Tiêu đề পরমাণু মডেল ও নিউক্লিয়ার পদার্থবিজ্ঞান (With Solve).pdf ✅

cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Varsity Practice Content 1 beg Aa ̈vq cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb Atomic Model & Nuclear Physics ACS Physics Department Gi g‡bvbxZ wjwLZ cÖkœmg~n 1| GKwU †ZRw ̄Œq c`v‡_©i †ZRw ̄ŒqZv hLb n1 ZLb Gi AÿZ cigvYy msL ̈v n2| †ZRw ̄Œq c`v_©wUi Aa©vqy wbY©q K‡iv| [Easy] mgvavb: – dN dt = N  n1 = n2  n1 = ln2 T1/2 n2  T1/2 = n2 n1 ln2 (Ans.) 2| †Kv‡bv †ZRw ̄Œq bgybvi Aa©vqy T| t1 I t2 mg‡q bgybvwUi mwμqZv h_vμ‡g R1 Ges R2| G mg‡qi e ̈eav‡b ÿqcÖvß cigvYy msL ̈v wbY©q K‡iv| [Medium] mgvavb: t1 mg‡q, R1 = N1  R1 = ln2 T N1  N1 = R1T ln2 t2 mg‡q, N2 = R2T ln2 t1 n‡Z t2 mg‡q ÿqcÖvß cigvYy msL ̈v = N1 – N2 = (R1 – R2) T ln2 (Ans.) 3| †Kv‡bv †ZRw ̄Œq c`v‡_©i mwμqZv 9 w`b e ̈eav‡b 8000 Bq n‡Z 1000 Bq G cwiYZ n‡jv| c`v_©wUi Aa©vqy I Mo Avqy? [Easy] mgvavb: mwμqZv 1 2 n Ask nq nT1/2 mg‡q  1000 8000 = 1 8 = 1 2 3  cÖ‡qvRbxq mgq, 3  T1/2 = 9  T1/2 = 3 days (Ans.)  Mo Avqy,  = T1/2 ln2 = 3 ln2 days (Ans.) 4| GKwU †ZRw ̄Œq bgybvq 1015 wU AÿZ cigvYy i‡q‡Q bgybvi H g~û‡Z© †ZRw ̄ŒqZv 6  1011 Bq n‡j bgybvi Aa©vqy? [Easy] mgvavb: – dN dt = N  6  1011 = ln2 T1/2  1015  T1/2 = ln2  1015 6  1011 = ln2 6  104 s (Ans.) 5| hw` jvBg ̈vb wmwi‡Ri me©wb¤œ Zi1⁄2‰`N© ̈ 912 A  nq Z‡e evgvi wmwi‡Ri me©wb¤œ Zi1⁄2‰`N© ̈ wbY©q K‡iv| [Medium] mgvavb: 1 1 = RH     1 1 2 – 1  = RH  RH = 1 912 A  Avevi, 1 2 = RH     1 2 2 – 1  = RH 4  2 = 4 RH = 4  912 A   2 = 3648 A  (Ans.) 6| nvB‡Wav‡Rb cigvYyi B‡jKUabwU 4_© Kÿc_ n‡Z 2q Kÿc‡_ Avm‡j wbM©Z †dvU‡bi K¤úv1⁄4 wbY©q K‡iv| [RH = 105 cm–1 ] [Easy] mgvavb: 1  = RH       1 n 2 1 – 1 n 2 2  f c = 107     1 4 – 1 16 = 107  3 16  f = 107  3  3  108 16  f = 9 16 1015 Hz (Ans.) 7| nvB‡Wav‡Rb cigvYyi B‡jKUab n = n1 †_‡K n = n2 Kÿc‡_ ̄’vbvšÍwiZ n‡jv| n1 Kÿc‡_ B‡jKUab Gi ch©vqKvj, n2 Kÿc‡_i ch©vqKv‡ji 8 ̧Y n‡j n1 I n2 Gi g‡a ̈ m¤úK© wbY©q K‡iv| [Medium] mgvavb: T = 2r v = 2r0  n 2 z v0  z n = kn3  T1 T2 =     n1 n2 3      8T2 T2 1 3 = n1 n2  n1 = 2n2 (Ans.)
2  Physics 2nd Paper Chapter-9 8| 1, 2, 3 h_vμ‡g jvBg ̈vb wmwi‡Ri m‡e©v”P K¤úv1⁄4, jvBg ̈vb wmwi‡Ri cÖ_g jvB‡bi K¤úv1⁄4 Ges evgvi wmwi‡Ri m‡e©v”P K¤úv1⁄4 n‡j 3 †K 1 I 2 Gi gva ̈‡g cÖKvk K‡iv| [Medium] mgvavb: 1  = RH       1 n 2 1 – 1 n 2 2   = cRH       1 n 2 1 – 1 n 2 2  1 = cRH     1 1 2 – 1  = cRH 2 = cRH     1 1 2 – 1 2 2 = cRH  3 4 3 = cRH     1 2 2 – 1  = cRH  1 4  3 = cRH     1 – 3 4 = cRH – cRH  3 4  3 = 1 – 2 (Ans.) 9| GKwU D‡ËwRZ nvB‡Wav‡Rb cigvYy  Zi1⁄2‰`‡N© ̈i †dvUb wewKwiZ K‡i f‚wg Ae ̄’vq wd‡i Av‡m| D‡ËwRZ ͇̄ii cÖavb †Kvqv›Uvg msL ̈vÑ [Medium] mgvavb: 1  = RH       1 n 2 1 – 1 n 2 2  1  = RH     1 1 2 – 1 n 2 = RH     1 – 1 n 2  1 RH = 1 – 1 n 2  1 n 2 = 1 – 1 RH = RH – 1 RH  n 2 = RH RH – 1  n = RH RH – 1 (Ans.) 10| f‚wg Ae ̄’vq GKK abvZ¥K wnwjqvg Avqb I wØ-abvZ¥K wjw_qvg Avq‡bi B‡jKUab؇qi Z¡i‡Yi AbycvZ? [Hard] mgvavb: F = 1 40 . (ze)e r 2 = 1 40 . ze2 r 2  Z¡iY, a = 1 40m . ze2 r 2  a  z r 2  aHe+ aLi2+ =     z1 z2      r2 r1 2 = 2 3       r0   1 2 3 r0  1 2 2 2  aHe+ aLi2+ = 2 3  4 3 = 8 27 (Ans.) 11| nvB‡Wav‡Rb cigvYyi B‡jKUabwU wØZxq Kÿc_ n‡Z cÖ_g Kÿc‡_ Avm‡j 2.7  1015 Hz K¤úv‡1⁄4i †dvUb wewKiY K‡i| B‡jKUabwU Z...Zxq Kÿc_ n‡Z cÖ_g Kÿc‡_ †b‡g Avm‡j wewKwiZ †dvU‡bi K¤úv1⁄4? [Hard] mgvavb: 1  = RH       1 n 2 1 – 1 n 2 2  f c = RH       1 n 2 1 – 1 n 2 2  f = cRH       1 n 2 1 – 1 n 2 2 cÖ_g †ÿ‡Î, f1 = cRH     1 1 2 – 1 2 2 = 3 4 cRH  cRH = 3 4  2.7  1015 = 3.6  1015 Hz wØZxq †ÿ‡Î, f2 = cRH     1 1 2 – 1 3 2 = 8 9 cRH  f2 = 8 9  3.6  1015 = 3.2  1015 Hz (Ans.) 12| evgvi I jvBg ̈vb wmwi‡Ri me©‡kl jvB‡bi Zi1⁄2‰`‡N© ̈i AbycvZ wbY©q K‡iv| [Easy] mgvavb: 1 min = RH     1 n 2 – 1  = 1 n 2 RH  min  n 2  Balmer Lyman =     n1 n2 2 =     2 1 2 = 4 : 1 (Ans.) 13| GKwU †ZRw ̄Œq bgybvi t = 0 mg‡q †ZRw ̄ŒqZv cÖwZ wgwb‡U 5400 decay| 5 min ci Gi †ZRw ̄ŒqZv n«vm †c‡q cÖwZ wgwb‡U 600 decay n‡j bgybvi Aa©vqy? [Easy] mgvavb: t = T1/2 ln2 ln     R0 R  5 = T1/2 ln2 ln     5400 600 = T1/2  ln3 ln2  T1/2 = 5ln2 ln3 (Ans.) 14| GKLÐ cÖvPxb Kv‡V 14C I 12C Gi AbycvZ eZ©gvbKv‡ji RxweZ Mv‡Qi Kv‡V H Abycv‡Zi 1 12 Ask| 14C Gi Aa©vqy T years n‡j cÖvPxb KvVwUi eqm wbY©q K‡iv| [Medium] mgvavb: N = N0e –t  N N0 = e–t  1 12 = e–t  e t = 12  t = ln(12)  = T ln(12) ln(2) (Ans.)
cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Varsity Practice Content 3 15| cÖwZ wdk‡b 200 MeV kw3 wbM©Z n‡j, 32 MW ÿgZv Drcv`‡b cÖwZ †m‡K‡Û KZwU wdkb n‡Z n‡e? [Easy] mgvavb: Pt = E  32  106  1 = n  200  106  1.6  10–19  n = 32 200  1.6  10–19 = 1 10–18 = 1018 wU 16| 210Po Gi Aa©vqy 140 d n‡j 70 w`b ci Gi †ZRw ̄ŒqZv kZKiv KZ n«vm cv‡e? [Medium] mgvavb: R = R0e –t = R0e – ln2 T1/2 t  R R0 = e – ln2 140  70 = e – 1 2 ln2  R R0 = e ln(2 ) – 1 2 = 1 2 [∵ e lnx = x]  R0 – R R0  100 =     1 – 1 2  100% = 2 – 1 2  100% (Ans.) 17| GKwU w ̄’i wbDwK¬qvm nVvr `yB UzK‡ivq wef3 n‡jv| UzK‡iv `ywUi †e‡Mi AbycvZ 1 : 27 n‡j Zv‡`i e ̈vmv‡a©i AbycvZ KZ? [Hard] mgvavb: fi‡e‡Mi msiÿYkxjZv Abyhvqx, m1v1 = m2v2  m1 m2 = 27  V1 V2 = 27      r1 r2 3 = 27  r1 r2 = 3 : 1 (Ans.) 18| z cvigvYweK msL ̈v wewkó nvB‡Wav‡Rb m`„k †Kv‡bv Avq‡bi n Zg Kÿc‡_i †gvU kw3i m~Î cÖwZcv`b K‡iv| [Hard] mgvavb: n Zg Kÿc‡_ B‡jUa‡bi †eM, vn = e 2 z 20nh Kÿc‡_i e ̈vmva©, rn = 0n 2 h 2 me2 z †gvU kw3 = MwZkw3 + wefekw3 = 1 2 mv2 – 1 40  (ze) e r = 1 2 m e 4 z 2 4 2 0 n 2 h 2 – ze2 40      0n 2 h 2 me2 z = me4 z 2 8 2 0 n 2 h 2 – me4 z 2 40n 2 h 2 En = – me4 z 2 8 2 0 n 2 h 2 (Ans.) 19| †evi cigvYy g‡Wj Abymv‡i wiWevM© aaæeK Gi m~Î cÖwZcv`b K‡iv| [Easy] mgvavb: n Zg Kÿc‡_i †gvU kw3, En = – me4 z 2 8 2 0 n 2 h 2  E2 – E1 = hc   hc  = me4 z 2 8 2 0 h 2       1 n 2 1 – 1 n 2 2  1  = me4 z 2 8 2 0 h 3 c       1 n 2 1 – 1 n 2 2  RH = me4 z 2 8 2 0 h 3 c (Ans.) 20| GKwU †ZRw ̄Œq bgybvi †ZRw ̄ŒqZv 20 hrs mg‡q 4  106 dps n‡Z 1  106 dps G n«vm cvq| mgq MYbvi 100 hrs c‡i Gi †ZRw ̄ŒqZv? [Medium] mgvavb: †ZRw ̄ŒqZv 1 2 2 ̧Y nq 20 hrs mg‡q,  2  T1/2 = 20  T1/2 = 10 hrs  100 hrs = 10 T1/2 mgq c‡i †ZRw ̄ŒqZv, 4  106 2 10 = 4  106 1024  4  106 1000  4  103 dps (cÖvq) (Ans.) 21| nvB‡Wav‡Rb cigvYyi evgvi wmwiR n‡Z wbM©Z Av‡jv †Kv‡bv avZe cv‡Zi Dci AvcwZZ n‡j, cvZ n‡Z d‡UvB‡jKUab wbM©Z nq| avZzi m¤¢ve ̈ m‡e©v”P Kvh©v‡cÿ‡Ki gvb wbY©q K‡iv| [Medium] mgvavb: evgvi wmwiR n‡Z m‡e©v”P kw3m¤úbœ †dvU‡bi kw3, E = 13.6       1 n 2 1 – 1 n 2 2 = 13.6     1 2 2 – 1   E = 13.6 4 = 3.4 eV  m¤¢ve ̈ m‡e©v”P Kvh©v‡cÿK = 3.4 eV (Ans.) 22| N = N0e –t mgxKiYwU cÖwZcv`b K‡iv| [Medium] mgvavb: †ZRw ̄ŒqZvi m~Î Abyhvqx, – dN dt = N  dN N = – dt   N N0 dN N =  t 0 – dt  ln     N N0 = – t  N N0 = e –t  N = N0e –t (Ans.)
4  Physics 2nd Paper Chapter-9 23| `ywU †ZRw ̄Œq c`v_© A I B Gi Aa©vqy h_vμ‡g 20 min I 40 min| `ywU bgybvi cÖviw¤¢K wbDwK¬qvm msL ̈v mgvb| 80 min ci bgybv `ywU‡Z AÿZ wbDwK¬qvm msL ̈vi AbycvZ? [Easy] mgvavb: awi, cÖviw¤¢K wbDwK¬qvm msL ̈v N0 A Gi Rb ̈, NA = N0 2 4 [4  T1/2 mgq] = N0 16 B Gi Rb ̈, NB = N0 2 2 [2  T1/2 mgq] = N0 4  NA NB = 1 16  4 = 1 : 4 (Ans.) 24| GKwU bgybvq `ywU AvB‡mv‡Uvc A Ges B Gi cwigvY h_vμ‡g 50 g I 30 g| A †ZRw ̄Œq wKš‘ B †ZRw ̄Œq bq| hw` A Gi Aa©vqy 2 hrs nq Z‡e 4 hrs c‡i bgybvi fi? [Medium] mgvavb: 2T1/2 mgq ci A Gi Aewkó cwigvY = W0 2 2 = 50 4 = 12.5 g  bgybvi †gvU fi = 12.5 + 30 = 42.5 g (Ans.) 25| GKwU †ZRw ̄Œq wbDwK¬qvm X (Aa©vqy 2 hours) †ZRw ̄Œq wewKi‡Yi ci ̄’vqx wbDwK¬qvm Y G cwiYZ nq, KZ NÈv mgq ci X Ges Y wbDwK¬qvm msL ̈vi AbycvZ n‡e 1 : 7? (cÖv_wgK Ae ̄’vq bgybvq ïay X wbDwK¬qvm wQj)| [Hard] mgvavb: X Gi Rb ̈ cÖviw¤¢K Ges t mgq ci wbDwK¬qvm msL ̈v N0 Ges N awi|  X Y = 1 7  N N0 – N = 1 7  N0 – N N = 7  N0 N = 8  N = N0 2 3  cÖ‡qvRbxq mgq = 3  T1/2 = 3  2 = 6 hours (Ans.) 26| 1 year mgq ci GKwU †ZRw ̄Œq bgybvi mwμqZv cÖv_wgK mwμqZvi (A0) 1 10 ̧Y nq| Av‡iv 9 years ci bgybvwUi mwμqZvÑ [Medium] mgvavb:  = 1 t ln A0 A = ln(10)  †gvU t = 10 years ci, A = A0e –t = A0e –ln(10)  10  A = A0e ln(10–10) = A0  10–10  A = A0 1010 (Ans.) 27| `ywU †ZRw ̄Œq c`v_© A Ges B †bqv n‡jv| B Gi ÿq aaæeK A Gi wØ ̧Y| cÖv_wgK Ae ̄’vq Dfq c`v‡_©i GKB msL ̈K AÿZ wbDwK¬qvm i‡q‡Q| A Gi n wU Aa©vqy mgq ci `ywU c`v‡_©i †ZRw ̄ŒqZv mgvb n‡j, n = ? [Medium] mgvavb: †ZRw ̄ŒqZv mgvb n‡j, ANA = BNB  NA NB = B A = 2  NA NB = N0e –At N0e –Bt  2 = et(B – A) = et(2A – A) = etA  ln2 = At  ln2 = ln2 T1/2  nT1/2  n = 1 (Ans.) 28| 30 min Aa©vqy wewkó †Kv‡bv †ZRw ̄Œq bgybvi mwμqZv 2 hours mgq ci 5 dps kbv3 Kiv n‡jv| bgybvwUi cÖv_wgK mwμqZv? [Medium] mgvavb: t = T1/2 ln2 ln     R0 R  4T1/2 = T1/2 ln2 ln     R0 5  4ln2 = ln     R0 5  R0 = 5  2 4 = 16  5 = 80 dps (Ans.) 29| †Kv‡bv †ZRw ̄Œq bgybvi Aa©vqy 10 min Ges cÖviw¤¢K cigvYy msL ̈v 108 wU| 5 min mgq ci ÿqcÖvß cigvYy msL ̈v? [Easy] mgvavb: t = T1/2 ln2 ln     108 N  5 = 10 ln2 ln     108 N  1 2 ln2 = ln     108 N  N = 108 2  ÿqcÖvß cigvYy msL ̈v = N0 – N = 108     1 – 1 2 = 108 ( 2 – 1) 2 wU (Ans.) 30| †Kv‡bv †ZRw ̄Œq c`v‡_©i Aa©vqy 12.5 hours Ges cÖviw¤¢K cwigvY 256 gm| KZ mgq ci c`v_©wUi 1 gm Aewkó _vK‡e? [Easy] mgvavb: W W0 = 1 256 = 1 2 8  cÖ‡qvRbxq mgq = 8  T1/2 = 8  12.5 = 100.0 hours (Ans.)
cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Varsity Practice Content 5 ACS Physics Department Gi g‡bvbxZ eûwbe©vPwb cÖkœmg~n 1. wb‡¤œi †Kvb AvB‡mv‡UvcwU K ̈vbmvi wPwKrmvq e ̈eüZ nq? [Easy] 40K 60Co 90Sr 131I DËi: 60Co 2. GK Mo Avqyi mgcwigvY mg‡q cÖviw¤¢K wbDwK¬qvm msL ̈viÑ [Medium] A‡a©K ÿqcÖvß n‡e A‡a©‡Ki Kg ÿqcÖvß n‡e A‡a©‡Ki †ewk ÿqcÖvß n‡e m¤ú~Y© cwigvY ÿqcÖvß n‡e DËi: A‡a©‡Ki †ewk ÿqcÖvß n‡e e ̈vL ̈v:  = 1  = T1/2 ln2 > T1/2   mg‡q A‡a©‡Ki †ewk wbDwK¬qvm ÿqcÖvß n‡e| 3. †Kv‡bv †ZRw ̄Œq c`v‡_©i ÿqaaæeK 1.5  10–9 s –1 n‡j Gi Mo AvqyÑ [Easy] 3.33  108 1.5  109 10.35  108 6.67  108 DËi: 6.67  108 e ̈vL ̈v:  = 1.5  10–9 s –1 = 3 2  10–9 s –1   = 1  = 2 3  109 s   = 0.667  109 s   = 6.67  108 s (Ans.) 4. wb‡¤œi wewμqvq Q = ? [Medium] A + B  C + D + Q MeV 1.002 amu 1.004 amu 1.001 amu 1.003 amu 1.234 MeV 0.931 MeV 0.465 MeV 1.863 MeV DËi: 1.863 MeV e ̈vL ̈v: m = (1.002 + 1.004) – (1.001 + 1.003)  m = 0.002 amu  Q = mc2 = 0.002  931 MeV  Q = 1.862 MeV  1.863 MeV 5. wb‡¤œi wewμqvq A, B, C, D, E = ? [Medium] α 92U 238  BThA E DPaC 92U 234 A = 234, B = 90, C = 234, D = 91, E =  A = 234, B = 90, C = 238, D = 94, E =  A = 238, B = 93, C = 234, D = 91, E =  A = 234, B = 90, C = 234, D = 93, E =  DËi: A = 234, B = 90, C = 234, D = 93, E =  e ̈vL ̈v: A = 238 – 4 = 234 B = 92 – 2 = 90 C = A = 234 D = B + 1 = 91 Ges E =  6. 15 cvigvYweK fi I 7 cvigvYweK msL ̈vwewkó †Kv‡bv cigvYy GKwU -KYv MÖnY K‡i Ges GKwU †cÖvUb wbM©Z K‡i| cigvYyi bZzb cvigvYweK msL ̈v I fi h_vμ‡gÑ [Easy] 18, 8 16, 4 8, 18 4, 16 DËi: 8, 18 e ̈vL ̈v: cvigvYweK msL ̈v = 7 + 2 – 1 = 8 fi = 15 + 4 – 1 = 18 7. wb‡¤œi wbDK¬xq wewμqvq wbm„Z  I  KYvi msL ̈vi mgwóÑ 92U 238  82Pb214 [Medium] 6 7 8 10 DËi: 8 e ̈vL ̈v:  KYv = 238 – 214 4 = 6 wU   KYv n wU n‡j, 82 = 92 – 6  2 + n  n = 2 wU mgwó = 6 + 2 = 8 wU 8. †Kv‡bv †ZRw ̄Œq c`v‡_©i t mgq c‡i 90% cigvYy AÿZ _v‡K| 2t mgq c‡i KZ Ask cigvYy ÿqcÖvß n‡e? [Hard] 20% 19% 40% 38% DËi: 19% e ̈vL ̈v: t = 1  ln     N0 N1 = 1  ln     100 90  2t = 1  ln     N0 N2  2  1  ln     10 9 = 1  ln     N0 N2      10 9 2 = N0 N  N N0 = 81 100  N0 – N N0 = 100 – 81 100 = 19 100  N0 – N N0  100% = 19 100  100% = 19%