(a) Interpretation: At higher temperature rate, transition of atoms of sodium occurs from 4s to 3p state and the average wavelength of sodium due to the emission is 1139 nm. Ratio of excited state 4 seconds to ground state 3seconds needs to be calculated for acetylene- oxygen flame of 3000 o C. Concept introduction: Calculation of ratio is done by using the Boltzmann equation, given as- N j N o = g j g o exp ( − E j k T ) Where, N j = no. of ions in excited state N o = no. of ions in ground state Ej = energy difference of excited state and ground state g j = statistical weight for excited state g o = statistical weight for ground state k= Boltzmann constant T = absolute temperature Energy of atom is calculated by the following formula- E j = h c λ Where, h= Planck’s constant c = light velocity λ= wavelength Ej= energy difference

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 9.15QAP
Interpretation Introduction

(a)Interpretation:At higher temperature rate, transition of atoms of sodium occurs from 4s to 3p state and the average wavelength of sodium due to the emission is 1139 nm. Ratio of excited state 4 seconds to ground state 3seconds needs to be calculated for acetylene- oxygen flame of 3000oC.Concept introduction:Calculation of ratio is done by using the Boltzmann equation, given as-NjNo=gjgoexp(−EjkT)Where,Nj = no. of ions in excited stateNo = no. of ions in ground stateEj = energy difference of excited state and ground stategj = statistical weight for excited statego = statistical weight for ground statek= Boltzmann constantT = absolute temperatureEnergy of atom is calculated by the following formula-Ej=hcλWhere,h= Planck’s constantc = light velocityλ= wavelengthEj= energy difference

Interpretation Introduction

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