The frequency of oscillation of several diatomic molecules isas follows: a. HF 1
ID: 684942 • Letter: T
Question
The frequency of oscillation of several diatomic molecules isas follows: a. HF 1.24*1014s-1 b. HCl 89.665*1013s-1 c. HBr 7.9414*1013s-1 d.Br2 9.7528*1012s-1 Find the force constant for the bond in each molecule. frequency v=1/2(pi)*(k/m)1/2 reduced mass=(m1m2)/(m1+m2) I've redone this problem several times and I'm off by oneorder of magnitude every time with each. I need to find out wherein the process I'm going wrong. The frequency of oscillation of several diatomic molecules isas follows: a. HF 1.24*1014s-1 b. HCl 89.665*1013s-1 c. HBr 7.9414*1013s-1 d.Br2 9.7528*1012s-1 Find the force constant for the bond in each molecule. frequency v=1/2(pi)*(k/m)1/2 reduced mass=(m1m2)/(m1+m2) I've redone this problem several times and I'm off by oneorder of magnitude every time with each. I need to find out wherein the process I'm going wrong.Explanation / Answer
Frequency , v = ( 1/ 2 ) ( k / m ) a ) Frequency of HF is 1.24*1014s-1 m 1 = mass of F = 19 g m 2 = mass of H = 1 g Reduced mass ,m = ( m 1 * m 2 ) / (m1 + m2 ) = 19 * 1 / ( 19 + 1 ) = 19 / 20 g v = ( 1/ 2 ) ( k / m ) From this Force constant , k = v 2 *4 2 * m 2 = ( 1.24*1014 ) 2 * 4 * 3.142 * ( 19 / 20 ) 2 = 54.7280 * 10 28 dynes / cm 3 b ) Frequency of HCl is89.665*1013s-1 m 1 = mass of Cl= 35.5 g m 2 = mass of H = 1 g Reduced mass ,m = ( m 1 * m 2 ) / (m1 + m2 ) = 35.5 * 1 / ( 35.5 + 1 ) = 35.5 / 36.5 g v = ( 1/ 2 ) ( k / m ) From this Force constant , k = v 2 *4 2 * m 2 = 299941.23 * 1026 dynes / cm 3 c )Frequencyof HBr 7.9414*1013s-1 m1 = mass of Br= 80 g m2 = mass of H = 1 g Reduced mass ,m = ( m1 * m 2 ) / ( m1 + m2) = 80 / 81 g v = ( 1/ 2 )( k / m ) From thisForce constant , k = v 2 * 4 2 * m2 = 2426.18 * 10 26 dynes/ cm 3 d)Frequency of Br29.7528*1012s-1 m 1 = massof Br= 80 g m 2 = massof Br = 80 g Reduced mass ,m = ( m 1 * m2 ) / ( m1 + m2) = 40g v = ( 1/ 2 ) ( k / m) From this Force constant , k= v 2 * 4 2 * m2 = 6002026.47* 10 24 dynes / cm3 = 54.7280 * 10 28 dynes / cm 3 b ) Frequency of HCl is89.665*1013s-1 m 1 = mass of Cl= 35.5 g m 2 = mass of H = 1 g Reduced mass ,m = ( m 1 * m 2 ) / (m1 + m2 ) = 35.5 * 1 / ( 35.5 + 1 ) = 35.5 / 36.5 g v = ( 1/ 2 ) ( k / m ) From this Force constant , k = v 2 *4 2 * m 2 = 299941.23 * 1026 dynes / cm 3 c )Frequencyof HBr 7.9414*1013s-1 m1 = mass of Br= 80 g m2 = mass of H = 1 g Reduced mass ,m = ( m1 * m 2 ) / ( m1 + m2) = 80 / 81 g v = ( 1/ 2 )( k / m ) From thisForce constant , k = v 2 * 4 2 * m2 = 2426.18 * 10 26 dynes/ cm 3 d)Frequency of Br29.7528*1012s-1 m 1 = massof Br= 80 g m 2 = massof Br = 80 g Reduced mass ,m = ( m 1 * m2 ) / ( m1 + m2) = 40g v = ( 1/ 2 ) ( k / m) From this Force constant , k= v 2 * 4 2 * m2 = 6002026.47* 10 24 dynes / cm3 c )Frequencyof HBr 7.9414*1013s-1 m1 = mass of Br= 80 g m2 = mass of H = 1 g Reduced mass ,m = ( m1 * m 2 ) / ( m1 + m2) = 80 / 81 g v = ( 1/ 2 )( k / m ) From thisForce constant , k = v 2 * 4 2 * m2 = 2426.18 * 10 26 dynes/ cm 3 d)Frequency of Br29.7528*1012s-1 m 1 = massof Br= 80 g m 2 = massof Br = 80 g Reduced mass ,m = ( m 1 * m2 ) / ( m1 + m2) = 40g v = ( 1/ 2 ) ( k / m) From this Force constant , k= v 2 * 4 2 * m2 = 6002026.47* 10 24 dynes / cm3 d)Frequency of Br29.7528*1012s-1 m 1 = massof Br= 80 g m 2 = massof Br = 80 g Reduced mass ,m = ( m 1 * m2 ) / ( m1 + m2) = 40g v = ( 1/ 2 ) ( k / m) From this Force constant , k= v 2 * 4 2 * m2 = 6002026.47* 10 24 dynes / cm3 m 1 = massof Br= 80 g m 2 = massof Br = 80 g Reduced mass ,m = ( m 1 * m2 ) / ( m1 + m2) = 40g v = ( 1/ 2 ) ( k / m) From this Force constant , k= v 2 * 4 2 * m2 = 6002026.47* 10 24 dynes / cm3Related Questions
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