Question

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• If the wave number of the rotational transition / -0 → 1 of 'H'Br is 16.93 cm. a- Calculate the rotational constant B (Hz) b- Calculate the bond length of HBr (8) C- Calculate the energy of 1=5 → transition (J) d- If we deuterate HBr without affecting the bond length, what will happen to the position of the absorption peak?

Answer 1

AD we foy gretatrioal transiti n Know enangy Es Ta Manent of inerta da neduled mass of B J (JtH) ব the metecue B rotahionaf constant where 2 (n term 2J m, mi B (in Hz 872F unit ot frequerny B lin om) wavenumbes of estationd trans Hion ( r EgI - C Kottae) 26 (1) unitof Navenumber BO-DT)-T - 16.93 cm Here J-O 16.93 cm 1 16.93 Cm - 8.465 crn C -3.0x10ms B tn H2)=C x8 C=3.0x1do cnls B lin H2) = 3.0xlomls 8.465 cm B Gn th) =25.395 X10 s 2.54 x lo" H2 B linth) Hence Eattunt the grotational con stant Bi 2.54x10 H2

(b) 2.54X10" Hz - 2.54xlo Hz he6.626X1634S 8A2J 6.626 Xl034 Js 3.304x lo Ka m2 =I T ura= 3.304x 10 lu for't'lBr 3.304 xl017 11 을지. 66x16회 kg 82 I.66 X10-2 1+81 3.304 x82 -20 66Xlo x l0 m2 81 X1.66 2 r=2.015x10 20, 1.42 x1om (9r= 1-42 A Al010m Hence, tere bond lengfh C) of HBr ts 1-42 A ce) B=8.465 Cm ES-8.465 cs (st1)) 8-465 em X30 = 253.95 cn 8.465 CMC6 (6+1)) 8.465 cmt x 42 =355.53 cm 1 DESg-EC-S - 35s.53 cmm -253-95 cm lo1.58 cm '3나 3.0x10m)s xlo1.58cm ACS96 3)he ses6 6.625xlo" S DES96 (J)-2.019 x l 0-2T

Alernatively Fi'rst calculate tee wave number A of 596 transition using S16.93 cm x (s+1) 16-93 x6 cm - l01.58 cm EhcT IJt 6-626X103TS 30xidmls 1o1.3g cun -21 2. 019 X102J ES6 2.019 x102 Hance the energy of l - s6 cd) As mylaed H D we deluteriam the reduced mass of tee HBr u) I fncrease And, Hence Hence, tee valua of seatational BJa It Constant decreases. between tere peak. Hence deerease tee Wwhich gap Lere peak positon of the absorption peak decreases,