Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 37, Problem 31P

(a)

To determine

The moment of inertia and rotation energy.

(a)

Expert Solution
Check Mark

Answer to Problem 31P

The moment of inertia is 1.454×1046kgm2 and rotational energy is 0.2385meV .

Explanation of Solution

Given:

The equilibrium separation is r=0.113nm .

Formula used:

The expression for moment of inertia is given by,

  I=mCmOr2mC+mO

The expression for rotational energy is given by,

  E0r=22I

Calculation:

The moment of inertiais calculated as,

  I=mCmOr2mC+mO=( 12u)( 16u)( 12u)+( 16u)(( 0.113nm)( 10 9 m 1nm ))2(1.661× 10 27kg/u)=1.454×1046kgm2

The rotational energy is calculated as,

  E0r=22I=( 6.58× 10 16 eVs)( 1.602× 10 19 J 1eV )2( 1.454× 10 46 kg m 2 )=(( 0.2385× 10 3 eV)( 10 3 meV 1eV ))=0.2385meV

Conclusion:

Therefore, the moment of inertia is 1.454×1046kgm2 and rotational energy is 0.2385meV .

(b)

To determine

The energy level diagram.

(b)

Expert Solution
Check Mark

Answer to Problem 31P

The energy level diagram is shown in figure 1.

Explanation of Solution

Calculation:

The energy level diagram for the rotational level from l=0tol=5 is shown below,

  Physics for Scientists and Engineers, Chapter 37, Problem 31P

Figure 1

Conclusion:

Therefore, the energy level diagram is shown in figure 1.

(c)

To determine

The wavelength for each transition of part (b).

(c)

Expert Solution
Check Mark

Answer to Problem 31P

The wavelengths for each transition from start are 2.6×103m , 1.3×103m , 0.867×103m , 0.650×103m and 0.52×103m .

Explanation of Solution

Formula used:

The expression for wavelength is given by,

  λl,l1=hc2lΔE0r

Calculation:

The first wavelength is calculated as,

  λl,l1=hc2lΔE 0rλ1,0=( 4.136× 10 15 eVs)( 3.0× 10 8 m/s )2(1)( ( 0.2385meV )( 10 3 eV 1meV ))=2.6×103m

The second wavelength is calculated as,

  λl,l1=hc2lΔE 0rλ2,1=( 4.136× 10 15 eVs)( 3.0× 10 8 m/s )2(2)( ( 0.2385meV )( 10 3 eV 1meV ))=1.3×103m

The third wavelength is calculated as,

  λl,l1=hc2lΔE 0rλ3,2=( 4.136× 10 15 eVs)( 3.0× 10 8 m/s )2(3)( ( 0.2385meV )( 10 3 eV 1meV ))=0.867×103m

The fourth wavelength is calculated as,

  λl,l1=hc2lΔE 0rλ4,3=( 4.136× 10 15 eVs)( 3.0× 10 8 m/s )2(4)( ( 0.2385meV )( 10 3 eV 1meV ))=0.650×103m

The fifth wavelength is calculated as,

  λl,l1=hc2lΔE 0rλ5,4=( 4.136× 10 15 eVs)( 3.0× 10 8 m/s )2(5)( ( 0.2385meV )( 10 3 eV 1meV ))=0.52×103m

Conclusion:

Therefore, the wavelengths for each transition from start are 2.6×103m , 1.3×103m , 0.867×103m , 0.650×103m and 0.52×103m .

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