# (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

### Principles of Instrumental Analysis

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

### 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=hcλAnd 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

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