(a) Interpretation: The excited-state and the ground-state population ratios for HCI: N ( v = 1 ) / N ( v = 0 ) is to be calculated. Concept introduction: The excited-state of an atom, molecule or electron is a state in which the electrons have sufficient energy to jump in to another orbital. The ground-state is the state of zero energy level. In this state the electron does not have sufficient energy to jump from the orbital. The Boltzmann equation is used to calculate the ratio of the exited-state and ground state. The concept of the energy difference between these states is also used to calculate the required ratio.

Question

Don't used Ai solution

Answer 1

Question

Chapter 16, Problem 16.11QAP

Interpretation Introduction

(a)

Interpretation:

The excited-state and the ground-state population ratios for HCI: N(v=1)/N(v=0) is to be calculated.

Concept introduction:

The excited-state of an atom, molecule or electron is a state in which the electrons have sufficient energy to jump in to another orbital. The ground-state is the state of zero energy level. In this state the electron does not have sufficient energy to jump from the orbital.

The Boltzmann equation is used to calculate the ratio of the exited-state and ground state. The concept of the energy difference between these states is also used to calculate the required ratio.

Interpretation Introduction

(b)

Interpretation:

The excited-state and the ground-state population ratios for HCI: N(v=2)/N(v=0) is to be calculated.

Concept introduction:

The excited-state of an atom, molecule or electron is a state in which the electrons have sufficient energy to jump in to another orbital. The ground-state is the state of zero energy level. In this state the electron does not have sufficient energy to jump from the orbital.

The Boltzmann equation is used to calculate the ratio of the exited-state and ground state. The concept of the energy difference between these states is also used to calculate the required ratio.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Students have asked these similar questions

Don't used Ai solution

Indicate the correct option.a) Graphite conducts electricity, being an isotropic materialb) Graphite is not a conductor of electricityc) Both are false

(f) SO: Best Lewis Structure 3 e group geometry:_ shape/molecular geometry:, (g) CF2CF2 Best Lewis Structure polarity: e group arrangement:_ shape/molecular geometry: (h) (NH4)2SO4 Best Lewis Structure polarity: e group arrangement: shape/molecular geometry: polarity: Sketch (with angles): Sketch (with angles): Sketch (with angles):