Tiêu đề 2. Engg. Practice Sheet-(1st Paper) With Solve.pdf ✅

̧YMZ imvqb  Engineering Practice Sheet ..................................................................................................................... 3 14. cÖavb †Kvqv›Uvg msL ̈v n = 3 n‡j Ab ̈vb ̈ †Kvqv›Uvg msL ̈v ̧‡jv wK wK? [BUET 14-15] [Easy] mgvavb: n = 3 l = 0, 1, 2 m = 0, (+ 1, 0, – 1), (+ 2, + 1, 0, – 1, – 2) s =    +  1 2  – 1 2 15. †ev‡ii cigvYy g‡Wj Abymv‡i wbDwK¬qv‡mi PZzw`©‡K N~Y©vqgvb B‡jKUa‡bi †K.wYK fi‡eM Gi mgxKiY wjL| [BUET 14-15] [Easy] mgvavb: mvr = nh 2 ; n = 1, 2, 3 ......... [h = cøv‡1⁄4i aaæeK; r = n Zg Kÿc‡_i e ̈vmva©] 16. -iwk¥ KYvi Av‡cwÿK fi k~b ̈| †Kvb †ZRw ̄...q †g.j †_‡K -iwk¥ wbM©Z n‡j †g.jwUi cvigvYweK msL ̈v GK GKK e„w× cvq wKš‧ cvigvYweK fi GKB _v‡K| GwU wKfv‡e m¤¢e? [BUET 13-14] [Medium] mgvavb: -KYv nj g~jZ B‡jKUa‡bi cÖevn|  KYvi PvR© – 1 aiv nq Ges Gi fi B‡jKUa‡bi b ̈vq AwZ bMY ̈| ZvB †Kv‡bv †g.j †_‡K  KYv wbtmwiZ n‡j 1wU wbDUab †f‡O †cÖvU‡b cwiYZ nq| d‡j †gvU fi msL ̈v GKB _v‡K wKš‘ cvigvYweK msL ̈v 1 GKK ev‡o| †hgb: 232 90 Th – 0 –1 e  232 91 Pa 17. GKwU B‡jKwUaK evj¦ njy` is‡qi Av‡jv (wewKiY) Qovq hvi Zi1⁄2‣`N ̈© 589 nm, wb‡ÿwcZ †dvU‡bi kw3 KZ Ryj? (†`Iqv Av‡Q h = 6.624  10–34 J s) [BUET 06-07] [Easy] mgvavb:  = 589 nm = 589  10–9 m E = h = h c  = 6.624  10–34  3  108 589  10–9 = 3.374  10–19 J 18. hLb mnKvix †Kvqv›Uvg msL ̈v, l = 3, ZLb Ab ̈vb ̈ †Kvqv›Uvg msL ̈v ̧wj wK wK? [BUET 06-07] [Medium] mgvavb: mnKvix †Kvqv›Uvg msL ̈v cÖavb †Kvqv›Uvg msL ̈v g ̈vM‡bwUK †Kvqv›Uvg msL ̈v w ̄úb †Kvqv›Uvg msL ̈v l = 3 n = 4, 5, 6,.... m = 0,  1,  2,  3 s =  1 2 19. û‡Ûi wbqg wjL| [BUET 06-07] [Easy] mgvavb: GKB kw3m¤úbœ B‡jKUab ̧‡jv wewfbœ AiweUv‡j me©vwaK msL ̈K we‡Rvo Ae ̄’vq _vK‡Z n‡e| GB me AhyM¥ B‡jKUa‡bi w ̄úb GKBgyLx n‡e| 20. k~b ̈ ̄ vb c~iY Ki| jyB wW-eaMwj cÖ`Ë  mgxKiYwU B‡jKUa‡bi KYv I Zi1⁄2 m¤úwK©Z| [BUET 04-05] [Easy] mgvavb:  = h mv  = B‡jKUa‡bi Zi1⁄2 •`N© ̈ h = cøv‡1⁄4i aaæeK m = B‡jKUa‡bi fi v = B‡jKUa‡bi †eM 21. 25C ZvcgvÎvq AgCl Gi 100 mL m¤ú„3 `ae‡Yi mv‡_ 0.03 M NaBr Gi 100 mL `aeY †hvM Kiv n‡jv| Aat‡ÿc •Zwi n‡e wK? [Ksp(AgCl) = 1  10–10; Ksp(AgBr) = 5  10–13] [BUET 04-05] [Medium] mgvavb: AgCl ⇌Ag+ + Cl – Ksp = S2 = 1 × 10–10  S = Ksp = 1  10–10 = 10–5 M  wewμqvi ïiæ‡Z, [Ag+ ] = [Cl – ] = 10–5 mol/L Avevi, NaBr  Na+ + Br–  wewμqvi ïiæ‡Z, [Br– ] = [Na+ ] = 0.03 mol/L Kip(AgBr) = [Ag+ ] × [Br– ] =     10–5 × 100 200 ×     0.03 × 100 200 = 7.5 × 10–8 †h‡nZzKip > Ksp myZivs, AgBr Gi Aat‡ÿc co‡e| 22. GKwU †g.‡ji cigvYy ̧wji wb¤œwjwLZ cÖvK...wZK cÖvPzh© I AvB‡mv‡UvwcK fi i‡q‡Q| †g.jwUi Mo cvigvYweK fi wnmve Ki| cÖvPzh© : 90.92% 0.26% 8.82% cvigvYweK fi : 19.99 amu 20.99 amu 21.99 amu [BUET 02-03] [Easy] mgvavb: Mo cvigvYweK fi = 90.92  19.99 + 0.26  20.99 + 8.82  21.99 100 = 20.169 amu 23. †μvwgqv‡gi B‡jKUab web ̈vm wjL| [BUET 01-02] [Easy] mgvavb: Cr(24) = 1s2 2s2 2p6 3s2 3p6 3d5 4s1