-13] AwZ †e ̧bx iwk¥ ivWvi gvB‡μvI‡qf Ae‡jvwnZ iwk¥ DËi: AwZ †e ̧bx iwk¥ e ̈vL ̈v: AwZ †e ̧bx iwk¥i Zi1⁄2‣`N© ̈ me‡P‡q Kg, ZvB Gi kw3 me©vwaK| 7. GKwU †ZRw ̄Œq c`v‡_©i †ZRw ̄ŒqZv 30 N›Uvq cÖviw¤¢K gv‡bi 15 16 Ask ÿqcÖvß n‡j c`v_©wUi Aa©vqy KZ n‡e? [BUET 10-11] 7.5 h 2.5 h 10 h 6 h DËi: 7.5 h e ̈vL ̈v: N = N0e –t  1 16 = e – ln2 T1/2  30  T1/2 = 7.5 h weKí: 1 16 = 1 2 4  4 wU Aa©vqy mgq ci 1 16 Ask Aewkó _vK‡e  4  T1/2 = 30 h  T1/2 = 7.5 h 8. hLb GKwU cigvYyi wbDwK¬qvm †_‡K GKwU weUv KYv wbM©Z nj, ZLb- [BUET 10-11] cvigvYweK msL ̈v K‡g hvq fi msL ̈v GK K‡g hvq cvigvYweK msL ̈v GK †e‡o hvq cvigvYweK msL ̈v `yB K‡g hvq DËi: cvigvYweK msL ̈v GK †e‡o hvq e ̈vL ̈v:  KYv wbM©gb A_© GKwU wbDUab †f‡1⁄2 hvIqv| n   – + p+  cvigvYweK msL ̈v 1 †e‡o hvq wKš‘ fi msL ̈v w ̄’i _v‡K| weMZ mv‡j CKRUET-G Avmv cÖkœvejx 1. cvigvYweK `~N©Ubvi d‡j †Kv‡bv GKwU K‡ÿ 20 d Aa©vqy m¤úbœ wKQz cwigvY †ZRw ̄Œq c`v_© cÖweó n‡jv| cixÿv K‡i †`Lv †Mj †h, †ZRw ̄Œq wewKi‡Yi gvÎv ̄^vfvweK gvÎvi †P‡q 70 ̧Y †ewk| KZ w`b c‡i NiwU wbivc‡` e ̈envi Kiv hv‡e?[CKRUET 23-24] 120.5 days 222.6 days 121.6 days 122.6 days 123.5 days DËi: 122.6 days e ̈vL ̈v: t = T1/2 ln2 ln     R0 R  t = 20 ln2 ln     70 1  t = 122.6 days 2. iƒccyi cvigvYweK we`y ̈r †K‡›`ai Drcv`b ÿgZv 2400 MW| G‡Z R¡vjvwb wnmv‡e U 235 e ̈envi Kiv n‡e| GK eQi ci GK †gvj R¡vjvwb n‡Z 9.2755 × 1011 wU cigvYy ÿq n‡q hvq| GKwU wdkb n‡Z 200 MeV kw3 cvIqv hvq| D3 we`y ̈r †K‡›`a GK eQ‡i wK cwigvY R¡vjvwb LiP n‡e? [CKRUET 23-24] 1536.1 kg 1436.1 kg 936.1 kg 921.6 kg 977.3 kg DËi: 921.6 kg e ̈vL ̈v: W = Pt = 2.4 × 109 × 365 × 24 × 3600 J = 7.57 × 1016 J awi, R¡vjvwb cÖ‡qvRb x kg  N NA = x 235 × 10–3  N = xNA 235 × 10–3 wU cigvYy  7.57 × 1016 = xNA 235 × 10–3 × 200 × 106 × 1.6 × 10–19  x = 923 kg  x  921.6 kg 3. GKLÐ †iW‡bi 70% ÿq n‡Z KZ mgq jvM‡e? †iW‡bi Aa©vqy 3.85 w`b| [CKRUET 22-23] 5.06 days 4.69 days 4.5 days 7.66 days 6.69 days DËi: 6.69 days e ̈vL ̈v: t = 1  ln     N0 N = 3.85 ln2 ln     100 30  t = 6.69 days 4. GK LÐ †iW‡bi 60% ÿq n‡Z KZ mgq jvM‡e? [†iW‡bi Aa©vqy 3.82 days] [CKRUET 20-21; BUET 18-19] 2.82 days 4.584 days 5.86 days 5.06 days None of them DËi: 5.06 days e ̈vL ̈v: t = T1/2 ln2 ln     N0 N = 3.82 ln2 ln     100 40  t = 5.05 days