Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780191092183
Author: ATKINS
Publisher: OXFORD UNIV.PRESS ACAD.UK (CC)
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Chapter 7, Problem 7A.4AE

(i)

Interpretation Introduction

Interpretation:

Under the given conditions, the energy of quantum has to be calculated.

Concept introduction:

The quantum of energy is called ass photon.  The transmission or absorption of energy takes place in small bundles; these bundles are collectively called energy packets or photons.  Photon is the fundamental particles of visible light.  The energy of quantum is directly proportional to the frequency of radiation.

(i)

Expert Solution
Check Mark

Answer to Problem 7A.4AE

The energy of quantum in joule and kilo joule per mole are 6.626×10-19J_ and 399kJmol-1_ respectively.

Explanation of Solution

The energy of quantum is directly proportional to the frequency of radiation

Therefore,

  E=hν                                                                                                          (1)

Where,

  • E is the energy of quantum.
  • ν is the frequency of radiation.
  • h is the Planck’s constant (6.626×1034Js).

The frequency in terms of time period is expressed as,

  ν=1T                                                                                                             (2)

Where,

  • T is the time period of  electronic oscillation.

Substitute the value of ν in equation (1).

  E=h×1T                                                                                                       (3)

It is given that,

The time period is 1.0fs.

The conversion of fs to s is done as,

  1fs=1015s

Substitute the value of h and T in equation (3).

  E=6.626×1034Js×11015s=6.626×10-19J_

Per mole energy (ENA) is calculated by using the formula,

  ENA=NA×E                                                                                                (4)

Where,

  • NA is Avogadro’s number (6.023×1023).

Substitute the value of NA and E in equation (4).

  ENA=6.023×1023mol1×6.626×1019J=39.9×104Jmol1=399×103Jmol1

The conversion of Jmol1 to kJmol1 is done as,

  1Jmol1=103kJmol1

Therefore, the conversion of 399×103Jmol1 to kJmol1 is done as,

  399×103Jmol1=399kJmol-1_

(ii)

Interpretation Introduction

Interpretation:

Under the given conditions, the energy of quantum has to be calculated.

Concept introduction:

The quantum of energy is called ass photon.  The transmission or absorption of energy takes place in small bundles; these bundles are collectively called energy packets or photons.  Photon is the fundamental particles of visible light.  The energy of quantum is directly proportional to the frequency of radiation.

(ii)

Expert Solution
Check Mark

Answer to Problem 7A.4AE

The energy of quantum in joule and kilo joule per mole are 6.626×10-20J_ and 39.9kJmol-1_ respectively.

Explanation of Solution

The energy of quantum is directly proportional to the frequency of radiation.

Therefore,

  E=hν                                                                                                           (1)

Where,

  • E is the energy of quantum.
  • ν is the frequency of radiation.
  • h is the Planck’s constant (6.626×1034Js)

The frequency in terms of time period is expressed as,

  ν=1T                                                                                                             (2)

Where,

  • T is the time period of  electronic oscillation.

Substitute the value of ν in equation (1).

  E=h×1T                                                                                                       (3)

It is given that,

The time period is 10fs.

The conversion of fs to s is done as,

  1fs=1015s

Therefore, the conversion of 10fs to s is done as,

  10fs=10×1015s=1014s

Substitute the value of h and T in equation (3).

  E=6.626×1034Js×11014s=6.626×10-20J_

Per mole energy (ENA) is calculated by using the formula,

  ENA=NA×E                                                                                                (4)

Where,

  • NA is Avogadro’s number (6.023×1023).

Substitute the value of NA and E in equation (4).

  ENA=6.023×1023mol1×6.626×1020J=39.9×103Jmol1

The conversion of Jmol1 to kJmol1 is done as,

  1Jmol1=103kJmol1

Therefore, the conversion of 39.9×103Jmol1 to kJmol1 is done as,

  39.9×103Jmol1=39.9kJmol-1_

(iii)

Interpretation Introduction

Interpretation:

Under the given conditions, the energy of quantum has to be calculated.

Concept introduction:

The quantum of energy is called ass photon.  The transmission or absorption of energy takes place in small bundles; these bundles are collectively called energy packets or photons.  Photon is the fundamental particles of visible light.  The energy of quantum is directly proportional to the frequency of radiation.

(iii)

Expert Solution
Check Mark

Answer to Problem 7A.4AE

The energy of quantum in joule and kilo joule per mole are 6.626×10-34J_ and 39.9×10-8kJmol-1_ respectively.

Explanation of Solution

The energy of quantum is directly proportional to the frequency of radiation.

Therefore,

  E=hν                                                                                                           (1)

Where,

  • E is the energy of quantum.
  • ν is the frequency of radiation.
  • h is the Planck’s constant (6.626×1034Js)

The frequency in terms of time period is expressed as,

  ν=1T                                                                                                             (2)

Where,

  • T is the time period of  electronic oscillation.

Substitute the value of ν in equation (1).

  E=h×1T                                                                                                       (3)

It is given that,

The time period is 1.0s.

Substitute the value of h and T in equation (3).

  E=6.626×1034Js×11.0s=6.626×10-34J_

Per mole energy (ENA) is calculated by using the formula,

  ENA=NA×E                                                                                                (4)

Where,

  • NA is Avogadro’s number (6.023×1023).

Substitute the value of NA and E in equation (4).

  ENA=6.023×1023mol1×6.626×1034J=39.9×1011Jmol1

The conversion of Jmol1 to kJmol1 is done as,

  1Jmol1=103kJmol1

Therefore, the conversion of 39.9×1011Jmol1 to kJmol1 is done as,

  39.9×1011Jmol1=39.9×10-8kJmol-1_

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Chapter 7 Solutions

Atkins' Physical Chemistry

Ch. 7 - Prob. 7D.1STCh. 7 - Prob. 7E.1STCh. 7 - Prob. 7E.2STCh. 7 - Prob. 7F.1STCh. 7 - Prob. 7A.1DQCh. 7 - Prob. 7A.2DQCh. 7 - Prob. 7A.3DQCh. 7 - Prob. 7A.4DQCh. 7 - Prob. 7A.1AECh. 7 - Prob. 7A.1BECh. 7 - Prob. 7A.2AECh. 7 - Prob. 7A.2BECh. 7 - Prob. 7A.3AECh. 7 - Prob. 7A.3BECh. 7 - Prob. 7A.4AECh. 7 - Prob. 7A.4BECh. 7 - Prob. 7A.5AECh. 7 - Prob. 7A.5BECh. 7 - Prob. 7A.6AECh. 7 - Prob. 7A.6BECh. 7 - Prob. 7A.7AECh. 7 - Prob. 7A.7BECh. 7 - Prob. 7A.8AECh. 7 - Prob. 7A.8BECh. 7 - Prob. 7A.9AECh. 7 - Prob. 7A.9BECh. 7 - Prob. 7A.10AECh. 7 - Prob. 7A.10BECh. 7 - Prob. 7A.11AECh. 7 - Prob. 7A.11BECh. 7 - Prob. 7A.12AECh. 7 - Prob. 7A.12BECh. 7 - Prob. 7A.13AECh. 7 - Prob. 7A.13BECh. 7 - Prob. 7A.1PCh. 7 - Prob. 7A.2PCh. 7 - Prob. 7A.3PCh. 7 - Prob. 7A.4PCh. 7 - Prob. 7A.5PCh. 7 - Prob. 7A.6PCh. 7 - Prob. 7A.7PCh. 7 - Prob. 7A.8PCh. 7 - Prob. 7A.9PCh. 7 - Prob. 7A.10PCh. 7 - Prob. 7B.1DQCh. 7 - Prob. 7B.2DQCh. 7 - Prob. 7B.3DQCh. 7 - Prob. 7B.1AECh. 7 - Prob. 7B.1BECh. 7 - Prob. 7B.2AECh. 7 - Prob. 7B.2BECh. 7 - Prob. 7B.3AECh. 7 - Prob. 7B.3BECh. 7 - Prob. 7B.4AECh. 7 - Prob. 7B.4BECh. 7 - Prob. 7B.5AECh. 7 - Prob. 7B.5BECh. 7 - Prob. 7B.6AECh. 7 - Prob. 7B.6BECh. 7 - Prob. 7B.7AECh. 7 - Prob. 7B.7BECh. 7 - Prob. 7B.8AECh. 7 - Prob. 7B.8BECh. 7 - Prob. 7B.1PCh. 7 - Prob. 7B.2PCh. 7 - Prob. 7B.3PCh. 7 - Prob. 7B.4PCh. 7 - Prob. 7B.5PCh. 7 - Prob. 7B.7PCh. 7 - Prob. 7B.8PCh. 7 - Prob. 7B.9PCh. 7 - Prob. 7B.11PCh. 7 - Prob. 7C.1DQCh. 7 - Prob. 7C.2DQCh. 7 - Prob. 7C.3DQCh. 7 - Prob. 7C.1AECh. 7 - Prob. 7C.1BECh. 7 - Prob. 7C.2AECh. 7 - Prob. 7C.2BECh. 7 - Prob. 7C.3AECh. 7 - Prob. 7C.3BECh. 7 - Prob. 7C.4AECh. 7 - Prob. 7C.4BECh. 7 - Prob. 7C.5AECh. 7 - Prob. 7C.5BECh. 7 - Prob. 7C.6AECh. 7 - Prob. 7C.6BECh. 7 - Prob. 7C.7AECh. 7 - Prob. 7C.7BECh. 7 - Prob. 7C.8AECh. 7 - Prob. 7C.8BECh. 7 - Prob. 7C.9AECh. 7 - Prob. 7C.9BECh. 7 - Prob. 7C.10AECh. 7 - Prob. 7C.10BECh. 7 - Prob. 7C.1PCh. 7 - Prob. 7C.2PCh. 7 - Prob. 7C.3PCh. 7 - Prob. 7C.4PCh. 7 - Prob. 7C.5PCh. 7 - Prob. 7C.6PCh. 7 - Prob. 7C.7PCh. 7 - Prob. 7C.8PCh. 7 - Prob. 7C.9PCh. 7 - Prob. 7C.11PCh. 7 - Prob. 7C.12PCh. 7 - Prob. 7C.13PCh. 7 - Prob. 7C.14PCh. 7 - Prob. 7C.15PCh. 7 - Prob. 7D.1DQCh. 7 - Prob. 7D.2DQCh. 7 - Prob. 7D.3DQCh. 7 - Prob. 7D.1AECh. 7 - Prob. 7D.1BECh. 7 - Prob. 7D.2AECh. 7 - Prob. 7D.2BECh. 7 - Prob. 7D.3AECh. 7 - Prob. 7D.3BECh. 7 - Prob. 7D.4AECh. 7 - Prob. 7D.4BECh. 7 - Prob. 7D.5AECh. 7 - Prob. 7D.5BECh. 7 - Prob. 7D.6AECh. 7 - Prob. 7D.6BECh. 7 - Prob. 7D.7AECh. 7 - Prob. 7D.7BECh. 7 - Prob. 7D.8AECh. 7 - Prob. 7D.8BECh. 7 - Prob. 7D.9AECh. 7 - Prob. 7D.9BECh. 7 - Prob. 7D.10AECh. 7 - Prob. 7D.10BECh. 7 - Prob. 7D.11AECh. 7 - Prob. 7D.11BECh. 7 - Prob. 7D.12AECh. 7 - Prob. 7D.12BECh. 7 - Prob. 7D.13AECh. 7 - Prob. 7D.13BECh. 7 - Prob. 7D.14AECh. 7 - Prob. 7D.14BECh. 7 - Prob. 7D.15AECh. 7 - Prob. 7D.15BECh. 7 - Prob. 7D.1PCh. 7 - Prob. 7D.2PCh. 7 - Prob. 7D.3PCh. 7 - Prob. 7D.4PCh. 7 - Prob. 7D.5PCh. 7 - Prob. 7D.6PCh. 7 - Prob. 7D.7PCh. 7 - Prob. 7D.8PCh. 7 - Prob. 7D.9PCh. 7 - Prob. 7D.11PCh. 7 - Prob. 7D.12PCh. 7 - Prob. 7D.14PCh. 7 - Prob. 7E.1DQCh. 7 - Prob. 7E.2DQCh. 7 - Prob. 7E.3DQCh. 7 - Prob. 7E.1AECh. 7 - Prob. 7E.1BECh. 7 - Prob. 7E.2AECh. 7 - Prob. 7E.2BECh. 7 - Prob. 7E.3AECh. 7 - Prob. 7E.3BECh. 7 - Prob. 7E.4AECh. 7 - Prob. 7E.4BECh. 7 - Prob. 7E.5AECh. 7 - Prob. 7E.5BECh. 7 - Prob. 7E.6AECh. 7 - Prob. 7E.6BECh. 7 - Prob. 7E.7AECh. 7 - Prob. 7E.7BECh. 7 - Prob. 7E.8AECh. 7 - Prob. 7E.8BECh. 7 - Prob. 7E.9AECh. 7 - Prob. 7E.9BECh. 7 - Prob. 7E.1PCh. 7 - Prob. 7E.2PCh. 7 - Prob. 7E.3PCh. 7 - Prob. 7E.4PCh. 7 - Prob. 7E.5PCh. 7 - Prob. 7E.6PCh. 7 - Prob. 7E.7PCh. 7 - Prob. 7E.8PCh. 7 - Prob. 7E.9PCh. 7 - Prob. 7E.12PCh. 7 - Prob. 7E.15PCh. 7 - Prob. 7E.16PCh. 7 - Prob. 7E.17PCh. 7 - Prob. 7F.1DQCh. 7 - Prob. 7F.2DQCh. 7 - Prob. 7F.3DQCh. 7 - Prob. 7F.1AECh. 7 - Prob. 7F.1BECh. 7 - Prob. 7F.2AECh. 7 - Prob. 7F.2BECh. 7 - Prob. 7F.3AECh. 7 - Prob. 7F.3BECh. 7 - Prob. 7F.4AECh. 7 - Prob. 7F.4BECh. 7 - Prob. 7F.5AECh. 7 - Prob. 7F.5BECh. 7 - Prob. 7F.6AECh. 7 - Prob. 7F.6BECh. 7 - Prob. 7F.7AECh. 7 - Prob. 7F.7BECh. 7 - Prob. 7F.8AECh. 7 - Prob. 7F.8BECh. 7 - Prob. 7F.9AECh. 7 - Prob. 7F.9BECh. 7 - Prob. 7F.10AECh. 7 - Prob. 7F.10BECh. 7 - Prob. 7F.11AECh. 7 - Prob. 7F.11BECh. 7 - Prob. 7F.12AECh. 7 - Prob. 7F.12BECh. 7 - Prob. 7F.13AECh. 7 - Prob. 7F.13BECh. 7 - Prob. 7F.14AECh. 7 - Prob. 7F.14BECh. 7 - Prob. 7F.1PCh. 7 - Prob. 7F.4PCh. 7 - Prob. 7F.6PCh. 7 - Prob. 7F.7PCh. 7 - Prob. 7F.8PCh. 7 - Prob. 7F.9PCh. 7 - Prob. 7F.10PCh. 7 - Prob. 7F.11PCh. 7 - Prob. 7.3IACh. 7 - Prob. 7.4IACh. 7 - Prob. 7.5IACh. 7 - Prob. 7.6IA
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