Question

The distance between the centers of the H and Cl atoms in the HCI molecule is approximately 0.127 nm. (Assume the molecule consists of hydrogen-1 and chlorine-35 isotopes.) (a) Find the angular velocity (in rad/s) of the molecule about its center of mass when l = 1 and l = 6. l = 1 rad/s f = 6 rad/s w = (b) What is the speed (in m/s) of the H atom in each of the cases in (a)? l = 1 V = m/s l = 6 m/s V = (c) What value of l is required for the H atom to have a speed of 0.140c? (d) Estimate the classical temperature (in K) associated with the l you found in (c). K

Answer 1

Solutions - (a) The angulis velocity of linecel diatomic mek cele is calculsekeel be equeating quantum mechanical rotectional energy of mokeuke to the classical rotational energy. The reduced mass m fos diatermic molecule * Rin lla m. m. mitm o o Here m mm are the masses of the tuo atems of diatemir mokcule The moment of inertia of diatemic molente Is UR ² Ray Intesatomic distance. The quantum - mechanical Rotectional Encegy is given by қ°4 (4+1) 0 21 The classical - Rotectional energy is given by € = /102 href) / 1 wh 21 Scanned by TapScanner

202 202f1) Alldal) - 21 2 Given that Intermokeulas distance for her is 0.127 mm R = 0.127 x 109m 2 all ? mm mifm 1835/C6-023 23) = 1.6 14 x 1024g (14 35)/6-0234123 = 16 14 %10 27 kg :: Da dip?= 1614856* (0-1974109)? = 2.60 x1067 ram? D fos lal Cross x1034120)(+1). 2 (2.60 x1067) 12 w = 8.7341012 rad (see fos da 6 wa J Rossxie-34)*x 6 x (6+1) (2.60 x1047)2 13 = 2.62x100 sadfree. Scanned by TapScanner

(6) The lineas relocity of the atom about centre of mass. v=w&cm (H) autem. The distance of centre of mass from the Hydrogen Hemer) minton m, fm ma X (m() R met ma 35 g 0.127x10 9 1+35 & m) - 1.234 X1010 m Scanned by TapScanner

Velocity of H aton for dit is V=W, Xcmore) 5-73410"x 1234 310 10 = 707.0 m/sec velocity of H atom for da6 Ve hemery 10 262710 x 1 234810 = 32 33.0 m/fle. (el as, w= x ²l (del) 12 w? blodfl) D2 w21² h2 ecdf1) = Com's t² dlef) = 0 1140434108 1:243 x1 Oto (2-60 xeb47)? Cross x1034)2 Ildf) e ²4 l - 6-93 x 18 = 6-93 x 109 Scanned by TapScanner

2 – bf R-Lac 2a e = - 1€ +291% 6-93x109 2xl I = 8.3246x106 of -8.3249x cot ectational quantum humbes cannot be negative l= 8.824x10s is required d) we know that Energy astoiated with per degree of freedom in E=% KT for dicatemic 2 degree of from :.£= 12x2xKTEKT compairing KT with herefly KT = merd fl) 21 T: mer8 !) 21k Scanned by TapScanner

T = (1-055 x 1734)?x 8.374 x 164 x (83748104+1) 2x (2-60 X104) (138 yio 39 T= 1.0767x10 T= 1.0767 yolk Scanned by TapScanner