Exploring Chemical Analysis
Exploring Chemical Analysis
5th Edition
ISBN: 9781429275033
Author: Daniel C. Harris
Publisher: Macmillan Higher Education
Question
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Chapter 20, Problem 20.6P

(a)

Interpretation Introduction

Interpretation:

Energy difference between the ground and excited state of Ca has to be calculated.

Concept Introduction:

Planck’s hypothesis states energy difference associated with photon is as follows:

  ΔE=hcλ

Here,

λ denotes wavelength.

h denotes Planck’s constant

c denotes speed of light.

ΔE denotes energy difference.

Boltzmann’s distribution law for relative population is as follows:

  N*N0=(g*g)eΔE/kT

Here,

N* denotes population of excited state.

N0 denotes population in ground state.

g* denotes degeneracy of excited state.

g denotes degeneracy of ground state.

k denotes Boltzmann’s distribution.

T denotes temperature.

ΔE denotes energy difference.

(a)

Expert Solution
Check Mark

Explanation of Solution

Conversion factor to convert nm to m is as follows:

  1 nm=109 m

Thus, 422.7 nm is converted to m as follows:

  Wavelength=(422.7 nm)(109 m1 nm)=4.227×107 m

Planck’s hypothesis states energy difference associated with photon is as follows:

  ΔE=hcλ        (1)

Substitute 4.227×107 m for λ , 6.626×1034 J s for h and 3×108 m/s for c in equation (3).

  E=(6.626×1034 J s)(3×108 m/s)(4.227×107 m)=4.7×1019 J/photon

(b)

Interpretation Introduction

Interpretation:

Ratio of N*/N0 at 2500 K has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Boltzmann’s distribution law for relative population is as follows:

  N*N0=(g*g)eΔE/kT        (2)

Substitute 3 for g*/g, 4.7×1019 J for ΔE , 1.381×1023 J/K for ΔE and 2500 K for T in equation (2).

  N*N0=(31)e(4.7×1019 J)/(1.381×1023)(2500 K)=3.672×1063.67×106

(c)

Interpretation Introduction

Interpretation:

Percent fraction change when temperature changes to 2515 K has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

Substitute 3 for g*/g, 4.7×1019 J for ΔE , 1.381×1023 J/K for ΔE and 2515 K for T in equation (2).

  N*N0=(31)e(4.7×1019 J)/(1.381×1023)(2515 K)=3.982×106

Thus percent change when temperature changes to 2515 K is calculated as follows:

  Percent change=(3.982×1063.672×1063.672×106)(100 %)=8.4 %

(d)

Interpretation Introduction

Interpretation:

Ratio of N*/N0 at 6000 K has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

Substitute 3 for g*/g, 4.7×1019 J for ΔE , 1.381×1023 J/K for ΔE and 6000 K for T in equation (2).

  N*N0=(31)e(4.7×1019 J)/(1.381×1023)(6000 K)=0.0103

Hence, ratio of N*/N0 is found as 0.0103.

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