(a) Interpretation: Relative intensity of potassium at flame center and height 2 cm above the flame needs to be calculated where the emission line of potassium is 766.5 nm. Concept introduction: For this calculation following formulas will be used- N j N o = g j g o exp ( − E j k T ) Energy of potassium is calculated by- E = h c λ And relative intensity formula- I X I Y = ( N j N o ) h 2 ( N j N o ) h 1

BuyFind

Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213
BuyFind

Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

Solutions

Chapter 9, Problem 9.12QAP
Interpretation Introduction

(a)

Interpretation:

Relative intensity of potassium at flame center and height 2 cm above the flame needs to be calculated where the emission line of potassium is 766.5 nm.

Concept introduction:

For this calculation following formulas will be used-

NjNo=gjgoexp(EjkT)

Energy of potassium is calculated by-

E=hcλ

And relative intensity formula-

IXIY=(NjNo)h2(NjNo)h1

Interpretation Introduction

(b)

Interpretation:

Relative intensity of potassium at flame center and height 3 cm above the flame needs to be calculated where the emission line of potassium is 766.5 nm.

Concept introduction:

Calculation of ratio of excited state and ground state is done by following formula-

NjNo=gjgoexp(EjkT)

And relative intensity formula-

IXIY=(NjNo)h2(NjNo)h1

Interpretation Introduction

(c)

Interpretation:

Relative intensity of potassium at flame center and height 4 cm above the flame needs to be calculated where the emission line of potassium is 766.5 nm.

Concept introduction:

Ratio of excited state and ground state is calculated by the following formula-

NjNo=gjgoexp(EjkT)

And relative intensity formula-

IXIY=(NjNo)h2(NjNo)h1

Interpretation Introduction

(d)

Interpretation:

Relative intensity of potassium at flame center and height 5 cm above the flame needs to be calculated where the emission line of potassium is 766.5 nm.

Concept introduction:

Ratio of excited state and ground state is calculated by the following formula-

NjNo=gjgoexp(EjkT)

And relative intensity formula-

IXIY(NjNo)h2(NjNo)h1

Want to see the full answer?

Check out a sample textbook solution.

Want to see this answer and more?

Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes!*

*Response times may vary by subject and question complexity. Median response time is 34 minutes for paid subscribers and may be longer for promotional offers.