The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

Question

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Answer 1

Question

Chapter 42, Problem 14P

(a)

To determine

The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

(a)

Expert Solution

Answer to Problem 14P

The moment of inertia is 1.88×10−45 kg⋅m2.

Explanation of Solution

Write the equation to find the moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

I=6mcrc2+6mhrh2

Here, I is the moment of inertia, mc is the mass of carbon atom, rc is the distance from carbon atom to point O, mh is the mass of hydrogen atom, and rh is the distance from hydrogen atom to point O.

Conclusion:

Substitute 1.99×10−26 kg for mc, 0.110 nm for rc, 1.67×10−27 kg for mh, and 0.210 nm for rh in the above equation to find I.

I=6(1.99×10−26 kg)(0.110 nm(10−9 m1 nm))2+6(1.67×10−27 kg)(0.210 nm(10−9 m1 nm))2=1.44×10−45 kg⋅m2+0.442×10−45 kg⋅m2=1.88×10−45 kg⋅m2

Therefore, the moment of inertia is 1.88×10−45 kg⋅m2.

(b)

To determine

The possible rotational energies about the axis.

(b)

Expert Solution

Answer to Problem 14P

The possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

Explanation of Solution

Write the equation to calculate the rotational energy.

Erot=ℏ22IJ(J+1) (I)

Here, Erot is the rotational energy, ℏ is the reduced Plank’s constant, I is the rotational moment of inertia, and J is the rotational quantum umber.

Conclusion:

Substitute 1.055×10−34 J⋅s for ℏ and 1.88×10−45 kg⋅m2 for I in the equation for Erot.

Erot=(1.055×10−34 J⋅s)22(1.88×10−45 kg⋅m2)J(J+1)=(2.95×10−24 J(1.6×10−19 eV1 J))J(J+1)=(18.4×10−6 eV)J(J+1)

The possible values of J are 0,1,2,3,4......

Substitute 0 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(0(0+1))=0 eV

Substitute 1 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(1(1+1))=36.9×10−6 eV(1 μeV10−6 eV)=36.9 μeV

Substitute 2 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(2(2+1))=111×10−6 eV(1 μeV10−6 eV)=111 μeV

Substitute 3 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(3(3+1))=221×10−6 eV(1 μeV10−6 eV)=221 μeV

Substitute 4 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(4(4+1))=369×10−6 eV(1 μeV10−6 eV)=369 μeV

Therefore, the possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

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Answer 2

Question

Chapter 42, Problem 14P

(a)

To determine

The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

(a)

Expert Solution

Answer to Problem 14P

The moment of inertia is 1.88×10−45 kg⋅m2.

Explanation of Solution

Write the equation to find the moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

I=6mcrc2+6mhrh2

Here, I is the moment of inertia, mc is the mass of carbon atom, rc is the distance from carbon atom to point O, mh is the mass of hydrogen atom, and rh is the distance from hydrogen atom to point O.

Conclusion:

Substitute 1.99×10−26 kg for mc, 0.110 nm for rc, 1.67×10−27 kg for mh, and 0.210 nm for rh in the above equation to find I.

I=6(1.99×10−26 kg)(0.110 nm(10−9 m1 nm))2+6(1.67×10−27 kg)(0.210 nm(10−9 m1 nm))2=1.44×10−45 kg⋅m2+0.442×10−45 kg⋅m2=1.88×10−45 kg⋅m2

Therefore, the moment of inertia is 1.88×10−45 kg⋅m2.

(b)

To determine

The possible rotational energies about the axis.

(b)

Expert Solution

Answer to Problem 14P

The possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

Explanation of Solution

Write the equation to calculate the rotational energy.

Erot=ℏ22IJ(J+1) (I)

Here, Erot is the rotational energy, ℏ is the reduced Plank’s constant, I is the rotational moment of inertia, and J is the rotational quantum umber.

Conclusion:

Substitute 1.055×10−34 J⋅s for ℏ and 1.88×10−45 kg⋅m2 for I in the equation for Erot.

Erot=(1.055×10−34 J⋅s)22(1.88×10−45 kg⋅m2)J(J+1)=(2.95×10−24 J(1.6×10−19 eV1 J))J(J+1)=(18.4×10−6 eV)J(J+1)

The possible values of J are 0,1,2,3,4......

Substitute 0 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(0(0+1))=0 eV

Substitute 1 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(1(1+1))=36.9×10−6 eV(1 μeV10−6 eV)=36.9 μeV

Substitute 2 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(2(2+1))=111×10−6 eV(1 μeV10−6 eV)=111 μeV

Substitute 3 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(3(3+1))=221×10−6 eV(1 μeV10−6 eV)=221 μeV

Substitute 4 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(4(4+1))=369×10−6 eV(1 μeV10−6 eV)=369 μeV

Therefore, the possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

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Answer 3

Question

Chapter 42, Problem 14P

(a)

To determine

The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

(a)

Expert Solution

Answer to Problem 14P

The moment of inertia is 1.88×10−45 kg⋅m2.

Explanation of Solution

Write the equation to find the moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

I=6mcrc2+6mhrh2

Here, I is the moment of inertia, mc is the mass of carbon atom, rc is the distance from carbon atom to point O, mh is the mass of hydrogen atom, and rh is the distance from hydrogen atom to point O.

Conclusion:

Substitute 1.99×10−26 kg for mc, 0.110 nm for rc, 1.67×10−27 kg for mh, and 0.210 nm for rh in the above equation to find I.

I=6(1.99×10−26 kg)(0.110 nm(10−9 m1 nm))2+6(1.67×10−27 kg)(0.210 nm(10−9 m1 nm))2=1.44×10−45 kg⋅m2+0.442×10−45 kg⋅m2=1.88×10−45 kg⋅m2

Therefore, the moment of inertia is 1.88×10−45 kg⋅m2.

(b)

To determine

The possible rotational energies about the axis.

(b)

Expert Solution

Answer to Problem 14P

The possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

Explanation of Solution

Write the equation to calculate the rotational energy.

Erot=ℏ22IJ(J+1) (I)

Here, Erot is the rotational energy, ℏ is the reduced Plank’s constant, I is the rotational moment of inertia, and J is the rotational quantum umber.

Conclusion:

Substitute 1.055×10−34 J⋅s for ℏ and 1.88×10−45 kg⋅m2 for I in the equation for Erot.

Erot=(1.055×10−34 J⋅s)22(1.88×10−45 kg⋅m2)J(J+1)=(2.95×10−24 J(1.6×10−19 eV1 J))J(J+1)=(18.4×10−6 eV)J(J+1)

The possible values of J are 0,1,2,3,4......

Substitute 0 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(0(0+1))=0 eV

Substitute 1 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(1(1+1))=36.9×10−6 eV(1 μeV10−6 eV)=36.9 μeV

Substitute 2 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(2(2+1))=111×10−6 eV(1 μeV10−6 eV)=111 μeV

Substitute 3 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(3(3+1))=221×10−6 eV(1 μeV10−6 eV)=221 μeV

Substitute 4 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(4(4+1))=369×10−6 eV(1 μeV10−6 eV)=369 μeV

Therefore, the possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

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Answer 4

Question

Chapter 42, Problem 14P

(a)

To determine

The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

(a)

Expert Solution

Answer to Problem 14P

The moment of inertia is 1.88×10−45 kg⋅m2.

Explanation of Solution

Write the equation to find the moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

I=6mcrc2+6mhrh2

Here, I is the moment of inertia, mc is the mass of carbon atom, rc is the distance from carbon atom to point O, mh is the mass of hydrogen atom, and rh is the distance from hydrogen atom to point O.

Conclusion:

Substitute 1.99×10−26 kg for mc, 0.110 nm for rc, 1.67×10−27 kg for mh, and 0.210 nm for rh in the above equation to find I.

I=6(1.99×10−26 kg)(0.110 nm(10−9 m1 nm))2+6(1.67×10−27 kg)(0.210 nm(10−9 m1 nm))2=1.44×10−45 kg⋅m2+0.442×10−45 kg⋅m2=1.88×10−45 kg⋅m2

Therefore, the moment of inertia is 1.88×10−45 kg⋅m2.

(b)

To determine

The possible rotational energies about the axis.

(b)

Expert Solution

Answer to Problem 14P

The possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

Explanation of Solution

Write the equation to calculate the rotational energy.

Erot=ℏ22IJ(J+1) (I)

Here, Erot is the rotational energy, ℏ is the reduced Plank’s constant, I is the rotational moment of inertia, and J is the rotational quantum umber.

Conclusion:

Substitute 1.055×10−34 J⋅s for ℏ and 1.88×10−45 kg⋅m2 for I in the equation for Erot.

Erot=(1.055×10−34 J⋅s)22(1.88×10−45 kg⋅m2)J(J+1)=(2.95×10−24 J(1.6×10−19 eV1 J))J(J+1)=(18.4×10−6 eV)J(J+1)

The possible values of J are 0,1,2,3,4......

Substitute 0 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(0(0+1))=0 eV

Substitute 1 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(1(1+1))=36.9×10−6 eV(1 μeV10−6 eV)=36.9 μeV

Substitute 2 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(2(2+1))=111×10−6 eV(1 μeV10−6 eV)=111 μeV

Substitute 3 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(3(3+1))=221×10−6 eV(1 μeV10−6 eV)=221 μeV

Substitute 4 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(4(4+1))=369×10−6 eV(1 μeV10−6 eV)=369 μeV

Therefore, the possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

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Answer 5

Chapter 42, Problem 14P

(a)

To determine

The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

(a)

Expert Solution

Answer to Problem 14P

The moment of inertia is 1.88×10−45 kg⋅m2.

Explanation of Solution

Write the equation to find the moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

I=6mcrc2+6mhrh2

Here, I is the moment of inertia, mc is the mass of carbon atom, rc is the distance from carbon atom to point O, mh is the mass of hydrogen atom, and rh is the distance from hydrogen atom to point O.

Conclusion:

Substitute 1.99×10−26 kg for mc, 0.110 nm for rc, 1.67×10−27 kg for mh, and 0.210 nm for rh in the above equation to find I.

I=6(1.99×10−26 kg)(0.110 nm(10−9 m1 nm))2+6(1.67×10−27 kg)(0.210 nm(10−9 m1 nm))2=1.44×10−45 kg⋅m2+0.442×10−45 kg⋅m2=1.88×10−45 kg⋅m2

Therefore, the moment of inertia is 1.88×10−45 kg⋅m2.

(b)

To determine

The possible rotational energies about the axis.

(b)

Expert Solution

Answer to Problem 14P

The possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

Explanation of Solution

Write the equation to calculate the rotational energy.

Erot=ℏ22IJ(J+1) (I)

Here, Erot is the rotational energy, ℏ is the reduced Plank’s constant, I is the rotational moment of inertia, and J is the rotational quantum umber.

Conclusion:

Substitute 1.055×10−34 J⋅s for ℏ and 1.88×10−45 kg⋅m2 for I in the equation for Erot.

Erot=(1.055×10−34 J⋅s)22(1.88×10−45 kg⋅m2)J(J+1)=(2.95×10−24 J(1.6×10−19 eV1 J))J(J+1)=(18.4×10−6 eV)J(J+1)

The possible values of J are 0,1,2,3,4......

Substitute 0 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(0(0+1))=0 eV

Substitute 1 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(1(1+1))=36.9×10−6 eV(1 μeV10−6 eV)=36.9 μeV

Substitute 2 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(2(2+1))=111×10−6 eV(1 μeV10−6 eV)=111 μeV

Substitute 3 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(3(3+1))=221×10−6 eV(1 μeV10−6 eV)=221 μeV

Substitute 4 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(4(4+1))=369×10−6 eV(1 μeV10−6 eV)=369 μeV

Therefore, the possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

Answer 6

Chapter 42, Problem 14P

(a)

To determine

The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

(a)

Expert Solution

Answer to Problem 14P

The moment of inertia is 1.88×10−45 kg⋅m2.

Explanation of Solution

Write the equation to find the moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

I=6mcrc2+6mhrh2

Here, I is the moment of inertia, mc is the mass of carbon atom, rc is the distance from carbon atom to point O, mh is the mass of hydrogen atom, and rh is the distance from hydrogen atom to point O.

Conclusion:

Substitute 1.99×10−26 kg for mc, 0.110 nm for rc, 1.67×10−27 kg for mh, and 0.210 nm for rh in the above equation to find I.

I=6(1.99×10−26 kg)(0.110 nm(10−9 m1 nm))2+6(1.67×10−27 kg)(0.210 nm(10−9 m1 nm))2=1.44×10−45 kg⋅m2+0.442×10−45 kg⋅m2=1.88×10−45 kg⋅m2

Therefore, the moment of inertia is 1.88×10−45 kg⋅m2.

Answer 7

Chapter 42, Problem 14P

(a)

To determine

The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

Answer 8

(a)

Expert Solution

Answer to Problem 14P

The moment of inertia is 1.88×10−45 kg⋅m2.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the equation to find the moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

I=6mcrc2+6mhrh2

Here, I is the moment of inertia, mc is the mass of carbon atom, rc is the distance from carbon atom to point O, mh is the mass of hydrogen atom, and rh is the distance from hydrogen atom to point O.

Conclusion:

Substitute 1.99×10−26 kg for mc, 0.110 nm for rc, 1.67×10−27 kg for mh, and 0.210 nm for rh in the above equation to find I.

I=6(1.99×10−26 kg)(0.110 nm(10−9 m1 nm))2+6(1.67×10−27 kg)(0.210 nm(10−9 m1 nm))2=1.44×10−45 kg⋅m2+0.442×10−45 kg⋅m2=1.88×10−45 kg⋅m2

Therefore, the moment of inertia is 1.88×10−45 kg⋅m2.

Answer 13

(b)

To determine

The possible rotational energies about the axis.

(b)

Expert Solution

Answer to Problem 14P

The possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

Explanation of Solution

Write the equation to calculate the rotational energy.

Erot=ℏ22IJ(J+1) (I)

Here, Erot is the rotational energy, ℏ is the reduced Plank’s constant, I is the rotational moment of inertia, and J is the rotational quantum umber.

Conclusion:

Substitute 1.055×10−34 J⋅s for ℏ and 1.88×10−45 kg⋅m2 for I in the equation for Erot.

Erot=(1.055×10−34 J⋅s)22(1.88×10−45 kg⋅m2)J(J+1)=(2.95×10−24 J(1.6×10−19 eV1 J))J(J+1)=(18.4×10−6 eV)J(J+1)

The possible values of J are 0,1,2,3,4......

Substitute 0 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(0(0+1))=0 eV

Substitute 1 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(1(1+1))=36.9×10−6 eV(1 μeV10−6 eV)=36.9 μeV

Substitute 2 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(2(2+1))=111×10−6 eV(1 μeV10−6 eV)=111 μeV

Substitute 3 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(3(3+1))=221×10−6 eV(1 μeV10−6 eV)=221 μeV

Substitute 4 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(4(4+1))=369×10−6 eV(1 μeV10−6 eV)=369 μeV

Therefore, the possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

Answer 14

(b)

To determine

The possible rotational energies about the axis.

Answer 15

(b)

Expert Solution

Answer to Problem 14P

The possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the equation to calculate the rotational energy.

Erot=ℏ22IJ(J+1) (I)

Here, Erot is the rotational energy, ℏ is the reduced Plank’s constant, I is the rotational moment of inertia, and J is the rotational quantum umber.

Conclusion:

Substitute 1.055×10−34 J⋅s for ℏ and 1.88×10−45 kg⋅m2 for I in the equation for Erot.

Erot=(1.055×10−34 J⋅s)22(1.88×10−45 kg⋅m2)J(J+1)=(2.95×10−24 J(1.6×10−19 eV1 J))J(J+1)=(18.4×10−6 eV)J(J+1)

The possible values of J are 0,1,2,3,4......

Substitute 0 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(0(0+1))=0 eV

Substitute 1 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(1(1+1))=36.9×10−6 eV(1 μeV10−6 eV)=36.9 μeV

Substitute 2 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(2(2+1))=111×10−6 eV(1 μeV10−6 eV)=111 μeV

Substitute 3 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(3(3+1))=221×10−6 eV(1 μeV10−6 eV)=221 μeV

Substitute 4 for J in the above equation to find Erot.

Erot=(18.4×10−6 eV)(4(4+1))=369×10−6 eV(1 μeV10−6 eV)=369 μeV

Therefore, the possible rotational energies about the axis are 0 μeV,36.9 μeV, 111 μeV, 221 μeV, and 369 μeV.

Answer 20

Ch. 42.1 - For each of the following atoms or molecules,...Ch. 42.2 - Prob. 42.2QQ Ch. 42.2 - Prob. 42.3QQ Ch. 42 - Prob. 1P Ch. 42 - Prob. 2P Ch. 42 - Prob. 3P Ch. 42 - Prob. 4P Ch. 42 - Prob. 5P Ch. 42 - The photon frequency that would be absorbed by the...Ch. 42 - Prob. 8P

The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

Concept explainers

The moment of inertia of benzene molecule about an axis passing through the centre and perpendicular to the plane.

Answer to Problem 14P

Explanation of Solution

The possible rotational energies about the axis.

Answer to Problem 14P

Explanation of Solution

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