The ratio of distance to the radius of a hydrogen atom.

Question

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Answer 1

Question

Chapter 16, Problem 86E

(a)

To determine

The ratio of distance to the radius of a hydrogen atom.

(a)

Expert Solution

Answer to Problem 86E

The ratio of distance to the radius of a hydrogen atom is 0.72_.

Explanation of Solution

Given that the charge of an electron is 1.6×10−19 C and the radius of hydrogen atom is 5.29×10−11 C.

Write the expression for the dipole moment.

p=dq (I)

Here, p is the dipole moment, d is the distance, and q is the charge.

Solve Equation (I) for d.

d=pq (II)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 1.6×10−19 C for q in Equation (II) to find the ratio of distance to the radius of a hydrogen atom.

d=6.13×10−30 C⋅m1.6×10−19 C=3.8×10−11 m

The ratio of the distance to the radius of a hydrogen atom is,

dR=3.8×10−11 m5.29×10−11 C=0.72

Therefore, the ratio of distance to the radius of a hydrogen atom is 0.72_.

(b)

To determine

The energy needed to flip the dipole.

(b)

Expert Solution

Answer to Problem 86E

The energy needed to flip the dipole is 7.66×10−5 eV_.

Explanation of Solution

Given that the electric field is 106 V⋅m−1.

The energy required to flip the dipole can be measured by using |−pEcosθ+pEcos180°|.

|−pEcosθ+pEcos180°|=2pE (III)

Here, E is the electric field.

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (III) to find the energy needed to flip the dipole.

|−pEcosθ+pEcos180°|=2pE=2(6.13×10−30 C⋅m)(106 V⋅m−1)=1.22×10−23 J×1 eV1.6×10−19 J=7.66×10−5 eV

Therefore, the energy needed to flip the dipole is 7.66×10−5 eV_.

(c)

To determine

The implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

(c)

Expert Solution

Answer to Problem 86E

The value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

Explanation of Solution

Given that the average kinetic energy of the molecule is 0.04 eV.

Let the angle θ the potential energy of water molecule is equal to kinetic energy.

0.04 eV=pEcosθ (IV)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (IV) to find the implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

0.04 eV=(6.13×10−30 C⋅m)(106 V⋅m−1)cosθcosθ=0.04 eV×1.6×10−19 J(6.13×10−30 C⋅m)(106 V⋅m−1)=1044

Therefore, the value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

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Answer 2

Question

Chapter 16, Problem 86E

(a)

To determine

The ratio of distance to the radius of a hydrogen atom.

(a)

Expert Solution

Answer to Problem 86E

The ratio of distance to the radius of a hydrogen atom is 0.72_.

Explanation of Solution

Given that the charge of an electron is 1.6×10−19 C and the radius of hydrogen atom is 5.29×10−11 C.

Write the expression for the dipole moment.

p=dq (I)

Here, p is the dipole moment, d is the distance, and q is the charge.

Solve Equation (I) for d.

d=pq (II)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 1.6×10−19 C for q in Equation (II) to find the ratio of distance to the radius of a hydrogen atom.

d=6.13×10−30 C⋅m1.6×10−19 C=3.8×10−11 m

The ratio of the distance to the radius of a hydrogen atom is,

dR=3.8×10−11 m5.29×10−11 C=0.72

Therefore, the ratio of distance to the radius of a hydrogen atom is 0.72_.

(b)

To determine

The energy needed to flip the dipole.

(b)

Expert Solution

Answer to Problem 86E

The energy needed to flip the dipole is 7.66×10−5 eV_.

Explanation of Solution

Given that the electric field is 106 V⋅m−1.

The energy required to flip the dipole can be measured by using |−pEcosθ+pEcos180°|.

|−pEcosθ+pEcos180°|=2pE (III)

Here, E is the electric field.

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (III) to find the energy needed to flip the dipole.

|−pEcosθ+pEcos180°|=2pE=2(6.13×10−30 C⋅m)(106 V⋅m−1)=1.22×10−23 J×1 eV1.6×10−19 J=7.66×10−5 eV

Therefore, the energy needed to flip the dipole is 7.66×10−5 eV_.

(c)

To determine

The implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

(c)

Expert Solution

Answer to Problem 86E

The value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

Explanation of Solution

Given that the average kinetic energy of the molecule is 0.04 eV.

Let the angle θ the potential energy of water molecule is equal to kinetic energy.

0.04 eV=pEcosθ (IV)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (IV) to find the implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

0.04 eV=(6.13×10−30 C⋅m)(106 V⋅m−1)cosθcosθ=0.04 eV×1.6×10−19 J(6.13×10−30 C⋅m)(106 V⋅m−1)=1044

Therefore, the value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

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Answer 3

Question

Chapter 16, Problem 86E

(a)

To determine

The ratio of distance to the radius of a hydrogen atom.

(a)

Expert Solution

Answer to Problem 86E

The ratio of distance to the radius of a hydrogen atom is 0.72_.

Explanation of Solution

Given that the charge of an electron is 1.6×10−19 C and the radius of hydrogen atom is 5.29×10−11 C.

Write the expression for the dipole moment.

p=dq (I)

Here, p is the dipole moment, d is the distance, and q is the charge.

Solve Equation (I) for d.

d=pq (II)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 1.6×10−19 C for q in Equation (II) to find the ratio of distance to the radius of a hydrogen atom.

d=6.13×10−30 C⋅m1.6×10−19 C=3.8×10−11 m

The ratio of the distance to the radius of a hydrogen atom is,

dR=3.8×10−11 m5.29×10−11 C=0.72

Therefore, the ratio of distance to the radius of a hydrogen atom is 0.72_.

(b)

To determine

The energy needed to flip the dipole.

(b)

Expert Solution

Answer to Problem 86E

The energy needed to flip the dipole is 7.66×10−5 eV_.

Explanation of Solution

Given that the electric field is 106 V⋅m−1.

The energy required to flip the dipole can be measured by using |−pEcosθ+pEcos180°|.

|−pEcosθ+pEcos180°|=2pE (III)

Here, E is the electric field.

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (III) to find the energy needed to flip the dipole.

|−pEcosθ+pEcos180°|=2pE=2(6.13×10−30 C⋅m)(106 V⋅m−1)=1.22×10−23 J×1 eV1.6×10−19 J=7.66×10−5 eV

Therefore, the energy needed to flip the dipole is 7.66×10−5 eV_.

(c)

To determine

The implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

(c)

Expert Solution

Answer to Problem 86E

The value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

Explanation of Solution

Given that the average kinetic energy of the molecule is 0.04 eV.

Let the angle θ the potential energy of water molecule is equal to kinetic energy.

0.04 eV=pEcosθ (IV)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (IV) to find the implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

0.04 eV=(6.13×10−30 C⋅m)(106 V⋅m−1)cosθcosθ=0.04 eV×1.6×10−19 J(6.13×10−30 C⋅m)(106 V⋅m−1)=1044

Therefore, the value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

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Answer 4

Question

Chapter 16, Problem 86E

(a)

To determine

The ratio of distance to the radius of a hydrogen atom.

(a)

Expert Solution

Answer to Problem 86E

The ratio of distance to the radius of a hydrogen atom is 0.72_.

Explanation of Solution

Given that the charge of an electron is 1.6×10−19 C and the radius of hydrogen atom is 5.29×10−11 C.

Write the expression for the dipole moment.

p=dq (I)

Here, p is the dipole moment, d is the distance, and q is the charge.

Solve Equation (I) for d.

d=pq (II)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 1.6×10−19 C for q in Equation (II) to find the ratio of distance to the radius of a hydrogen atom.

d=6.13×10−30 C⋅m1.6×10−19 C=3.8×10−11 m

The ratio of the distance to the radius of a hydrogen atom is,

dR=3.8×10−11 m5.29×10−11 C=0.72

Therefore, the ratio of distance to the radius of a hydrogen atom is 0.72_.

(b)

To determine

The energy needed to flip the dipole.

(b)

Expert Solution

Answer to Problem 86E

The energy needed to flip the dipole is 7.66×10−5 eV_.

Explanation of Solution

Given that the electric field is 106 V⋅m−1.

The energy required to flip the dipole can be measured by using |−pEcosθ+pEcos180°|.

|−pEcosθ+pEcos180°|=2pE (III)

Here, E is the electric field.

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (III) to find the energy needed to flip the dipole.

|−pEcosθ+pEcos180°|=2pE=2(6.13×10−30 C⋅m)(106 V⋅m−1)=1.22×10−23 J×1 eV1.6×10−19 J=7.66×10−5 eV

Therefore, the energy needed to flip the dipole is 7.66×10−5 eV_.

(c)

To determine

The implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

(c)

Expert Solution

Answer to Problem 86E

The value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

Explanation of Solution

Given that the average kinetic energy of the molecule is 0.04 eV.

Let the angle θ the potential energy of water molecule is equal to kinetic energy.

0.04 eV=pEcosθ (IV)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (IV) to find the implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

0.04 eV=(6.13×10−30 C⋅m)(106 V⋅m−1)cosθcosθ=0.04 eV×1.6×10−19 J(6.13×10−30 C⋅m)(106 V⋅m−1)=1044

Therefore, the value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

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Answer 5

Chapter 16, Problem 86E

(a)

To determine

The ratio of distance to the radius of a hydrogen atom.

(a)

Expert Solution

Answer to Problem 86E

The ratio of distance to the radius of a hydrogen atom is 0.72_.

Explanation of Solution

Given that the charge of an electron is 1.6×10−19 C and the radius of hydrogen atom is 5.29×10−11 C.

Write the expression for the dipole moment.

p=dq (I)

Here, p is the dipole moment, d is the distance, and q is the charge.

Solve Equation (I) for d.

d=pq (II)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 1.6×10−19 C for q in Equation (II) to find the ratio of distance to the radius of a hydrogen atom.

d=6.13×10−30 C⋅m1.6×10−19 C=3.8×10−11 m

The ratio of the distance to the radius of a hydrogen atom is,

dR=3.8×10−11 m5.29×10−11 C=0.72

Therefore, the ratio of distance to the radius of a hydrogen atom is 0.72_.

(b)

To determine

The energy needed to flip the dipole.

(b)

Expert Solution

Answer to Problem 86E

The energy needed to flip the dipole is 7.66×10−5 eV_.

Explanation of Solution

Given that the electric field is 106 V⋅m−1.

The energy required to flip the dipole can be measured by using |−pEcosθ+pEcos180°|.

|−pEcosθ+pEcos180°|=2pE (III)

Here, E is the electric field.

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (III) to find the energy needed to flip the dipole.

|−pEcosθ+pEcos180°|=2pE=2(6.13×10−30 C⋅m)(106 V⋅m−1)=1.22×10−23 J×1 eV1.6×10−19 J=7.66×10−5 eV

Therefore, the energy needed to flip the dipole is 7.66×10−5 eV_.

(c)

To determine

The implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

(c)

Expert Solution

Answer to Problem 86E

The value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

Explanation of Solution

Given that the average kinetic energy of the molecule is 0.04 eV.

Let the angle θ the potential energy of water molecule is equal to kinetic energy.

0.04 eV=pEcosθ (IV)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (IV) to find the implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

0.04 eV=(6.13×10−30 C⋅m)(106 V⋅m−1)cosθcosθ=0.04 eV×1.6×10−19 J(6.13×10−30 C⋅m)(106 V⋅m−1)=1044

Therefore, the value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

Answer 6

Chapter 16, Problem 86E

(a)

To determine

The ratio of distance to the radius of a hydrogen atom.

(a)

Expert Solution

Answer to Problem 86E

The ratio of distance to the radius of a hydrogen atom is 0.72_.

Explanation of Solution

Given that the charge of an electron is 1.6×10−19 C and the radius of hydrogen atom is 5.29×10−11 C.

Write the expression for the dipole moment.

p=dq (I)

Here, p is the dipole moment, d is the distance, and q is the charge.

Solve Equation (I) for d.

d=pq (II)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 1.6×10−19 C for q in Equation (II) to find the ratio of distance to the radius of a hydrogen atom.

d=6.13×10−30 C⋅m1.6×10−19 C=3.8×10−11 m

The ratio of the distance to the radius of a hydrogen atom is,

dR=3.8×10−11 m5.29×10−11 C=0.72

Therefore, the ratio of distance to the radius of a hydrogen atom is 0.72_.

Answer 7

Chapter 16, Problem 86E

(a)

To determine

The ratio of distance to the radius of a hydrogen atom.

Answer 8

(a)

Expert Solution

Answer to Problem 86E

The ratio of distance to the radius of a hydrogen atom is 0.72_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given that the charge of an electron is 1.6×10−19 C and the radius of hydrogen atom is 5.29×10−11 C.

Write the expression for the dipole moment.

p=dq (I)

Here, p is the dipole moment, d is the distance, and q is the charge.

Solve Equation (I) for d.

d=pq (II)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 1.6×10−19 C for q in Equation (II) to find the ratio of distance to the radius of a hydrogen atom.

d=6.13×10−30 C⋅m1.6×10−19 C=3.8×10−11 m

The ratio of the distance to the radius of a hydrogen atom is,

dR=3.8×10−11 m5.29×10−11 C=0.72

Therefore, the ratio of distance to the radius of a hydrogen atom is 0.72_.

Answer 13

(b)

To determine

The energy needed to flip the dipole.

(b)

Expert Solution

Answer to Problem 86E

The energy needed to flip the dipole is 7.66×10−5 eV_.

Explanation of Solution

Given that the electric field is 106 V⋅m−1.

The energy required to flip the dipole can be measured by using |−pEcosθ+pEcos180°|.

|−pEcosθ+pEcos180°|=2pE (III)

Here, E is the electric field.

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (III) to find the energy needed to flip the dipole.

|−pEcosθ+pEcos180°|=2pE=2(6.13×10−30 C⋅m)(106 V⋅m−1)=1.22×10−23 J×1 eV1.6×10−19 J=7.66×10−5 eV

Therefore, the energy needed to flip the dipole is 7.66×10−5 eV_.

Answer 14

(b)

To determine

The energy needed to flip the dipole.

Answer 15

(b)

Expert Solution

Answer to Problem 86E

The energy needed to flip the dipole is 7.66×10−5 eV_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given that the electric field is 106 V⋅m−1.

The energy required to flip the dipole can be measured by using |−pEcosθ+pEcos180°|.

|−pEcosθ+pEcos180°|=2pE (III)

Here, E is the electric field.

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (III) to find the energy needed to flip the dipole.

|−pEcosθ+pEcos180°|=2pE=2(6.13×10−30 C⋅m)(106 V⋅m−1)=1.22×10−23 J×1 eV1.6×10−19 J=7.66×10−5 eV

Therefore, the energy needed to flip the dipole is 7.66×10−5 eV_.

Answer 20

(c)

To determine

The implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

(c)

Expert Solution

Answer to Problem 86E

The value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

Explanation of Solution

Given that the average kinetic energy of the molecule is 0.04 eV.

Let the angle θ the potential energy of water molecule is equal to kinetic energy.

0.04 eV=pEcosθ (IV)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (IV) to find the implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

0.04 eV=(6.13×10−30 C⋅m)(106 V⋅m−1)cosθcosθ=0.04 eV×1.6×10−19 J(6.13×10−30 C⋅m)(106 V⋅m−1)=1044

Therefore, the value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

Answer 21

(c)

To determine

The implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.

Answer 22

(c)

Expert Solution

Answer to Problem 86E

The value of the cosθ<1 since, at this field and kinetic energy orientation of water molecule is not possible.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given that the average kinetic energy of the molecule is 0.04 eV.

Let the angle θ the potential energy of water molecule is equal to kinetic energy.

0.04 eV=pEcosθ (IV)

Conclusion:

Substitute 6.13×10−30 C⋅m for p and 106 V⋅m−1 for E in Equation (IV) to find the implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106 V⋅m−1.