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 8, Problem 8A.1ST

(a)

Interpretation Introduction

Interpretation:

The ionization energy of the lower state of the deuterium atom has to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength of the lines present in the atomic spectrum of an element.  The Rydberg equation for a deuterium atom is given by the expression as shown below.

    1λ=Z2RD(1n121n22)

(a)

Expert Solution
Check Mark

Answer to Problem 8A.1ST

The ionization energy of the lower state of the deuterium atom is 328.109 kJ mol1_.

Explanation of Solution

The given data is shown below.

n=3n=215238 cm1
n=4n=220571 cm1
n=5n=223039 cm1
n=6n=224380 cm1

The Rydberg equation for a deuterium atom is given by the expression as shown below.

    v¯=Z2RD(1n121n22)        (1)

Where,

  • RD is Rydberg constant for deuterium.
  • n1 is the lower energy level.
  • n2 is the higher energy level.
  • v¯ is the wavenumber of the photon released.

The atomic number of deuterium is 1.

Rearrange the equation (1) for the value of RD.

  RD=v¯Z2(1n221n12)1        (2)

Substitute the value of an atomic number of a deuterium atom, v¯=15238 cm1, n1=2, Z=1 and n1=3 in the equation (2).

  RD=15238 cm1(1)2(1(2)21(3)2)1=(15238 cm1)(0.13889)1 cm1=109712.7223 cm1

Therefore, the value of RD for deuterium atom is 109712.7223 cm1.

The lower energy state is n=2.

The ionization energy per mole of the deuterium atom is given by the formula as shown below.

  I=hcZ2RDNA(1n2)        (3)

Where,

  • h is Plank’s constant with a value 6.62608×1034Js.
  • c is the speed of light with value 2.997945×1010cms1.
  • NA is the Avogadro’s number (6.022×1023atoms mol1).

Substitute the values of Z, h, c, RD, and n=2 in the equation (3).

  I=((1)2(6.62608×1034J s)(2.997945×1010cms1)(109712.7223 cm1)(6.022×1023mol1)(1(2)2))=(328108.9577 J mol1)(1 kJ103 J)=328.109 kJ mol1_

Therefore, the ionization energy of the lower state of the deuterium atom is 328.109 kJ mol1_.

(b)

Interpretation Introduction

Interpretation:

The ionization energy of the ground state of the deuterium atom has to be calculated.

Concept introduction:

The energy required to remove an electron from an ion or isolated atom is known as its ionization energy.  The ionization energy of metals is generally lower than the ionization energy of nonmetals.  The ionization energy per mole of the deuterium atom is given by the formula as shown below.

  I=hcZ2RDNA(1n2)

(b)

Expert Solution
Check Mark

Answer to Problem 8A.1ST

The ionization energy of the ground state of the deuterium atom is 1312.4 kJ mol1_.

Explanation of Solution

The atomic number of deuterium is 1.

The value of RD for deuterium atom is 109712.7223 cm1.

The ground energy level is n=1.

The ionization energy per mole of the deuterium atom is given by the formula as shown below.

  I=hcZ2RDNA(1n2)        (3)

Where,

  • h is Plank’s constant with a value 6.62608×1034Js.
  • c is the speed of light with value 2.997945×1010cms1.
  • NA is the Avogadro’s number (6.022×1023atoms mol1).

Substitute the values of Z, h, c, RD, and n=2 in the equation (3).

  I=((1)2(6.62608×1034J s)(2.997945×1010cms1)(109712.7223 cm1)(6.022×1023mol1)(1(1)2))=(1312435.831 J mol1)(1 kJ103 J)=1312.4 kJ mol1_

Therefore, the ionization energy of the ground state of the deuterium atom is 1312.4 kJ mol1_.

(c)

Interpretation Introduction

Interpretation:

The mass of deuteron has to be stated.

Concept introduction:

An atom is made up of three subatomic particles-neutrons, protons, and electrons. Neutron and protons are present in the nucleus of the atom, whereas electrons are revolving outside the nucleus in an atom.  The ground state energy of atom corresponds to the energy level for which the value of the principal quantum number is 1.

(c)

Expert Solution
Check Mark

Answer to Problem 8A.1ST

The mass of deuteron is 3.7037×1027kg_.

Explanation of Solution

The value of RD for deuterium atom is 109712.7223 cm1.

The value of Rydberg constant for an atom is given by the expression as shown below.

  R˜N=μme×R˜

Where,

  • R˜ is the Rydberg constant for an electron with a value of 109737cm1.
  • me is the mass of an electron with a value of 9.10938×1031kg.
  • μ is the reduced mass of the atom.

Substitute the value of R˜, Rydberg constant for deuterium atom and me in the above equation.

    109712.7223 cm1=μ9.10938×1031kg(109737cm1)109712.7223 cm1=1.2047×1035kgcm1μ

Rearrange the above equation for the value of μ.

    μ=109712.7223 cm11.2047×1035kgcm1=9.1071×1031kg

The reduced mass of a deuterium atom is given by the expression as shown below.

  1μ=1me+1mD

Where,

  • me is the mass of an electron with a value of 9.10938×1031kg.
  • mD is the mass of the deuterium nucleus (deuteron).

Substitute the values of me and μ in the above equation.

  19.1071×1031kg=19.10938×1031kg+1mD1.09804×1030kg1=1.09777×1030kg1+1mD

Rearrange the above equation for the value of mD.

    mD=11.09804×10301.09777×1030kg=3.7037×1027kg_

Therefore, the mass of deuteron is 3.7037×1027kg_.

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Chapter 8 Solutions

Atkins' Physical Chemistry

Ch. 8 - Prob. 8A.2AECh. 8 - Prob. 8A.2BECh. 8 - Prob. 8A.3AECh. 8 - Prob. 8A.3BECh. 8 - Prob. 8A.4AECh. 8 - Prob. 8A.4BECh. 8 - Prob. 8A.5AECh. 8 - Prob. 8A.5BECh. 8 - Prob. 8A.6AECh. 8 - Prob. 8A.6BECh. 8 - Prob. 8A.7AECh. 8 - Prob. 8A.7BECh. 8 - Prob. 8A.9AECh. 8 - Prob. 8A.10AECh. 8 - Prob. 8A.10BECh. 8 - Prob. 8A.11AECh. 8 - Prob. 8A.11BECh. 8 - Prob. 8A.12AECh. 8 - Prob. 8A.12BECh. 8 - Prob. 8A.1PCh. 8 - Prob. 8A.2PCh. 8 - Prob. 8A.3PCh. 8 - Prob. 8A.4PCh. 8 - Prob. 8A.6PCh. 8 - Prob. 8A.7PCh. 8 - Prob. 8A.8PCh. 8 - Prob. 8A.9PCh. 8 - Prob. 8A.10PCh. 8 - Prob. 8A.11PCh. 8 - Prob. 8B.1DQCh. 8 - Prob. 8B.2DQCh. 8 - Prob. 8B.3DQCh. 8 - Prob. 8B.4DQCh. 8 - Prob. 8B.1AECh. 8 - Prob. 8B.1BECh. 8 - Prob. 8B.2AECh. 8 - Prob. 8B.2BECh. 8 - Prob. 8B.3AECh. 8 - Prob. 8B.3BECh. 8 - Prob. 8B.4AECh. 8 - Prob. 8B.4BECh. 8 - Prob. 8B.5AECh. 8 - Prob. 8B.5BECh. 8 - Prob. 8B.1PCh. 8 - Prob. 8B.2PCh. 8 - Prob. 8B.3PCh. 8 - Prob. 8B.4PCh. 8 - Prob. 8B.5PCh. 8 - Prob. 8C.1DQCh. 8 - Prob. 8C.2DQCh. 8 - Prob. 8C.3DQCh. 8 - Prob. 8C.4DQCh. 8 - Prob. 8C.1AECh. 8 - Prob. 8C.1BECh. 8 - Prob. 8C.2AECh. 8 - Prob. 8C.2BECh. 8 - Prob. 8C.3AECh. 8 - Prob. 8C.3BECh. 8 - Prob. 8C.4AECh. 8 - Prob. 8C.4BECh. 8 - Prob. 8C.5AECh. 8 - Prob. 8C.5BECh. 8 - Prob. 8C.6AECh. 8 - Prob. 8C.6BECh. 8 - Prob. 8C.7AECh. 8 - Prob. 8C.7BECh. 8 - Prob. 8C.8AECh. 8 - Prob. 8C.8BECh. 8 - Prob. 8C.9AECh. 8 - Prob. 8C.9BECh. 8 - Prob. 8C.10AECh. 8 - Prob. 8C.10BECh. 8 - Prob. 8C.11AECh. 8 - Prob. 8C.11BECh. 8 - Prob. 8C.12AECh. 8 - Prob. 8C.12BECh. 8 - Prob. 8C.13AECh. 8 - Prob. 8C.13BECh. 8 - Prob. 8C.14AECh. 8 - Prob. 8C.14BECh. 8 - Prob. 8C.1PCh. 8 - Prob. 8C.2PCh. 8 - Prob. 8C.3PCh. 8 - Prob. 8C.4PCh. 8 - Prob. 8C.5PCh. 8 - Prob. 8C.6PCh. 8 - Prob. 8C.7PCh. 8 - Prob. 8C.8PCh. 8 - Prob. 8C.9PCh. 8 - Prob. 8C.10PCh. 8 - Prob. 8C.11PCh. 8 - Prob. 8C.12PCh. 8 - Prob. 8.1IACh. 8 - Prob. 8.2IACh. 8 - Prob. 8.3IA
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