Chapter 6, Problem 11PS

### 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

# An energy of 3.3 × 10–19 J/atom is required to cause a cesium atom on a metal surface to lose an electron. Calculate the longest possible wavelength of light that can ionize a cesium atom. In what region of the electromagnetic spectrum is that radiation found?

Interpretation Introduction

Interpretation:

The longest possible wavelength that can ionize a cesium atom and the region in which this radiation belongs to has to be determined.

Concept introduction:

• Electromagnetic radiations are a type of energy surrounding us. They are of different types like radio waves, IR, UV, X-ray etc.
• The wavelength of visible light lies between 400nm to 700nm
• Photoelectric effect: electrons get ejected if a ray of light hits on the surface of metal when the light has adequately high frequency.
• Planck’s equation,

E=where, 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

The longest possible wavelength that can ionize a cesium atom and the region in which this radiation belongs to is calculated below.

Given,

The energy required to lose an electron is 3.3â€‰Ã—â€‰10âˆ’19â€‰J/atom601.96â€‰nm

â€‚Â Planck'sâ€‰constant,hâ€‰=â€‰6.626â€‰Ã—â€‰10â€‰âˆ’34â€‰J.sspeedâ€‰ofâ€‰light,câ€‰=â€‰2.998â€‰Ã—â€‰10â€‰8â€‰m/s

The energy per photon is,

Â Â Â Â Eâ€‰=â€‰hÎ½â€‰Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (a)

The frequency per mole of photons is,

Â Â Â Â Î½â€‰=â€‰cÎ»Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b)

Combining equation (a) and (b),

The energy per photon is,

â€‚Â Eâ€‰=â€‰hcÎ»â€‰

Therefore,

â€‚Â Î»â€‰=hcEâ€‰=â€‰6

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