Tiêu đề P2C9-Nuclear Physics (Engg.)_With Solve.pdf ✅

cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Engineering Practice Content 1 cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb Atomic Model & Nuclear Physics beg Aa ̈vq ACS Physics Department Gi g‡bvbxZ cÖkœmg~n 1| cÖwZ NÈvq wK cwigvY fi kw3‡Z iƒcvšÍwiZ n‡q 500 MW we`y ̈r kw3 •Zwi Ki‡Z cvi‡e? [Easy] mgvavb: P = 500 MW = 500 × 106 W = 5 × 108 W t = 1 hour = 3600 s  E = Pt = 5 × 108 × 3600 J = 1.8 × 1012 J  m = E c 2 = 1.8 × 1012 (3 × 108 ) 2 = 2 × 105 kg = 20 mg (Ans.) 2| 238 92 U cigvYy 206 82 Pb cigvYy‡Z cwiYZ n‡Z KqwU Avjdv KYv I KqwU weUv KYv wbtmiY K‡i? [Medium] mgvavb: awi, m msL ̈K  KYv I n msL ̈K  KYv wbtm„Z nq|   I  wbtmi‡Yi d‡j fimsL ̈v n«vm = 4m Ges cvigvYweK msL ̈v n«vm = 2m – n  238 – 4m = 206  m = 8 Avevi, 92 – 82 = 2m – n  n = 6  8 wU  KYv I 6 wU  KYv wbtm„Z nq| (Ans.) 3| 222 86 A cigvYyi wbDwK¬qvm †_‡K 3wU  KYv I 4wU  KYv wbtm„Z n‡j †h bZzb †g.wjK c`v_© m„wó nq Zvi fimsL ̈v, cvigvYweK msL ̈v I cÖZxK wbY©q Ki| [Medium] mgvavb: 3wU  KYv I 4wU  KYv wbtmi‡Y †gvU fimsL ̈v n«vm = 3 × 4 + 0 = 12 Ges cigvYweK msL ̈v n«vm = (3 × 2) – (4 × 1) = 2  bZzb †gŠjwUi fimsL ̈v 210, cvigvYweK msL ̈v 84 Ges cÖZxK 210 84 Po| (Ans.) 4| †Kv‡bv †ZR ̄Œxq †g.‡ji 5wU Aa©vqy c‡i H †g.‡ji KZ Ask Aewkó _vK‡e? [Medium] mgvavb: t = 5T n‡j, N N0 = 1 2 t T = 1 2 5T T = 1 32 Ask (Ans.) 5| 5 wgwjKzwi kw3m¤úbœ  iwk¥i Drm •Zwi Ki‡Z wK cwigvY 210 84 Po Gi cÖ‡qvRb n‡e? †cv‡jvwbqv‡gi Aa©vqy 138 w`b| [Medium] mgvavb:  = 0.693 T1 2 = 0.693 138 × 24 × 3600 s –1 = 5.812 × 10–8 s –1 GLb, dN dt = N  5 × 10–3 × 3.7 × 1010 = 5.812 × 10–8 × N  N = 3.183 × 1015  †cv‡jvwbqv‡gi cwigvY = 3.183 × 1015 × 210 6.023 × 1023 g = 1.11 × 10–6 g (Ans.) 6| wbDK¬xq wewμqv ̧‡jv m¤ú~Y© Ki: [Medium] (i) 14 7 N + 4 2He  17 8 O + ? mgvavb: k~b ̈ ̄’v‡b wbDwK¬qvmwUi fi msL ̈v = 14 + 4 – 17 = 1 k~b ̈ ̄’v‡b wbDwK¬qvmwUi cvigvYweK msL ̈v = 7 + 2 – 8 = 1  †gŠjwU 1 1H  m¤ú~Y© mgxKiY 14 7 N + 4 2He  17 8 O + 1 1H (Ans.) (ii) 19 9 F + 1 1H  16 8 O + ? mgvavb: k~b ̈ ̄’v‡b wbDwK¬qvmwUi fi msL ̈v = 19 + 1 – 16 = 4 k~b ̈ ̄’v‡b wbDwK¬qvmwUi cvigvYweK msL ̈v = 9 + 1 – 8 = 2  †gŠjwU 4 2He m¤ú~Y© mgxKiY 19 9 F + 1 1H  16 8 O + 4 2He (Ans.) (iii) 25 12Mg + ?  22 11Na  + 4 2He mgvavb: k~b ̈ ̄’v‡b wbDwK¬qvmwUi fi msL ̈v = 22 + 4 – 25 = 1 k~b ̈ ̄’v‡b wbDwK¬qvmwUi cvigvYweK msL ̈v = 11 + 2 – 12 = 1  †gŠjwU 1 1H m¤ú~Y© wewμqv 25 12Mg + 1 1H  22 11Na + 4 2He (Ans.) (iv) 27 13Al + 4 2He  30 15P + ? mgvavb: k~b ̈ ̄’v‡b wbDwK¬qvmwUi fi msL ̈v = 27 + 4 – 30 = 1 k~b ̈ ̄’v‡b wbDwK¬qvmwUi cvigvYweK msL ̈v = 13 + 2 – 15 = 0  KYvwU 1 0 n m¤ú~Y© wewμqv 27 13Al + 4 2He  30 15P + 1 0 n (Ans.)
2  Physics 2nd Paper Chapter-9 7| wbDwK¬qvm X GKwU  KYv wbtmiY K‡i wbDwK¬qvm Y m„wó K‡i| hw` Zv‡`i cvigvYweK fi h_vμ‡g MX I MY nq Z‡e wbtm„Z  KYvi m‡e©v”P kw3 wbY©q Ki| (B‡jKUa‡bi fi me I Av‡jvi †eM c) [Medium] mgvavb: †ZR ̄Œxq wewKi‡Yi mgxKiY: A Z X  A Z+1 X +  X I Y wbDwK¬qvm؇qi fi mX I mY n‡j mX = MX – Zme mY = MY – (Z + 1)me  wbtm„Z  KYvi m‡e©v”P kw3 Emax = [MX – Zme – MY + (Z + 1)me]c2 = (MX – MY + me)c2 (Ans.) 8| 2kg wWD‡Uwiqvg Gi wbqK¬xq ms‡hvRb Øviv GKwU 100 W ÿgZvi •e`y ̈wZK evj¦ KZ mgqe ̈vcx R¡j‡Z _vK‡e? ms‡hvRb cÖwμqvwU nj: 2 1H + 2 1H  3 2He + n + 3.27 MeV [Medium] mgvavb: 2kg wWD‡Uwiqv‡g cigvYy msL ̈v = 6.023 × 1023 × 2000 2 = 6.023 × 1026 wU GLb, 2wU wWD‡Uwiqvg ms‡hvR‡b wbM©Z kw3 = 3.27 MeV  2kg wWD‡Uwiqv‡gi ms‡hvR‡b wbM©Z kw3 = 3.27 2 × 6.023 × 1026 × 106 × 1.6 × 10–19 J = 1.576 × 1014 J •e`y ̈wZK evj¦wU t mgq R¡j‡j, t = W P = 1.576 × 1014 100 s = 1.576 × 1012 s = 1.576 × 1012 24 × 3600 × 365 y = 49974.63 y (Ans.) 9| X †ZR ̄Œxq AvB‡mvU‡ci Aa©vqy 20 eQi| GwU ÿq n‡q GKwU ̄’vqx †g.j Y †Z cwiYZ nj| GKwU cv_‡ii bgybvq X I Y Gi AbycvZ 1:7 n‡j cv_iwUi AvbygvwbK eqm KZ? [Medium] mgvavb: cÖv_wgK Ae ̄’vq, X Gi cwigvY N0 Y Gi cwigvY k~b ̈ t mgq c‡i, X Gi cwigvY N Y Gi cwigvY N0 – N  N N0 – N = 1 7  N N0 = 1 8  N N0 =     1 2 3  t T1 2 = 3  t = 3 × 20 = 60 eQi (Ans.) 10| GKwU wgkÖY `ywU †ZR ̄Œxq †g.j A1 I A2 Gi mgš^‡q MwVZ hv‡`i Aa©vqy h_vμ‡g 20s I 10s| cÖv_wgKfv‡e wgkÖ‡Y 40g A1 I 160g A2 Av‡Q| KZ mgq ci wgkÖYwU‡Z †g.j`ywUi cwigvY mgvb n‡e? [Medium] mgvavb: A1 Gi Rb ̈ N1 = N01     1 2 t 20 A2 Gi Rb ̈ N2 = N02     1 2 t 10 cÖkœvbyhvqx, N1 = N2  40 2 t 20 = 160 2 t 10  1 4 = 2t ( ) 1 20 – 1 10  – t 20 = – 2  t = 40 s (Ans.) 11| `yBwU †ZR ̄Œxq †g.j A I B Gi ÿq aaæeK h_vμ‡g 5 I | t = 0 †Z Zv‡`i wbDwK¬qv‡mi msL ̈v mgvb| KZ mg‡q A I B Gi wbDwK¬qvm msL ̈vi AbycvZ e –2 n‡e? [Medium] mgvavb: A †gŠ‡ji Rb ̈, NA = N0Ae –5t  N0A = NAe 5t B †gŠ‡ji Rb ̈, N0B = NBe t cÖkœg‡Z, N0A = N0B  NAe 5t = NBe t  e –2 = e–4t  4t = 2  t = 1 2 (Ans.) 12| GKwU †ZR ̄Œxq c`v‡_©i cÖ_g 2s G x wU cigvYy weNwUZ nq Ges c‡ii 2s G 3x 4 wU cigvYy weNwUZ nq| cv_iwUi Mo Avqy wbY©q Ki| [Medium] mgvavb: awi, cÖv_wgK cigvYy msL ̈v, N0 cÖ_g 2s Gi Rb ̈ N0 – x N0 = e–2  1– x N0 = e–2  x N0 = 1 – e –2 ..... (i) c‡ii 2s Gi Rb ̈ N0 – x – 3x 4 N0 – x = e–2  1 – 0.75 x N0 e –2t = e–2  1 – e –2 = 0.75 × 1 – e –2 e –2  e –2 = 0.75 [∵ 1  e –2 ]   = 0.1438   = 6.952 s (Ans.)
cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Engineering Practice Content 3 13| 2 N›Uv Aa©vqy m¤úbœ GKwU m` ̈ cÖ ̄‘Z †ZR ̄Œxq Drm †_‡K wbtmwiZ wewKi‡Yi ZxeaZv Aby‡gvw`Z gvÎvi 64 ̧Y| me©wb¤œ KZ mgh c‡i H Drm w`‡q wbivc‡` KvR Kiv hv‡e? [Medium] mgvavb: N = N0e –t  1 64 = e– tln2 2  t = 12 hr (Ans.) 14| †Kv‡bv wbw`©ó mg‡q, GKwU bgybvq Dcw ̄’Z wbDwK¬qv‡mi 25% †ZR ̄Œxq| 40s ci †ZR ̄Œxq wbDwK¬qv‡mi cwigvY nq 12.5%| weNUb ïiæ nIqvi KZ mgq c‡i bgybvwU‡Z gvÎ 0.78% †ZR ̄Œxq wbDwK¬qvm Aewkó _vK‡e? [Medium] mgvavb: N N0 = e–t  12.5 25 = e–40   = 0.0173 s–1 t = – 1  ln     0.78 25 = 200.09 s (Ans.) 15| †Kv‡bv †ZR ̄Œxq bgybvq `y-ai‡bi †g.j Av‡Q, hv‡`i fvOb aaæeK 1 I 2| cÖ_g cÖKvi †g.‡ji cÖv_wgK mwμqZv wØZxq cÖKvi †g.‡ji f ̧Y n‡j KZ mgq c‡i †g.j `ywUi mwμqZv mgvb n‡e? [Medium] mgvavb: cÖkœg‡Z, 1N01 = f 2N02 ..... (i) Ges N11 = N22  N01 e –1t 1 = N02 e –2t 2  f 2 N02 e –1t = N02 e –2t 2  f e–1t = e–2t  e 1 t – 2 t = f  lnf = 1t – 2t  t = lnf 1 – 2 (Ans.) 16| GKwU 1000 MW ÿgZvi wefvRb wiA ̈v±i 5 eQ‡i Zv‡Z _vKv R¡vjvbxi A‡a©K cwigvY e ̈q K‡i| ïiæ‡Z KZ cwigvY 235 92 U wQj? a‡i bvI wiA ̈v±iwU 80% mgq KvR K‡i| mg ̄Í kw3 m„wó nq 235 92 U Gi wefvRb †_‡K Ges wbDwK¬qvm ̧‡jv ïaygvÎ wefvRb cÖwμqv‡ZB e ̈q nq| [Medium] mgvavb: 235 92 U Gi T1 2 = 5 eQi = 5 × 365 × 24 × 3600 s = 1.5768 × 108 s 1g 235 92 U = 6.023 × 1023 235 wU cigvYy = 2.5629 × 1021 wU cigvYy  cÖwZ MÖvg 235 92 U †_‡K wbM©Z kw3i cwigvY = 2.5629 × 1021 × 200 MeV = 5.1259 × 1023 MeV = 8.202 × 1010 J wiA ̈v±iwU 80% mgq KvR Ki‡j, 5 eQ‡i e ̈wqZ 235 92 U Gi cwigvY = 1.5768 × 108 × 1000 × 106 8.202 × 1010 × 0.8 g = 1537.96 kg  ïiæ‡Z 235 92 U Gi cwigvY = 2 × 1537.96 kg = 3075 kg (Ans.) 17| †Mvì AvB‡mv‡Uvc 197 79 Au Ges wmjfvi AvB‡mv‡Uvc 107 47 Ag Gi wbDwK¬qv‡mi e ̈vmv‡a©i AbycvZ wbY©q Ki| [Easy] mgvavb: RAu RAg =     AAu AAg 1 3 =     197 107 1 3 = 1.2256  RAu : AAg = 1.2256 : 1 (Ans.) 18| Co60 Gi Aa©vqy 5.3 eQi| GB AvB‡mvU‡ci wKQz cwigvY c`v‡_©i eZ©gvb mwμqZv 0.877 mCi n‡j GK eQi c~‡e© AvB‡mvUcwUi mwμqZv KZ wQj? [Easy] mgvavb: t = – 1  ln           dN dt f     dN dt i  1 = – 5.3 ln2 ln       0.877     dN dt i      dN dt i = 1 mCi (Ans.) 19| GKwU †ZR ̄Œxq c`v‡_©i Aa©RxebKvj 30 wgwbU| c`v_©wUi 40% weNUb †_‡K 85% weNU‡bi Rb ̈ KZ wgwbU mgq jvM‡e? [Medium] mgvavb: awi, 40% Ges 85% weNU‡bi Rb ̈ h_vμ‡g t1 I t2 wgwbU mgq jv‡M|  N0 × 0.6 = N0e –t1 .... (i) N0 × 0.15 = N0e –t2 .... (ii)  0.6 0.15 = e–(t2 – t1 )  ln4 =  (t2 – t1 )  t2 – t1 = ln4 ln2 × 30 = 60 wgwbU (Ans.) 20| GKRb Wv3vi wUDgv‡i AvμvšÍ GK †ivMxi wPwKrmvi Rb ̈ wUDgv‡ii gv‡S 9 × 1015 wU †ZR ̄Œxq Av‡qvwWb cigvYy wewkó GK UzKiv Av‡qvwWb ivL‡jb| Av‡qvwWb cigvYyi fvO‡bi cwigvY cÖwZw`b 15 × 106 wU ev Zvi †ewk n‡j wUDgvi Qvov kix‡ii Ab ̈vb ̈ m~2 †Kvl ̧‡jvi Ici ÿwZKi cÖfve co‡Z cv‡i| Av‡qvwW‡bi Aa©vqy 6 hr n‡j wPwKrmv c×wZwU wbivc` wKbv hvPvB Ki| [Easy]
4  Physics 2nd Paper Chapter-9 mgvavb: cÖ_gw`‡b m‡e©v”P msL ̈K cigvYy †f‡1⁄2 hv‡e|  = ln2 T1 2 = ln2 6 = 0.1155 hr–1  N = N0e –t = 9 × 1015 e –0.1155 × 24 = 5.625 × 1014 wU  cÖ_g w`‡b †f‡1⁄2 hvIqv cigvYy msL ̈v = 9 × 1015 – 5.625 × 1014 = 8.4375 × 1015 wU  wPwKrmv c×wZwU wbivc` bq| (Ans.) 21| 3 × 10–5 g 239Pu Drm cÖwZ †m‡K‡Û 2000 wU  KYv wbtmiY K‡i| cøy‡Uvwbqv‡gi Aa©vqy KZ? [Medium] mgvavb: N0 = 3 × 10–5 × 6.023 × 1023 239 = 7.56 × 1016 wU dN dt = N  dN dt = N0 × ln2 T1 2  2000 = 7.56 × 1016 × ln2 T1 2  T1 2 = 2.62 × 1013 s = 8.308 × 105 yrs (Ans.) 22| cÖvPxb mf ̈Zvi GKwU Kv‡Vi †Ljbvi †ZR ̄ŒxqZvi gvb 18 count/gm| Kv‡V †LjbvwUi bZzb Ae ̄’vq †ZR ̄ŒxqZvi gvb 24 count/gm| †ZR ̄Œxq Kve©‡bi Aa©vqy 5600 eQi n‡j Kv‡Vi †LjbvwUi eqm KZ? [Medium] mgvavb: t = – 1  ln     N N0 = – T1 2 ln2     18 24 = 2324.2 years (Ans.) 23| 28 13 Al wbDUa‡bi mv‡_ msN‡l© †iwWI †mvwWqvg 24 11 Na G cwiYZ nq Ges GKwU KYv wbtmiY K‡i| 24 11 Na Avevi GKwU KYv wbtmiY K‡i 24 12 Mg †Z iƒcvšÍwiZ nq| wewμqvi mgxKiY `ywU wj‡L KYvØq kbv3 Ki| [Easy] mgvavb: 28 13 Al  24 11 Na + 4 2 He 24 11 Na  24 12 Mg +   KYvØq h_vμ‡g 4 2 He, 0 –1  (Ans.) 24| M‡elYvMv‡i e ̈envi Dc‡hvMxZvi Rb ̈ †Kv‡bv bgybv c`v‡_©i b~ ̈bZg fi 80 g nIqv cÖ‡qvRb| M‡elYvi Rb ̈ 250 g 238 92 U msMÖn Kiv n‡j 20 eQi ci Zv M‡elYv‡hvM ̈ _vK‡e wK? [ 238 92 U Gi Aa©vqy 10 eQi] [Medium] mgvavb: w = w0e –t = 250 e– 20 × ln2 10 = 62.5 g  20 eQi ci bgybvwU M‡elYv‡hvM ̈ _vK‡e bv| (Ans.) 25| †ZR ̄Œxq Av‡qvwW‡bi 5.8 Ci cwigvY †WvR GKRb †ivMx‡K B‡ÄKk‡bi gva ̈‡g cÖ`vb Kiv n‡jv| AvB‡mvUcwUi Aa©Rxeb 5h n‡j AvB‡mvUcwUi KZ ̧‡jv Avw` wbDwK¬qvm B‡ÄKk‡bi gva ̈‡g cÖ`vb Kiv n‡qwQj? [Medium] mgvavb: dN dt = N  5.8 × 10–6 × 3.7 × 1010 = ln2 5 × 3600 × N  N = 5.5728 × 109 wU (Ans.) 26| GKwU w ̄’i †_wiqvg wbDwK¬qvm (A = 220, Z = 90) n‡Z Ed MwZkw3i GKwU  KYv wbM©Z nq| wewμqvq †iwWqvg wbDwK¬qv‡mi (A = 216, Z = 88) MwZkw3 KZ? [Medium] [DU 19-20] mgvavb: cÖkœg‡Z, Ed = 1 2 × 4 v2   v = Ed 2 ..... (i) fi‡eM msiÿY m~Îvbymv‡i, MRa vRa = Mv  216 vRa = 4 Ed 2  vRa = 1 54 Ed 2  †iwWqvg wbDwK¬qv‡mi MwZkw3 = 1 2 × 216 ×    1  54 Ed 2 2 = Ed 54 (Ans.) 27| GKwU cigvYyi μgea©gvb kw3i wZbwU ̄Íi h_vμ‡g A, B I C| B‡jKUab C †_‡K B Ges B †_‡K A †Z Avmvq wewKi‡Yi Zi1⁄2‣`N© ̈ h_vμ‡g 1 I 2| C †_‡K GKwU B‡jKUab A †Z Avm‡j wewKi‡Yi Zi1⁄2‣`N© ̈ KZ n‡e? [Medium] mgvavb: EC – EA = hc   (EC – EB) + (EB – EA) = hc   hc 1 + hc 2 = hc   1  = 1 1 + 1 2   = 12 1 + 2 (Ans.)
cigvYyi g‡Wj I wbDwK¬qvi c`v_©weÁvb  Engineering Practice Content 5 28| GKwU †ZR ̄Œxq bgybvi  weNUb n‡”Q| t1 I t2 mg‡qi gyn~‡Z© Gi mwμqZv h_vμ‡g A Ges A 5 | c`v_©wUi Mo RxebKvj KZ? [Medium] mgvavb: c`v_©wUi cÖv_wgK mwμqZv A0 n‡j, A = A0 e –t1 Ges A 5 = A0 e –t2  A A 5 = A0 e –t1 A0 e –t2  5 = e(t2 – t1)  ln5 = (t2 – t1)   = ln5 t2 – t1   = t2 – t1 ln5 (Ans.) 29| GKwU †ZR ̄Œxq c`v_©  KYvi wbtmiY K‡i| cÖ_g 2s G n msL ̈K  KYv Ges c‡ii 2s G 0.75 n msL ̈K  KYv wbtmiY K‡i| †ZR ̄Œxq c`v_©wUi Mo Avqy wbY©q Ki| [Medium] mgvavb: cÖ_g 2s G weNwUZ cigvYy msL ̈v = N0 – N = N0 – N0 e –t  n = N0(1 – e –t ) .... (i) cÖ_g 4s G weNwUZ cigvYy msL ̈v = (n + 0.75n) = N0(1 – e –4 )  1.75n = N0(1 – e –4 ) .... (ii) (ii)  (i) K‡i cvB, 1.75 = 1 – e –4 1 – e –2  1.75 = (1 + e –2 ) (1 – e –2 ) (1 – e –2 )  1 + e–2 = 1.75  e –2 = 0.75   = 0.1438 s–1   = 6.95 s (Ans.) 30| t = 0 †Z `yBwU †ZR ̄Œxq c`v_© A I B Gi mwμqZv mgvb| Zv‡`i mwμqZvi nv‡ii AbycvZ RB RA , mgq t Gi mv‡_ e –3t wbq‡g n«vm cvq| A Gi Aa©vqy ln2 n‡j B Gi Aa©vqy KZ? [Hard] mgvavb: kZ©vbymv‡i, (RB)0 = (RA)0 .... (i) t mgq c‡i, (RB)t = (RB)0 e –Bt Ges (RA)t = (RA)0 e –At cÖkœvbymv‡i, (RB)t (RA)t = e–3t  (RB)0 e –Bt (RA)0 e –At = e–3t  e (A – B)t = e–3t  B – A = 3  ln2 TB – ln2 TA = 3  ln2 TB – 1 = 3  TB = ln2 4 (Ans.) 31| wbDwK¬qv‡mi NbZ¡ Ges cvwbi Nb‡Z¡i AbycvZ wbY©q Ki| [Easy] mgvavb:  = wbDwK¬qv‡mi fi wbDwK¬qv‡mi AvqZb = Amp 4 3 R 3 = Amp 4 3  (r0A ) 1 3 3 = 3mp 4r 3 0 = 3 × 1.67 × 10–27 4 × (1.2 × 10–15) 3 kg m–3 = 2.3 × 1017 kg m–3   w = 2.3 × 1014  wbDwK¬qv‡mi NbZ¡ : cvwbi NbZ¡ = 2.3 × 1014 : 1 (Ans.) 32| `yBwU †ZR ̄Œxq c`v_© X I Y Gi cÖv_wgK wbDwK¬qvm msL ̈v h_vμ‡g N1 I N2| X Gi Aa©vqy Y Gi Aa©vqyi A‡a©K| Y Gi wZb Aa©RxebKvj c‡i Df‡qi wbDwK¬qvm msL ̈v mgvb nq| N1 I N2 Gi AbycvZ wbY©q Ki| [Medium] mgvavb: awi, Y Gi Aa©vqy, Ty = 2T  X Gi Aa©vqy, Tx = T kZ©vbymv‡i 6T mgq ci X I Y Gi wbDwK¬qvm msL ̈v mgvb nq|  N1 e –6Tx = N2 e –6Ty  N1 e –6T ln2 T = N2 e –6T ln2 2T  N1 × 1 64 = N2 × 1 8  N1 N2 = 8 (Ans.) 33| 14 7 N wbDwK¬qv‡mi eÜbkw3i gvb MeV GK‡K wbY©q Ki| †`Iqv Av‡Q, m ( ) 14 7 N = 14.00307 u; mp = 1.007825 u; mn = 1.008665 u [Easy] mgvavb: fiÎæwU, m = 7mp + 7mn – m = {(7 × 1.007825) + (7 × 1.008665) – 14.00307} u = 0.11236 u  eÜbkw3 = mc2 = 0.11236 × 931.5 MeV = 104.66334 MeV (Ans.) 34| 90 38 Sr Gi Aa©RxebKvj 28 eQi| GB AvB‡mvU‡ci 15mg Gi fvObnvi KZ? [Easy] mgvavb: dN dt = N = ln2 T1 2 × w M × NA