EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 37, Problem 24P

(a)

To determine

The reduced mass of a diatomic molecule is always smaller than the mass of the molecule and to calculate the reduced mass for H2 .

(a)

Expert Solution
Check Mark

Answer to Problem 24P

Thereduced mass of H2 is 0.500u .

Explanation of Solution

Formula used:

The expression for reduced mass of a two body system is given by,

  μ=m1m2m1+m2

Calculation:

Dividing the expression for reduced mass of a two body systemby m2 ,

  μ=m1m2m1+m2=m11+ m 1 m 2

Again, dividing the expression for reduced mass of a two body system by m1

  μ=m1m2m1+m2=m21+ m 2 m 1

In both case, the denominator is greater than 1 which implies

  μ<m1 and μ<m2

The reduced mass of H2 is calculated as,

  μ=m1m2m1+m2=( 1u)( 1u)( 1u)+( 1u)=0.500u

Conclusion:

Therefore, thereduced mass of H2 is 0.500u .

(b)

To determine

The reduced mass of N2 .

(b)

Expert Solution
Check Mark

Answer to Problem 24P

The reduced mass of N2 is 7.00u .

Explanation of Solution

Calculation:

The reduced mass of N2 is calculated as,

  μ=m1m2m1+m2=( 14u)( 14u)( 14u)+( 14u)=7.00u

Conclusion:

Therefore, the reduced mass of N2 is 7.00u .

(c)

To determine

The reduced mass of CO .

(c)

Expert Solution
Check Mark

Answer to Problem 24P

The reduced mass of CO is 6.86u .

Explanation of Solution

Calculation:

The reduced mass of CO is calculated as,

  μ=m1m2m1+m2=( 12u)( 16u)( 12u)+( 16u)=6.86u

Conclusion:

Therefore, the reduced mass of CO is 6.86u .

(d)

To determine

The reduced mass of HCl .

(d)

Expert Solution
Check Mark

Answer to Problem 24P

The reduced mass of HCl is 0.973u .

Explanation of Solution

Calculation:

The reduced mass of HCl is calculated as,

  μ=m1m2m1+m2=( 1u)( 35.5u)( 1u)+( 35.5u)=0.973u

Conclusion:

Therefore, the reduced mass of HCl is 0.973u .

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