   # 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? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 6, Problem 69GQ
Textbook Problem
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## 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 of Solution

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

Given,

The transition of electron of the excited H atom is from n=5 to n=2

R=1.097×107m1h=6.626×10-34J.sc=2.998×108m/sninitial=5nfinal=2

The energy difference between states while emitting photons is,

EnergybetweenthestatesΔE=EfinalEinitial=Rhc(1nfinal21ninitial2)=1.097×107m1×6.626×1034J.s×2.998×108m/s(122152)=4.576×1019J

Therefore,

The energy difference between states while emitting photons is 4.576×1019J

The frequency of electromagnetic radiation emitted during n=5 to n=2 transition of the excited H atom is calculated,

Ephoton=

Therefore,

ν=Ephotonh=4

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