General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 16, Problem 86E

(a)

To determine

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

(a)

Expert Solution
Check Mark

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 1.6×1019C and the radius of hydrogen atom is 5.29×1011C.

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 6.13×1030Cm for p and 1.6×1019C for q in Equation (II) to find the ratio of distance to the radius of a hydrogen atom.

  d=6.13×1030Cm1.6×1019C=3.8×1011m

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

  dR=3.8×1011m5.29×1011C=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
Check Mark

Answer to Problem 86E

The energy needed to flip the dipole is 7.66×105eV_.

Explanation of Solution

Given that the electric field is 106Vm1.

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 6.13×1030Cm for p and 106Vm1 for E in Equation (III) to find the energy needed to flip the dipole.

  |pEcosθ+pEcos180°|=2pE=2(6.13×1030Cm)(106Vm1)=1.22×1023J×1eV1.6×1019J=7.66×105eV

Therefore, the energy needed to flip the dipole is 7.66×105eV_.

(c)

To determine

The implication average kinetic energy of a molecule for the orientation of water molecule in the relatively strong field of 106Vm1.

(c)

Expert Solution
Check Mark

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 0.04eV.

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

  0.04eV=pEcosθ        (IV)

Conclusion:

Substitute 6.13×1030Cm for p and 106Vm1 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 106Vm1.

  0.04eV=(6.13×1030Cm)(106Vm1)cosθcosθ=0.04eV×1.6×1019J(6.13×1030Cm)(106Vm1)=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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Chapter 16 Solutions

General Physics, 2nd Edition

Ch. 16 - Prob. 11RQCh. 16 - Prob. 12RQCh. 16 - Prob. 13RQCh. 16 - Prob. 1ECh. 16 - Prob. 2ECh. 16 - Prob. 3ECh. 16 - Prob. 4ECh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8ECh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Prob. 28ECh. 16 - Prob. 29ECh. 16 - Prob. 30ECh. 16 - Prob. 31ECh. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - Prob. 34ECh. 16 - Prob. 35ECh. 16 - Prob. 36ECh. 16 - Prob. 37ECh. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Prob. 40ECh. 16 - Prob. 41ECh. 16 - Prob. 42ECh. 16 - Prob. 43ECh. 16 - Prob. 44ECh. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48ECh. 16 - Prob. 49ECh. 16 - Prob. 50ECh. 16 - Prob. 51ECh. 16 - Prob. 52ECh. 16 - Prob. 53ECh. 16 - Prob. 54ECh. 16 - Prob. 55ECh. 16 - Prob. 56ECh. 16 - Prob. 57ECh. 16 - Prob. 58ECh. 16 - Prob. 59ECh. 16 - Prob. 60ECh. 16 - Prob. 61ECh. 16 - Prob. 62ECh. 16 - Prob. 63ECh. 16 - Prob. 64ECh. 16 - Prob. 65ECh. 16 - Prob. 66ECh. 16 - Prob. 67ECh. 16 - Prob. 68ECh. 16 - Prob. 69ECh. 16 - Prob. 70ECh. 16 - Prob. 72ECh. 16 - Prob. 73ECh. 16 - Prob. 74ECh. 16 - Prob. 75ECh. 16 - Prob. 76ECh. 16 - Prob. 78ECh. 16 - Prob. 81ECh. 16 - Prob. 82ECh. 16 - Prob. 83ECh. 16 - Prob. 84ECh. 16 - Prob. 85ECh. 16 - Prob. 86ECh. 16 - Prob. 87ECh. 16 - Prob. 88ECh. 16 - Prob. 89ECh. 16 - Prob. 90ECh. 16 - Prob. 91ECh. 16 - Prob. 92ECh. 16 - Prob. 93ECh. 16 - Prob. 94ECh. 16 - Prob. 95ECh. 16 - Prob. 96ECh. 16 - Prob. 97ECh. 16 - Prob. 98ECh. 16 - Prob. 99ECh. 16 - Prob. 100ECh. 16 - Prob. 101ECh. 16 - Prob. 102ECh. 16 - Prob. 103ECh. 16 - Prob. 104ECh. 16 - Prob. 105ECh. 16 - Prob. 106ECh. 16 - Prob. 107ECh. 16 - Prob. 108E
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