   Chapter 6, Problem 54GQ

Chapter
Section
Textbook Problem

Excited H atoms have many emission lines. One series of lines, called the Pfund series, occurs in the infrared region. It results when an electron changes from higher energy levels to a level with n = 5. Calculate the wavelength and frequency of the lowest energy line of this series.

Interpretation Introduction

Interpretation: The wavelength and frequency of the least energetic in Pfund series of the excited H atom is to be calculated.

Concept introduction:

• Electronic transitions that take place in excited H atom is,
1. 1. Lyman series: electronic transitions take place to the n=1 level and it is in ultraviolet region.
1. 2. Balmer series: electronic transitions take place from n>2 to the n=2 level and it is in visible region.
2. 3. Ritz-Paschen series: electronic transitions take place from n>3 to the n=3 level and it is in infrared region.
3. 4. Brackett series: electronic transitions take place from n>4 to the n=4 level.
4. 5. Pfund series: electronic transitions take place from n>5 to the n=5 level

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

As the energy gap between two transition states increases the wavelength of the radiation emitted decreases

• Planck’s equation,

E==hcλwhere, E=energyh=Planck'sconstantν=frequency

The energy increases as the wavelength of the light decrease. Also the energy increases as the frequency of the light increases.

• The frequency of the light is inversely proportional to its wavelength.

ν=cλwhere, c=speedoflightν=frequencyλ=wavelength

Explanation

As the energy gap between two transition states decreases the wavelength of the radiation emitted increases

Hence the line with highest wavelength is produced in Pfund series of H atom when the electronic transitions that take place from the n=6 to the n=5.

Since E=hcλ ,the energy decreases as the wavelength of the light increases. The electronic transitions that taking place from the n=6 to the n=5 has the least energy

The lowest energy line in Pfund series of the excited H atom forms if the transition of electron is from n=6 to the n=5

R=1.097×107m1h=6.626×10-34J.sc=2.998×108m/sninitial=6nfinal=5

Energy is determined,

EnergybetweenthestatesΔE=EfinalEinitial=Rhc(1nfinal21ninitial2)=1.097×107m1×6.626×1034J.s×2.998×108m/s(152162)=2.6634×1020J/photon

The energy emitted by the photon does not have any sign and absolute value is taken

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