Chapter 6, Problem 69GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# For an electron in a hydrogen atom, calculate the energy of the photon emitted when an electron falls in energy from the n = 5 level to the n = 2 state. What are the frequency and wavelength of this electromagnetic radiation?

Interpretation Introduction

Interpretation: The energy, frequency and wavelength of electromagnetic radiation emitted during n=5 to n=2 transition of the excited H atom have to be calculated.

Concept introduction:

EnergybetweenthestatesΔE=EfinalEinitial=Rhc(1nfinal21ninitial2)where,R=Rydbergconstanth=Planck'sconstantc=speedoflightn=Principalquantumnumber

Planck’s equation,

E==hcλwhere, ν=frequencyλ=wavelength

The energy increases as the wavelength of the light decreases.

Explanation

The energy, frequency and wavelength of electromagnetic radiation emitted during nâ€‰=â€‰5 to nâ€‰=â€‰2 transition of the excited H atom is calculated.

Given,

The transition of electron of the excited H atom is from nâ€‰=â€‰5 to nâ€‰=â€‰2

Â Â Râ€‰=â€‰1.097â€‰Ã—â€‰107â€‰mâ€‰âˆ’1hâ€‰=â€‰6.626â€‰Ã—â€‰10â€‰-34â€‰J.scâ€‰=â€‰2.998â€‰Ã—â€‰10â€‰8â€‰m/sninitialâ€‰=â€‰5nfinalâ€‰=â€‰2

The energy difference between states while emitting photons is,

Energyâ€‰betweenâ€‰theâ€‰statesâ€‰â€‰Î”Eâ€‰=â€‰Efinalâ€‰âˆ’â€‰Einitial=â€‰âˆ’Rhc(1nfinal2â€‰âˆ’â€‰1ninitial2)â€‰=â€‰âˆ’â€‰1.097â€‰Ã—â€‰107â€‰mâ€‰âˆ’1â€‰Ã—â€‰6.626â€‰Ã—â€‰10â€‰âˆ’34â€‰J.sâ€‰Ã—â€‰2.998â€‰Ã—â€‰10â€‰8â€‰m/sâ€‰(122â€‰âˆ’â€‰152)=â€‰âˆ’4.576â€‰Ã—â€‰10â€‰âˆ’19â€‰J

Therefore,

The energy difference between states while emitting photons is 4.576â€‰Ã—â€‰10â€‰âˆ’19â€‰J

The frequency of electromagnetic radiation emitted during nâ€‰=â€‰5 to nâ€‰=â€‰2 transition of the excited H atom is calculated,

â€‚Â Ephotonâ€‰=â€‰hÎ½

Therefore,

â€‚Â Î½â€‰=â€‰Ephotonh=â€‰4

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