Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
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Chapter 7, Problem 7D.3BE

(i)

Interpretation Introduction

Interpretation:

The energy separation in joules, kilojoules per mole, electronvolts and reciprocal centimetres between the levels n=3 and n=2 of an electron in a box of length 1.50nm has to be calculated.

Concept introduction:

The separation between the adjacent energy levels with quantum numbers n and n+1 is given by the expression,

    En+1En=(2n+1)h28mL2

(i)

Expert Solution
Check Mark

Answer to Problem 7D.3BE

The energy separation in joules, kilojoules per mole, electronvolts and reciprocal centimetres between the levels  n=3 and n=2 of an electron in a box of length 1.50nm is 1.33×10-19J_, 800.926kJmol1_, 0.8246eV_ and 6.695×103cm1_, respectively.

Explanation of Solution

The separation between the adjacent energy levels with quantum numbers n and n+1 is given by the expression,

    En+1En=(2n+1)h28mL2                                                                              (1)

Where,

  • nandn+1 are the quantum numbers of the energy levels.
  • h is the Planck’s constant.
  • m is the mass of the electron (9.109×1031kg).
  • L is the length of the box (1.50nm) .

The value of n is 2.

The value of n+1 is 3.

The value of h is 6.626×1034kgm2s1.

The conversion of nm to m is done as,

  1nm=109m

The conversion of 1.50nm to m is done as,

  1.50nm=1.50nm×109m1nm

The value of L is 1.5×109m

Substitute the values of n, n+1, h and L in equation (1).

    E2E1=(2×2+1)(6.626×1034kgm2s1)28×9.109×1031kg(1.5×109m)2=2.195×10661.64×1047=1.33×1019kgm2s1

The unit conversion of kgm2s1 into J is,

    1kgm2s1=1J

The conversion of 1.33×1019kgm2s1 into J is,

    1.33×1019kgm2s1=1.33×1019kgm2s1×1J1kgm2s1=1.33×10-19J_

Hence, the energy separation in joules between the levels n=2 and n=3 of an electron in a box of length 1.50nm is 1.33×10-19J_.

The conversion of J into kJ is,

    1kJ=103J

The value of Avogadro number is 6.022×1023mol1

The conversion of 1.33×1019J into kJmol1 is,

    1.33×1019J=1.33×1019J×103kJ1J×6.022×1023mol1=800.926kJmol1_

Hence, the energy separation in kilojoules per mole between the levels n=2 and n=3 of an electron in a box of length 1.50nm is 800.926kJmol1_.

The conversion of J into eV is,

    1J=6.2×1018eV

The conversion of 1.33×1019J into eV is,

    1.33×1019J=1.33×1019J×6.2×1018eV1J=0.8246eV_

Hence, the energy separation in electronvolts between the levels n=2 and n=3 of an electron in a box of length 1.50nm is 0.8246eV_.

The conversion of J to cm1 is,

    1J=5.034×1022cm1

The conversion of 1.33×1019J into cm1 is,

    1.33×1019J=1.33×1019J×5.034×1022cm11J=6.695×103cm1_

Hence, the energy separation in reciprocal centimetres between the levels n=2 and n=3 of an electron in a box of length 1.50nm is 6.695×103cm1_.

(ii)

Interpretation Introduction

Interpretation:

The energy separation in joules, kilojoules per mole, electronvolts and reciprocal centimetres between the levels n=7 and n=6 of an electron in a box of length 1.50nm has to be calculated.

Concept introduction:

The separation between the adjacent energy levels with quantum numbers n and n+1 is given by the expression,

    En+1En=(2n+1)h28mL2

(ii)

Expert Solution
Check Mark

Answer to Problem 7D.3BE

The energy separation in joules, kilojoules per mole, electronvolts and reciprocal centimetres between the levels n=7 and n=6 of an electron in a box of length 1.50nm is 3.49×10-19J_, 21.01×102kJmol1_, 2.16eV_ and 1.756×104cm1_, respectively.

Explanation of Solution

The separation between the adjacent energy levels with quantum numbers n and n+1 is given by the expression,

    En+1En=(2n+1)h28mL2                                                                              (1)

Where,

  • nandn+1 are the quantum numbers of the energy levels.
  • h is the Planck’s constant.
  • m is the mass of the electron (9.109×1031kg).
  • L is the length of the box (1.50nm).

The value of n is 6.

The value of n+1 is 7.

The value of h is 6.626×1034kgm2s1.

The conversion of nm to m is done as,

  1nm=109m

The value of L is 109m

Substitute the values of n, n+1, h and L in equation (1).

    E6E5=(2×6+1)(6.626×1034kgm2s1)28×9.109×1031kg(1.5×109m)2=5.70×10661.63×1047=3.49×1019kgm2s1

The unit conversion of kgm2s1 into J is,

    1kgm2s1=1J

The conversion of 3.49×1019kgm2s1 into J is,

    3.49×1019kgm2s1=3.49×1019kgm2s1×1J1kgm2s1=3.49×10-19J_

Hence, the energy separation in joules between the levels n=7 and n=6 of an electron in a box of length 1.50nm is 3.49×10-19J_.

The conversion of J into kJ is,

    1kJ=103J

The value of Avogadro number is 6.022×1023mol1

The conversion of 3.49×1019J into kJmol1 is,

    3.49×1019J=3.49×1019J×103kJ1J×6.022×1023mol1=21.01×102kJmol1_

Hence, the energy separation in kilojoules per mole between the levels n=7 and n=6 of an electron in a box of length 1.50nm is 21.01×102kJmol1_.

The conversion of J into eV is,

    1J=6.2×1018eV

The conversion of 3.49×1019J into eV is,

    3.49×1019J=3.49×1019J×6.2×1018eV1J=2.16eV_

Hence, the energy separation in electronvolts between the levels n=7 and n=6 of an electron in a box of length 1.50nm is 2.16eV_.

The conversion of J to cm1 is,

    1J=5.034×1022cm1

The conversion of 3.49×1019J into cm1 is,

    3.49×1019J=3.49×1019J×5.034×1022cm11J=1.756×104cm1_

Hence, the energy separation in reciprocal centimetres between the levels n=7 and n=6 of an electron in a box of length 1.50nm is 1.756×104cm1_

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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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