The values of the five transitions for HCl using equation for the energy of a Morse oscillator are to be calculated. Concept introduction: Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant.

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Answer 1

Question

Chapter 14, Problem 14.66E

Interpretation Introduction

Interpretation:

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are to be calculated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant.

Expert Solution & Answer

Answer to Problem 14.66E

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

Explanation of Solution

The energy of a Morse oscillator is calculated by the formula shown below.

E=hve(v+12)−hvexe(v+12)2 …(1)

Where,

• v is the vibrational quantum number.

• h is the Planck’s constant.

• xe is the anharmonicity constant.

• ve is the harmonic vibrational frequency.

The value of ve for HCl is 2989.74 cm−1.

The value of xeve for HCl is 52.05 cm−1.

Convert 2989.74 cm−1 to s−1.

2989.74 cm−1=2989.74 cm−1×2.9979×1010 cm s−1=8962.94×1010 s−1

Convert 52.05 cm−1 to s−1.

52.05 cm−1=52.05 cm−1×2.9979×1010 cm s−1=156.04×1010 s−1

Substitute v=0 in equation (1).

E=hve(0+12)−hvexe(0+12)2=12hve−14hvexe …(2)

Substitute v=1 in equation (1).

E=hve(1+12)−hvexe(1+12)2=32hve−94hvexe …(3)

Subtract equation (2) from equation (3).

ΔE(v=0→v=1)=(32hve−94hvexe)−(12hve−14hvexe)=hve−2hvexe …(4)

Substitute the value of Planck’s constant, ve and xeve in equation (4).

ΔE(v=0→v=1)=(6.626×10−34 J⋅s×8962.94×1010 s−1−2×6.626×10−34 J⋅s×156.04×1010 s−1)=59388.44×10−24 J−2067.84×10−24 J=5.732×10−20 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(5)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (5).

v=5.732×10−20 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=2885.61 cm−1

Substitute v=2 in equation (1).

E=hve(2+12)−hvexe(2+12)2=52hve−254hvexe …(6)

Subtract equation (2) from equation (6).

ΔE(v=0→v=2)=(52hve−254hvexe)−(12hve−14hvexe)=2hve−6hvexe …(7)

Substitute the value of Planck’s constant, ve and xeve in equation (7).

ΔE(v=0→v=2)=(2×6.626×10−34 J⋅s×8962.94×1010 s−1−6×6.626×10−34 J⋅s×156.04×1010 s−1)=118776.88×10−24 J−6203.53×10−24 J=1.126×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(8)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (8).

v=1.126×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=5668.52 cm−1

Substitute v=3 in equation (1).

E=hve(3+12)−hvexe(3+12)2=72hve−494hvexe …(9)

Subtract equation (2) from equation (9).

ΔE(v=0→v=3)=(72hve−494hvexe)−(12hve−14hvexe)=3hve−12hvexe …(10)

Substitute the value of Planck’s constant, ve and xeve in equation (10).

ΔE(v=0→v=3)=(3×6.626×10−34 J⋅s×8962.94×1010 s−1−12×6.626×10−34 J⋅s×156.04×1010 s−1)=178165.32×10−24 J−12407.05×10−24 J=1.657×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(11)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (11).

v=1.657×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=8341.69 cm−1

Substitute v=4 in equation (1).

E=hve(4+12)−hvexe(4+12)2=92hve−814hvexe …(12)

Subtract equation (2) from equation (12).

ΔE(v=0→v=4)=(92hve−814hvexe)−(12hve−14hvexe)=4hve−20hvexe …(13)

Substitute the value of Planck’s constant, ve and xeve in equation (13).

ΔE(v=0→v=4)=(4×6.626×10−34 J⋅s×8962.94×1010 s−1−20×6.626×10−34 J⋅s×156.04×1010 s−1)=237553.76×10−24 J−20678.42×10−24 J=2.169×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(14)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (14).

v=2.169×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=10919.20 cm−1

Substitute v=5 in equation (1).

E=hve(5+12)−hvexe(5+12)2=112hve−1214hvexe …(15)

Subtract equation (2) from equation (15).

ΔE(v=0→v=5)=(112hve−1214hvexe)−(12hve−14hvexe)=5hve−30hvexe …(16)

Substitute the value of Planck’s constant, ve and xeve in equation (16).

ΔE(v=0→v=5)=(5×6.626×10−34 J⋅s×8962.94×1010 s−1−30×6.626×10−34 J⋅s×156.04×1010 s−1)=296942.20×10−24 J−31017.63×10−24 J=2.659×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(17)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (17).

v=2.659×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=13385.96 cm−1

Conclusion

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

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Answer 2

Question

Chapter 14, Problem 14.66E

Interpretation Introduction

Interpretation:

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are to be calculated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant.

Expert Solution & Answer

Answer to Problem 14.66E

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

Explanation of Solution

The energy of a Morse oscillator is calculated by the formula shown below.

E=hve(v+12)−hvexe(v+12)2 …(1)

Where,

• v is the vibrational quantum number.

• h is the Planck’s constant.

• xe is the anharmonicity constant.

• ve is the harmonic vibrational frequency.

The value of ve for HCl is 2989.74 cm−1.

The value of xeve for HCl is 52.05 cm−1.

Convert 2989.74 cm−1 to s−1.

2989.74 cm−1=2989.74 cm−1×2.9979×1010 cm s−1=8962.94×1010 s−1

Convert 52.05 cm−1 to s−1.

52.05 cm−1=52.05 cm−1×2.9979×1010 cm s−1=156.04×1010 s−1

Substitute v=0 in equation (1).

E=hve(0+12)−hvexe(0+12)2=12hve−14hvexe …(2)

Substitute v=1 in equation (1).

E=hve(1+12)−hvexe(1+12)2=32hve−94hvexe …(3)

Subtract equation (2) from equation (3).

ΔE(v=0→v=1)=(32hve−94hvexe)−(12hve−14hvexe)=hve−2hvexe …(4)

Substitute the value of Planck’s constant, ve and xeve in equation (4).

ΔE(v=0→v=1)=(6.626×10−34 J⋅s×8962.94×1010 s−1−2×6.626×10−34 J⋅s×156.04×1010 s−1)=59388.44×10−24 J−2067.84×10−24 J=5.732×10−20 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(5)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (5).

v=5.732×10−20 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=2885.61 cm−1

Substitute v=2 in equation (1).

E=hve(2+12)−hvexe(2+12)2=52hve−254hvexe …(6)

Subtract equation (2) from equation (6).

ΔE(v=0→v=2)=(52hve−254hvexe)−(12hve−14hvexe)=2hve−6hvexe …(7)

Substitute the value of Planck’s constant, ve and xeve in equation (7).

ΔE(v=0→v=2)=(2×6.626×10−34 J⋅s×8962.94×1010 s−1−6×6.626×10−34 J⋅s×156.04×1010 s−1)=118776.88×10−24 J−6203.53×10−24 J=1.126×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(8)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (8).

v=1.126×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=5668.52 cm−1

Substitute v=3 in equation (1).

E=hve(3+12)−hvexe(3+12)2=72hve−494hvexe …(9)

Subtract equation (2) from equation (9).

ΔE(v=0→v=3)=(72hve−494hvexe)−(12hve−14hvexe)=3hve−12hvexe …(10)

Substitute the value of Planck’s constant, ve and xeve in equation (10).

ΔE(v=0→v=3)=(3×6.626×10−34 J⋅s×8962.94×1010 s−1−12×6.626×10−34 J⋅s×156.04×1010 s−1)=178165.32×10−24 J−12407.05×10−24 J=1.657×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(11)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (11).

v=1.657×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=8341.69 cm−1

Substitute v=4 in equation (1).

E=hve(4+12)−hvexe(4+12)2=92hve−814hvexe …(12)

Subtract equation (2) from equation (12).

ΔE(v=0→v=4)=(92hve−814hvexe)−(12hve−14hvexe)=4hve−20hvexe …(13)

Substitute the value of Planck’s constant, ve and xeve in equation (13).

ΔE(v=0→v=4)=(4×6.626×10−34 J⋅s×8962.94×1010 s−1−20×6.626×10−34 J⋅s×156.04×1010 s−1)=237553.76×10−24 J−20678.42×10−24 J=2.169×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(14)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (14).

v=2.169×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=10919.20 cm−1

Substitute v=5 in equation (1).

E=hve(5+12)−hvexe(5+12)2=112hve−1214hvexe …(15)

Subtract equation (2) from equation (15).

ΔE(v=0→v=5)=(112hve−1214hvexe)−(12hve−14hvexe)=5hve−30hvexe …(16)

Substitute the value of Planck’s constant, ve and xeve in equation (16).

ΔE(v=0→v=5)=(5×6.626×10−34 J⋅s×8962.94×1010 s−1−30×6.626×10−34 J⋅s×156.04×1010 s−1)=296942.20×10−24 J−31017.63×10−24 J=2.659×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(17)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (17).

v=2.659×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=13385.96 cm−1

Conclusion

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

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Answer 3

Question

Chapter 14, Problem 14.66E

Interpretation Introduction

Interpretation:

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are to be calculated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant.

Expert Solution & Answer

Answer to Problem 14.66E

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

Explanation of Solution

The energy of a Morse oscillator is calculated by the formula shown below.

E=hve(v+12)−hvexe(v+12)2 …(1)

Where,

• v is the vibrational quantum number.

• h is the Planck’s constant.

• xe is the anharmonicity constant.

• ve is the harmonic vibrational frequency.

The value of ve for HCl is 2989.74 cm−1.

The value of xeve for HCl is 52.05 cm−1.

Convert 2989.74 cm−1 to s−1.

2989.74 cm−1=2989.74 cm−1×2.9979×1010 cm s−1=8962.94×1010 s−1

Convert 52.05 cm−1 to s−1.

52.05 cm−1=52.05 cm−1×2.9979×1010 cm s−1=156.04×1010 s−1

Substitute v=0 in equation (1).

E=hve(0+12)−hvexe(0+12)2=12hve−14hvexe …(2)

Substitute v=1 in equation (1).

E=hve(1+12)−hvexe(1+12)2=32hve−94hvexe …(3)

Subtract equation (2) from equation (3).

ΔE(v=0→v=1)=(32hve−94hvexe)−(12hve−14hvexe)=hve−2hvexe …(4)

Substitute the value of Planck’s constant, ve and xeve in equation (4).

ΔE(v=0→v=1)=(6.626×10−34 J⋅s×8962.94×1010 s−1−2×6.626×10−34 J⋅s×156.04×1010 s−1)=59388.44×10−24 J−2067.84×10−24 J=5.732×10−20 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(5)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (5).

v=5.732×10−20 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=2885.61 cm−1

Substitute v=2 in equation (1).

E=hve(2+12)−hvexe(2+12)2=52hve−254hvexe …(6)

Subtract equation (2) from equation (6).

ΔE(v=0→v=2)=(52hve−254hvexe)−(12hve−14hvexe)=2hve−6hvexe …(7)

Substitute the value of Planck’s constant, ve and xeve in equation (7).

ΔE(v=0→v=2)=(2×6.626×10−34 J⋅s×8962.94×1010 s−1−6×6.626×10−34 J⋅s×156.04×1010 s−1)=118776.88×10−24 J−6203.53×10−24 J=1.126×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(8)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (8).

v=1.126×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=5668.52 cm−1

Substitute v=3 in equation (1).

E=hve(3+12)−hvexe(3+12)2=72hve−494hvexe …(9)

Subtract equation (2) from equation (9).

ΔE(v=0→v=3)=(72hve−494hvexe)−(12hve−14hvexe)=3hve−12hvexe …(10)

Substitute the value of Planck’s constant, ve and xeve in equation (10).

ΔE(v=0→v=3)=(3×6.626×10−34 J⋅s×8962.94×1010 s−1−12×6.626×10−34 J⋅s×156.04×1010 s−1)=178165.32×10−24 J−12407.05×10−24 J=1.657×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(11)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (11).

v=1.657×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=8341.69 cm−1

Substitute v=4 in equation (1).

E=hve(4+12)−hvexe(4+12)2=92hve−814hvexe …(12)

Subtract equation (2) from equation (12).

ΔE(v=0→v=4)=(92hve−814hvexe)−(12hve−14hvexe)=4hve−20hvexe …(13)

Substitute the value of Planck’s constant, ve and xeve in equation (13).

ΔE(v=0→v=4)=(4×6.626×10−34 J⋅s×8962.94×1010 s−1−20×6.626×10−34 J⋅s×156.04×1010 s−1)=237553.76×10−24 J−20678.42×10−24 J=2.169×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(14)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (14).

v=2.169×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=10919.20 cm−1

Substitute v=5 in equation (1).

E=hve(5+12)−hvexe(5+12)2=112hve−1214hvexe …(15)

Subtract equation (2) from equation (15).

ΔE(v=0→v=5)=(112hve−1214hvexe)−(12hve−14hvexe)=5hve−30hvexe …(16)

Substitute the value of Planck’s constant, ve and xeve in equation (16).

ΔE(v=0→v=5)=(5×6.626×10−34 J⋅s×8962.94×1010 s−1−30×6.626×10−34 J⋅s×156.04×1010 s−1)=296942.20×10−24 J−31017.63×10−24 J=2.659×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(17)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (17).

v=2.659×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=13385.96 cm−1

Conclusion

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

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Answer 4

Question

Chapter 14, Problem 14.66E

Interpretation Introduction

Interpretation:

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are to be calculated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant.

Expert Solution & Answer

Answer to Problem 14.66E

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

Explanation of Solution

The energy of a Morse oscillator is calculated by the formula shown below.

E=hve(v+12)−hvexe(v+12)2 …(1)

Where,

• v is the vibrational quantum number.

• h is the Planck’s constant.

• xe is the anharmonicity constant.

• ve is the harmonic vibrational frequency.

The value of ve for HCl is 2989.74 cm−1.

The value of xeve for HCl is 52.05 cm−1.

Convert 2989.74 cm−1 to s−1.

2989.74 cm−1=2989.74 cm−1×2.9979×1010 cm s−1=8962.94×1010 s−1

Convert 52.05 cm−1 to s−1.

52.05 cm−1=52.05 cm−1×2.9979×1010 cm s−1=156.04×1010 s−1

Substitute v=0 in equation (1).

E=hve(0+12)−hvexe(0+12)2=12hve−14hvexe …(2)

Substitute v=1 in equation (1).

E=hve(1+12)−hvexe(1+12)2=32hve−94hvexe …(3)

Subtract equation (2) from equation (3).

ΔE(v=0→v=1)=(32hve−94hvexe)−(12hve−14hvexe)=hve−2hvexe …(4)

Substitute the value of Planck’s constant, ve and xeve in equation (4).

ΔE(v=0→v=1)=(6.626×10−34 J⋅s×8962.94×1010 s−1−2×6.626×10−34 J⋅s×156.04×1010 s−1)=59388.44×10−24 J−2067.84×10−24 J=5.732×10−20 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(5)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (5).

v=5.732×10−20 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=2885.61 cm−1

Substitute v=2 in equation (1).

E=hve(2+12)−hvexe(2+12)2=52hve−254hvexe …(6)

Subtract equation (2) from equation (6).

ΔE(v=0→v=2)=(52hve−254hvexe)−(12hve−14hvexe)=2hve−6hvexe …(7)

Substitute the value of Planck’s constant, ve and xeve in equation (7).

ΔE(v=0→v=2)=(2×6.626×10−34 J⋅s×8962.94×1010 s−1−6×6.626×10−34 J⋅s×156.04×1010 s−1)=118776.88×10−24 J−6203.53×10−24 J=1.126×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(8)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (8).

v=1.126×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=5668.52 cm−1

Substitute v=3 in equation (1).

E=hve(3+12)−hvexe(3+12)2=72hve−494hvexe …(9)

Subtract equation (2) from equation (9).

ΔE(v=0→v=3)=(72hve−494hvexe)−(12hve−14hvexe)=3hve−12hvexe …(10)

Substitute the value of Planck’s constant, ve and xeve in equation (10).

ΔE(v=0→v=3)=(3×6.626×10−34 J⋅s×8962.94×1010 s−1−12×6.626×10−34 J⋅s×156.04×1010 s−1)=178165.32×10−24 J−12407.05×10−24 J=1.657×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(11)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (11).

v=1.657×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=8341.69 cm−1

Substitute v=4 in equation (1).

E=hve(4+12)−hvexe(4+12)2=92hve−814hvexe …(12)

Subtract equation (2) from equation (12).

ΔE(v=0→v=4)=(92hve−814hvexe)−(12hve−14hvexe)=4hve−20hvexe …(13)

Substitute the value of Planck’s constant, ve and xeve in equation (13).

ΔE(v=0→v=4)=(4×6.626×10−34 J⋅s×8962.94×1010 s−1−20×6.626×10−34 J⋅s×156.04×1010 s−1)=237553.76×10−24 J−20678.42×10−24 J=2.169×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(14)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (14).

v=2.169×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=10919.20 cm−1

Substitute v=5 in equation (1).

E=hve(5+12)−hvexe(5+12)2=112hve−1214hvexe …(15)

Subtract equation (2) from equation (15).

ΔE(v=0→v=5)=(112hve−1214hvexe)−(12hve−14hvexe)=5hve−30hvexe …(16)

Substitute the value of Planck’s constant, ve and xeve in equation (16).

ΔE(v=0→v=5)=(5×6.626×10−34 J⋅s×8962.94×1010 s−1−30×6.626×10−34 J⋅s×156.04×1010 s−1)=296942.20×10−24 J−31017.63×10−24 J=2.659×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(17)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (17).

v=2.659×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=13385.96 cm−1

Conclusion

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

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Answer 5

Chapter 14, Problem 14.66E

Interpretation Introduction

Interpretation:

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are to be calculated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant.

Expert Solution & Answer

Answer to Problem 14.66E

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

Explanation of Solution

The energy of a Morse oscillator is calculated by the formula shown below.

E=hve(v+12)−hvexe(v+12)2 …(1)

Where,

• v is the vibrational quantum number.

• h is the Planck’s constant.

• xe is the anharmonicity constant.

• ve is the harmonic vibrational frequency.

The value of ve for HCl is 2989.74 cm−1.

The value of xeve for HCl is 52.05 cm−1.

Convert 2989.74 cm−1 to s−1.

2989.74 cm−1=2989.74 cm−1×2.9979×1010 cm s−1=8962.94×1010 s−1

Convert 52.05 cm−1 to s−1.

52.05 cm−1=52.05 cm−1×2.9979×1010 cm s−1=156.04×1010 s−1

Substitute v=0 in equation (1).

E=hve(0+12)−hvexe(0+12)2=12hve−14hvexe …(2)

Substitute v=1 in equation (1).

E=hve(1+12)−hvexe(1+12)2=32hve−94hvexe …(3)

Subtract equation (2) from equation (3).

ΔE(v=0→v=1)=(32hve−94hvexe)−(12hve−14hvexe)=hve−2hvexe …(4)

Substitute the value of Planck’s constant, ve and xeve in equation (4).

ΔE(v=0→v=1)=(6.626×10−34 J⋅s×8962.94×1010 s−1−2×6.626×10−34 J⋅s×156.04×1010 s−1)=59388.44×10−24 J−2067.84×10−24 J=5.732×10−20 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(5)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (5).

v=5.732×10−20 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=2885.61 cm−1

Substitute v=2 in equation (1).

E=hve(2+12)−hvexe(2+12)2=52hve−254hvexe …(6)

Subtract equation (2) from equation (6).

ΔE(v=0→v=2)=(52hve−254hvexe)−(12hve−14hvexe)=2hve−6hvexe …(7)

Substitute the value of Planck’s constant, ve and xeve in equation (7).

ΔE(v=0→v=2)=(2×6.626×10−34 J⋅s×8962.94×1010 s−1−6×6.626×10−34 J⋅s×156.04×1010 s−1)=118776.88×10−24 J−6203.53×10−24 J=1.126×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(8)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (8).

v=1.126×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=5668.52 cm−1

Substitute v=3 in equation (1).

E=hve(3+12)−hvexe(3+12)2=72hve−494hvexe …(9)

Subtract equation (2) from equation (9).

ΔE(v=0→v=3)=(72hve−494hvexe)−(12hve−14hvexe)=3hve−12hvexe …(10)

Substitute the value of Planck’s constant, ve and xeve in equation (10).

ΔE(v=0→v=3)=(3×6.626×10−34 J⋅s×8962.94×1010 s−1−12×6.626×10−34 J⋅s×156.04×1010 s−1)=178165.32×10−24 J−12407.05×10−24 J=1.657×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(11)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (11).

v=1.657×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=8341.69 cm−1

Substitute v=4 in equation (1).

E=hve(4+12)−hvexe(4+12)2=92hve−814hvexe …(12)

Subtract equation (2) from equation (12).

ΔE(v=0→v=4)=(92hve−814hvexe)−(12hve−14hvexe)=4hve−20hvexe …(13)

Substitute the value of Planck’s constant, ve and xeve in equation (13).

ΔE(v=0→v=4)=(4×6.626×10−34 J⋅s×8962.94×1010 s−1−20×6.626×10−34 J⋅s×156.04×1010 s−1)=237553.76×10−24 J−20678.42×10−24 J=2.169×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(14)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (14).

v=2.169×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=10919.20 cm−1

Substitute v=5 in equation (1).

E=hve(5+12)−hvexe(5+12)2=112hve−1214hvexe …(15)

Subtract equation (2) from equation (15).

ΔE(v=0→v=5)=(112hve−1214hvexe)−(12hve−14hvexe)=5hve−30hvexe …(16)

Substitute the value of Planck’s constant, ve and xeve in equation (16).

ΔE(v=0→v=5)=(5×6.626×10−34 J⋅s×8962.94×1010 s−1−30×6.626×10−34 J⋅s×156.04×1010 s−1)=296942.20×10−24 J−31017.63×10−24 J=2.659×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(17)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (17).

v=2.659×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=13385.96 cm−1

Conclusion

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

Answer 6

Chapter 14, Problem 14.66E

Interpretation Introduction

Interpretation:

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are to be calculated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant.

Expert Solution & Answer

Answer to Problem 14.66E

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

Explanation of Solution

The energy of a Morse oscillator is calculated by the formula shown below.

E=hve(v+12)−hvexe(v+12)2 …(1)

Where,

• v is the vibrational quantum number.

• h is the Planck’s constant.

• xe is the anharmonicity constant.

• ve is the harmonic vibrational frequency.

The value of ve for HCl is 2989.74 cm−1.

The value of xeve for HCl is 52.05 cm−1.

Convert 2989.74 cm−1 to s−1.

2989.74 cm−1=2989.74 cm−1×2.9979×1010 cm s−1=8962.94×1010 s−1

Convert 52.05 cm−1 to s−1.

52.05 cm−1=52.05 cm−1×2.9979×1010 cm s−1=156.04×1010 s−1

Substitute v=0 in equation (1).

E=hve(0+12)−hvexe(0+12)2=12hve−14hvexe …(2)

Substitute v=1 in equation (1).

E=hve(1+12)−hvexe(1+12)2=32hve−94hvexe …(3)

Subtract equation (2) from equation (3).

ΔE(v=0→v=1)=(32hve−94hvexe)−(12hve−14hvexe)=hve−2hvexe …(4)

Substitute the value of Planck’s constant, ve and xeve in equation (4).

ΔE(v=0→v=1)=(6.626×10−34 J⋅s×8962.94×1010 s−1−2×6.626×10−34 J⋅s×156.04×1010 s−1)=59388.44×10−24 J−2067.84×10−24 J=5.732×10−20 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(5)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (5).

v=5.732×10−20 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=2885.61 cm−1

Substitute v=2 in equation (1).

E=hve(2+12)−hvexe(2+12)2=52hve−254hvexe …(6)

Subtract equation (2) from equation (6).

ΔE(v=0→v=2)=(52hve−254hvexe)−(12hve−14hvexe)=2hve−6hvexe …(7)

Substitute the value of Planck’s constant, ve and xeve in equation (7).

ΔE(v=0→v=2)=(2×6.626×10−34 J⋅s×8962.94×1010 s−1−6×6.626×10−34 J⋅s×156.04×1010 s−1)=118776.88×10−24 J−6203.53×10−24 J=1.126×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(8)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (8).

v=1.126×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=5668.52 cm−1

Substitute v=3 in equation (1).

E=hve(3+12)−hvexe(3+12)2=72hve−494hvexe …(9)

Subtract equation (2) from equation (9).

ΔE(v=0→v=3)=(72hve−494hvexe)−(12hve−14hvexe)=3hve−12hvexe …(10)

Substitute the value of Planck’s constant, ve and xeve in equation (10).

ΔE(v=0→v=3)=(3×6.626×10−34 J⋅s×8962.94×1010 s−1−12×6.626×10−34 J⋅s×156.04×1010 s−1)=178165.32×10−24 J−12407.05×10−24 J=1.657×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(11)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (11).

v=1.657×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=8341.69 cm−1

Substitute v=4 in equation (1).

E=hve(4+12)−hvexe(4+12)2=92hve−814hvexe …(12)

Subtract equation (2) from equation (12).

ΔE(v=0→v=4)=(92hve−814hvexe)−(12hve−14hvexe)=4hve−20hvexe …(13)

Substitute the value of Planck’s constant, ve and xeve in equation (13).

ΔE(v=0→v=4)=(4×6.626×10−34 J⋅s×8962.94×1010 s−1−20×6.626×10−34 J⋅s×156.04×1010 s−1)=237553.76×10−24 J−20678.42×10−24 J=2.169×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(14)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (14).

v=2.169×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=10919.20 cm−1

Substitute v=5 in equation (1).

E=hve(5+12)−hvexe(5+12)2=112hve−1214hvexe …(15)

Subtract equation (2) from equation (15).

ΔE(v=0→v=5)=(112hve−1214hvexe)−(12hve−14hvexe)=5hve−30hvexe …(16)

Substitute the value of Planck’s constant, ve and xeve in equation (16).

ΔE(v=0→v=5)=(5×6.626×10−34 J⋅s×8962.94×1010 s−1−30×6.626×10−34 J⋅s×156.04×1010 s−1)=296942.20×10−24 J−31017.63×10−24 J=2.659×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(17)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (17).

v=2.659×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=13385.96 cm−1

Conclusion

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

Answer 7

Chapter 14, Problem 14.66E

Interpretation Introduction

Interpretation:

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are to be calculated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. An electronic state of energy has its own vibrational states. The energy between the electronic states is large followed by vibrational states and then rotational states. During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant.

Answer 8

Expert Solution & Answer

Answer to Problem 14.66E

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The energy of a Morse oscillator is calculated by the formula shown below.

E=hve(v+12)−hvexe(v+12)2 …(1)

Where,

• v is the vibrational quantum number.

• h is the Planck’s constant.

• xe is the anharmonicity constant.

• ve is the harmonic vibrational frequency.

The value of ve for HCl is 2989.74 cm−1.

The value of xeve for HCl is 52.05 cm−1.

Convert 2989.74 cm−1 to s−1.

2989.74 cm−1=2989.74 cm−1×2.9979×1010 cm s−1=8962.94×1010 s−1

Convert 52.05 cm−1 to s−1.

52.05 cm−1=52.05 cm−1×2.9979×1010 cm s−1=156.04×1010 s−1

Substitute v=0 in equation (1).

E=hve(0+12)−hvexe(0+12)2=12hve−14hvexe …(2)

Substitute v=1 in equation (1).

E=hve(1+12)−hvexe(1+12)2=32hve−94hvexe …(3)

Subtract equation (2) from equation (3).

ΔE(v=0→v=1)=(32hve−94hvexe)−(12hve−14hvexe)=hve−2hvexe …(4)

Substitute the value of Planck’s constant, ve and xeve in equation (4).

ΔE(v=0→v=1)=(6.626×10−34 J⋅s×8962.94×1010 s−1−2×6.626×10−34 J⋅s×156.04×1010 s−1)=59388.44×10−24 J−2067.84×10−24 J=5.732×10−20 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(5)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (5).

v=5.732×10−20 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=2885.61 cm−1

Substitute v=2 in equation (1).

E=hve(2+12)−hvexe(2+12)2=52hve−254hvexe …(6)

Subtract equation (2) from equation (6).

ΔE(v=0→v=2)=(52hve−254hvexe)−(12hve−14hvexe)=2hve−6hvexe …(7)

Substitute the value of Planck’s constant, ve and xeve in equation (7).

ΔE(v=0→v=2)=(2×6.626×10−34 J⋅s×8962.94×1010 s−1−6×6.626×10−34 J⋅s×156.04×1010 s−1)=118776.88×10−24 J−6203.53×10−24 J=1.126×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(8)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (8).

v=1.126×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=5668.52 cm−1

Substitute v=3 in equation (1).

E=hve(3+12)−hvexe(3+12)2=72hve−494hvexe …(9)

Subtract equation (2) from equation (9).

ΔE(v=0→v=3)=(72hve−494hvexe)−(12hve−14hvexe)=3hve−12hvexe …(10)

Substitute the value of Planck’s constant, ve and xeve in equation (10).

ΔE(v=0→v=3)=(3×6.626×10−34 J⋅s×8962.94×1010 s−1−12×6.626×10−34 J⋅s×156.04×1010 s−1)=178165.32×10−24 J−12407.05×10−24 J=1.657×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(11)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (11).

v=1.657×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=8341.69 cm−1

Substitute v=4 in equation (1).

E=hve(4+12)−hvexe(4+12)2=92hve−814hvexe …(12)

Subtract equation (2) from equation (12).

ΔE(v=0→v=4)=(92hve−814hvexe)−(12hve−14hvexe)=4hve−20hvexe …(13)

Substitute the value of Planck’s constant, ve and xeve in equation (13).

ΔE(v=0→v=4)=(4×6.626×10−34 J⋅s×8962.94×1010 s−1−20×6.626×10−34 J⋅s×156.04×1010 s−1)=237553.76×10−24 J−20678.42×10−24 J=2.169×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(14)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (14).

v=2.169×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=10919.20 cm−1

Substitute v=5 in equation (1).

E=hve(5+12)−hvexe(5+12)2=112hve−1214hvexe …(15)

Subtract equation (2) from equation (15).

ΔE(v=0→v=5)=(112hve−1214hvexe)−(12hve−14hvexe)=5hve−30hvexe …(16)

Substitute the value of Planck’s constant, ve and xeve in equation (16).

ΔE(v=0→v=5)=(5×6.626×10−34 J⋅s×8962.94×1010 s−1−30×6.626×10−34 J⋅s×156.04×1010 s−1)=296942.20×10−24 J−31017.63×10−24 J=2.659×10−19 J

The frequency of a transition in cm−1 is calculated by the formula shown below.

v=ΔEhc …(17)

Substitute the value of ΔE, Planck’s constant and speed of light in equation (17).

v=2.659×10−19 J6.626×10−34 J⋅s×2.9979×1010 cm s−1=13385.96 cm−1

Answer 12

Conclusion

The values of the five transitions for HCl using equation for the energy of a Morse oscillator are 2885.61 cm−1, 5668.52 cm−1, 8341.69 cm−1, 10919.20 cm−1 and 13385.96 cm−1. The predicted values are very close to the experimental values.

Answer 13

Ch. 14 - Prob. 14.1E Ch. 14 - Determine if the following integrals can be...Ch. 14 - What is the frequency of light having the...Ch. 14 - What is the wavelength of light having the given...Ch. 14 - What is the energy of light having each...Ch. 14 - The Cu(H2O)62+ complex has octahedral symmetry. Is...Ch. 14 - What are the wavelength, speed, and energy of a...Ch. 14 - Prob. 14.8E Ch. 14 - Prob. 14.9E Ch. 14 - Prob. 14.10E

Answer to Problem 14.66E

Explanation of Solution

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Chapter 14 Solutions