The moment of inertia and rotation energy.

Question

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Answer 1

Question

Chapter 37, Problem 31P

(a)

To determine

The moment of inertia and rotation energy.

(a)

Expert Solution

Answer to Problem 31P

The moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

Explanation of Solution

Given:

The equilibrium separation is r=0.113 nm .

Formula used:

The expression for moment of inertia is given by,

I=mCmOr2mC+mO

The expression for rotational energy is given by,

E0r=ℏ22I

Calculation:

The moment of inertiais calculated as,

I=mCmOr2mC+mO=( 12 u)( 16 u)( 12 u)+( 16 u)(( 0.113 nm)( 10 −9 m 1 nm ))2(1.661× 10 −27 kg/u)=1.454×10−46 kg⋅m2

The rotational energy is calculated as,

E0r=ℏ22I=( 6.58× 10 −16 eV⋅s)( 1.602× 10 −19 J 1 eV )2( 1.454× 10 −46 kg⋅ m 2 )=(( 0.2385× 10 −3 eV)( 10 3 meV 1 eV ))=0.2385 meV

Conclusion:

Therefore, the moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

(b)

To determine

The energy level diagram.

(b)

Expert Solution

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=0 to l=5 is shown below,

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, 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

Answer to Problem 31P

The wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

Explanation of Solution

Formula used:

The expression for wavelength is given by,

λl,l−1=hc2lΔE0r

Calculation:

The first wavelength is calculated as,

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

The second wavelength is calculated as,

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

The third wavelength is calculated as,

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

The fourth wavelength is calculated as,

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

The fifth wavelength is calculated as,

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

Conclusion:

Therefore, the wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

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Answer 2

Question

Chapter 37, Problem 31P

(a)

To determine

The moment of inertia and rotation energy.

(a)

Expert Solution

Answer to Problem 31P

The moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

Explanation of Solution

Given:

The equilibrium separation is r=0.113 nm .

Formula used:

The expression for moment of inertia is given by,

I=mCmOr2mC+mO

The expression for rotational energy is given by,

E0r=ℏ22I

Calculation:

The moment of inertiais calculated as,

I=mCmOr2mC+mO=( 12 u)( 16 u)( 12 u)+( 16 u)(( 0.113 nm)( 10 −9 m 1 nm ))2(1.661× 10 −27 kg/u)=1.454×10−46 kg⋅m2

The rotational energy is calculated as,

E0r=ℏ22I=( 6.58× 10 −16 eV⋅s)( 1.602× 10 −19 J 1 eV )2( 1.454× 10 −46 kg⋅ m 2 )=(( 0.2385× 10 −3 eV)( 10 3 meV 1 eV ))=0.2385 meV

Conclusion:

Therefore, the moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

(b)

To determine

The energy level diagram.

(b)

Expert Solution

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=0 to l=5 is shown below,

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, 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

Answer to Problem 31P

The wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

Explanation of Solution

Formula used:

The expression for wavelength is given by,

λl,l−1=hc2lΔE0r

Calculation:

The first wavelength is calculated as,

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

The second wavelength is calculated as,

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

The third wavelength is calculated as,

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

The fourth wavelength is calculated as,

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

The fifth wavelength is calculated as,

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

Conclusion:

Therefore, the wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

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Answer 3

Question

Chapter 37, Problem 31P

(a)

To determine

The moment of inertia and rotation energy.

(a)

Expert Solution

Answer to Problem 31P

The moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

Explanation of Solution

Given:

The equilibrium separation is r=0.113 nm .

Formula used:

The expression for moment of inertia is given by,

I=mCmOr2mC+mO

The expression for rotational energy is given by,

E0r=ℏ22I

Calculation:

The moment of inertiais calculated as,

I=mCmOr2mC+mO=( 12 u)( 16 u)( 12 u)+( 16 u)(( 0.113 nm)( 10 −9 m 1 nm ))2(1.661× 10 −27 kg/u)=1.454×10−46 kg⋅m2

The rotational energy is calculated as,

E0r=ℏ22I=( 6.58× 10 −16 eV⋅s)( 1.602× 10 −19 J 1 eV )2( 1.454× 10 −46 kg⋅ m 2 )=(( 0.2385× 10 −3 eV)( 10 3 meV 1 eV ))=0.2385 meV

Conclusion:

Therefore, the moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

(b)

To determine

The energy level diagram.

(b)

Expert Solution

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=0 to l=5 is shown below,

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, 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

Answer to Problem 31P

The wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

Explanation of Solution

Formula used:

The expression for wavelength is given by,

λl,l−1=hc2lΔE0r

Calculation:

The first wavelength is calculated as,

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

The second wavelength is calculated as,

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

The third wavelength is calculated as,

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

The fourth wavelength is calculated as,

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

The fifth wavelength is calculated as,

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

Conclusion:

Therefore, the wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

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Answer 4

Question

Chapter 37, Problem 31P

(a)

To determine

The moment of inertia and rotation energy.

(a)

Expert Solution

Answer to Problem 31P

The moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

Explanation of Solution

Given:

The equilibrium separation is r=0.113 nm .

Formula used:

The expression for moment of inertia is given by,

I=mCmOr2mC+mO

The expression for rotational energy is given by,

E0r=ℏ22I

Calculation:

The moment of inertiais calculated as,

I=mCmOr2mC+mO=( 12 u)( 16 u)( 12 u)+( 16 u)(( 0.113 nm)( 10 −9 m 1 nm ))2(1.661× 10 −27 kg/u)=1.454×10−46 kg⋅m2

The rotational energy is calculated as,

E0r=ℏ22I=( 6.58× 10 −16 eV⋅s)( 1.602× 10 −19 J 1 eV )2( 1.454× 10 −46 kg⋅ m 2 )=(( 0.2385× 10 −3 eV)( 10 3 meV 1 eV ))=0.2385 meV

Conclusion:

Therefore, the moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

(b)

To determine

The energy level diagram.

(b)

Expert Solution

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=0 to l=5 is shown below,

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, 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

Answer to Problem 31P

The wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

Explanation of Solution

Formula used:

The expression for wavelength is given by,

λl,l−1=hc2lΔE0r

Calculation:

The first wavelength is calculated as,

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

The second wavelength is calculated as,

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

The third wavelength is calculated as,

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

The fourth wavelength is calculated as,

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

The fifth wavelength is calculated as,

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

Conclusion:

Therefore, the wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

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Answer 5

Chapter 37, Problem 31P

(a)

To determine

The moment of inertia and rotation energy.

(a)

Expert Solution

Answer to Problem 31P

The moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

Explanation of Solution

Given:

The equilibrium separation is r=0.113 nm .

Formula used:

The expression for moment of inertia is given by,

I=mCmOr2mC+mO

The expression for rotational energy is given by,

E0r=ℏ22I

Calculation:

The moment of inertiais calculated as,

I=mCmOr2mC+mO=( 12 u)( 16 u)( 12 u)+( 16 u)(( 0.113 nm)( 10 −9 m 1 nm ))2(1.661× 10 −27 kg/u)=1.454×10−46 kg⋅m2

The rotational energy is calculated as,

E0r=ℏ22I=( 6.58× 10 −16 eV⋅s)( 1.602× 10 −19 J 1 eV )2( 1.454× 10 −46 kg⋅ m 2 )=(( 0.2385× 10 −3 eV)( 10 3 meV 1 eV ))=0.2385 meV

Conclusion:

Therefore, the moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

(b)

To determine

The energy level diagram.

(b)

Expert Solution

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=0 to l=5 is shown below,

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, 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

Answer to Problem 31P

The wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

Explanation of Solution

Formula used:

The expression for wavelength is given by,

λl,l−1=hc2lΔE0r

Calculation:

The first wavelength is calculated as,

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

The second wavelength is calculated as,

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

The third wavelength is calculated as,

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

The fourth wavelength is calculated as,

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

The fifth wavelength is calculated as,

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

Conclusion:

Therefore, the wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

Answer 6

Chapter 37, Problem 31P

(a)

To determine

The moment of inertia and rotation energy.

(a)

Expert Solution

Answer to Problem 31P

The moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

Explanation of Solution

Given:

The equilibrium separation is r=0.113 nm .

Formula used:

The expression for moment of inertia is given by,

I=mCmOr2mC+mO

The expression for rotational energy is given by,

E0r=ℏ22I

Calculation:

The moment of inertiais calculated as,

I=mCmOr2mC+mO=( 12 u)( 16 u)( 12 u)+( 16 u)(( 0.113 nm)( 10 −9 m 1 nm ))2(1.661× 10 −27 kg/u)=1.454×10−46 kg⋅m2

The rotational energy is calculated as,

E0r=ℏ22I=( 6.58× 10 −16 eV⋅s)( 1.602× 10 −19 J 1 eV )2( 1.454× 10 −46 kg⋅ m 2 )=(( 0.2385× 10 −3 eV)( 10 3 meV 1 eV ))=0.2385 meV

Conclusion:

Therefore, the moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

Answer 7

Chapter 37, Problem 31P

(a)

To determine

The moment of inertia and rotation energy.

Answer 8

(a)

Expert Solution

Answer to Problem 31P

The moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The equilibrium separation is r=0.113 nm .

Formula used:

The expression for moment of inertia is given by,

I=mCmOr2mC+mO

The expression for rotational energy is given by,

E0r=ℏ22I

Calculation:

The moment of inertiais calculated as,

I=mCmOr2mC+mO=( 12 u)( 16 u)( 12 u)+( 16 u)(( 0.113 nm)( 10 −9 m 1 nm ))2(1.661× 10 −27 kg/u)=1.454×10−46 kg⋅m2

The rotational energy is calculated as,

E0r=ℏ22I=( 6.58× 10 −16 eV⋅s)( 1.602× 10 −19 J 1 eV )2( 1.454× 10 −46 kg⋅ m 2 )=(( 0.2385× 10 −3 eV)( 10 3 meV 1 eV ))=0.2385 meV

Conclusion:

Therefore, the moment of inertia is 1.454×10−46 kg⋅m2 and rotational energy is 0.2385 meV .

Answer 13

(b)

To determine

The energy level diagram.

(b)

Expert Solution

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=0 to l=5 is shown below,

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 37, Problem 31P

Figure 1

Conclusion:

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

Answer 14

(b)

To determine

The energy level diagram.

Answer 15

(b)

Expert Solution

Answer to Problem 31P

The energy level diagram is shown in figure 1.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

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

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 37, Problem 31P

Figure 1

Conclusion:

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

Answer 20

(c)

To determine

The wavelength for each transition of part (b).

(c)

Expert Solution

Answer to Problem 31P

The wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

Explanation of Solution

Formula used:

The expression for wavelength is given by,

λl,l−1=hc2lΔE0r

Calculation:

The first wavelength is calculated as,

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

The second wavelength is calculated as,

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

The third wavelength is calculated as,

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

The fourth wavelength is calculated as,

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

The fifth wavelength is calculated as,

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

Conclusion:

Therefore, the wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

Answer 21

(c)

To determine

The wavelength for each transition of part (b).

Answer 22

(c)

Expert Solution

Answer to Problem 31P

The wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Formula used:

The expression for wavelength is given by,

λl,l−1=hc2lΔE0r

Calculation:

The first wavelength is calculated as,

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

The second wavelength is calculated as,

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

The third wavelength is calculated as,

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

The fourth wavelength is calculated as,

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

The fifth wavelength is calculated as,

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

Conclusion:

Therefore, the wavelengths for each transition from start are 2.6×10−3 m , 1.3×10−3 m , 0.867×10−3 m , 0.650×10−3 m and 0.52×10−3 m .

Answer 27

Ch. 37 - Prob. 1P Ch. 37 - Prob. 2P Ch. 37 - Prob. 3P Ch. 37 - Prob. 4P Ch. 37 - Prob. 5P Ch. 37 - Prob. 6P Ch. 37 - Prob. 7P Ch. 37 - Prob. 8P Ch. 37 - Prob. 9P Ch. 37 - Prob. 10P

The moment of inertia and rotation energy.

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The moment of inertia and rotation energy.

Answer to Problem 31P

Explanation of Solution

The energy level diagram.

Answer to Problem 31P

Explanation of Solution

The wavelength for each transition of part (b).

Answer to Problem 31P

Explanation of Solution

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