Tiêu đề 9. P2C9. পরমানুর মডেল ও নিউক্লিয়ার পদার্থবিজ্ঞান Merge Ok_Sha 25.4.24_Mahee - Ok.pdf ✅

cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Engineering Question Bank 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 92U Gi GKwU wdkb wewμqvq 200 MeV kw3 Drcbœ nq| Giƒc 2 gm 235 92U 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  mr 2 = 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:
2  Physics 2nd Paper Chapter-9 N0 N = 2 – time T1 2  10 = 2 t 1000  t 1000 ln2 = ln10  t = ln10 × 1000 ln2 = 3321.928 years. (Ans.) 7| †iwWqv‡gi Aa©vqy 1620 eQi| 1-gram f‡ii †iwWqvg n‡Z cÖwZ †m‡K‡Û KZ ̧‡jv †iwWqvg cigvYy wbM©Z n‡e? [†iwWqv‡gi cvigvYweK fi = 226 kg/k mol, A ̈v‡fvMv‡Wav b¤^i = 6.02 × 1026 atoms/k mol] [BUET 16-17] mgvavb: dN dt = N  dN dt = ln2 T1 2 × N = ln2 1620 × 365 × 86400 × 1 × 6.023 × 1026 × 10–3 226 = 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 × 1023mol–1 ] [BUET 14-15] mgvavb: dN dt = N = ln2 T1 2 N  dN dt = ln2 1.42 × 1017 × 1 × 6.023 × 1023 238 = 1.235 × 104 atoms–1 = 1.235 × 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‡j±abwUi †K.wYK Z¡iY wbY©q Ki| [BUET 07-08] [me = 9.1 × 10–31 kg, e = 1.6 × 10–19 C and 0 = 8.85 × 10–12 C 2Nm–2 ] mgvavb: cÖ‡Z ̈K kw3 ͇̄i _vKv B‡jKUab Gi Rb ̈ •iwLK Ges †KŠwYK †eM aaæe _v‡K| ZvB †KŠwYK 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 = 0.2324 × 100% = 23.248% (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: A1 = A0e –t  4750 = A0e –t .............(i) A2 = A0e –t 2  2700 = A0e –(t + 5) .............(ii) (i)  (ii)  4750 2700 = e –t e –(t + 5)  4750 2700 = e –t + t+ 5 = e 5   = 0.1129 min–1 T1 2 = ln2  = 6.134 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 †ewWqvg 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 ) = 1e – ln2 T1/2  5  T1 2 = 1648.27 years (Ans.) 14| GKwU †ZRw ̄Œq e ̄‘‡Z 1018 cigvYy Av‡Q| e ̄‘wUi Aa©vqy n‡”Q 2000 w`b| 5000 w`b ci KZ fMœvsk Ae‡kl _vK‡e? [BUET 01-02] mgvavb: N0 N = 2 – T T1/2  N0 N = 2 5000 2000 = 5.6568  Aewkó _vK‡e N0 N = 0.1767 (Ans.) Traditional way †Z: N0 N = e – ln2 T1 2  t = e – ln2 2000 × 500 = 0.17677 (Ans.) 15| 214P82 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 atom Av‡Q, m 214 = N 6.023 × 1023  N = m × 6.023 × 1023 214
cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Engineering Question Bank 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.)
4  Physics 2nd Paper Chapter-9 (N) Nuclear Fission (Ans.) weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv eûwbe©vPbx cÖkœmg~n 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 Kÿ c‡_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 5. wdkb wewμqvq cÖwZwU wbDwK¬qvm †_‡K wbtm„Z kw3i cwigvY- [BUET 13-14] 200 MeV 931 MeV 200 eV 200 GeV DËi: 200 MeV 6. 83Bi210 †ZRw ̄Œq wewKi‡Yi ci 84Po210 †g.jwU •Zix K‡i| GLv‡b wewKi‡Yi aiY n‡”Q- [BUET 13-14] -decay -decay  and -decay -decay DËi: -decay e ̈vL ̈v: 210 83Bi –   210 84Po 7. me‡P‡q kw3kvjx bbAv‡qvbvwRs †iwW‡qkb nj- [BUET 12-13] AwZ †e ̧bx iwk¥ ivWvi gvB‡μvI‡qf Ae‡jvwnZ iwk¥ DËi: Ae‡jvwnZ iwk¥ e ̈vL ̈v: Infrared Light 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| 9. 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: A = dN dt = – N  A1 A2 = N1 N2  70A2 A2 = N1 N2  N2 N1 = 1 70      1 2 t 20 = 1 70  t = 122.6 d 10. 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 †g.j 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