   # 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? ### 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 11PS
Textbook Problem
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## 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 of Solution

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 3.3×1019J/atom601.96nm

Planck'sconstant,h=6.626×1034J.sspeedoflight,c=2.998×108m/s

The energy per photon is,

E=                                (a)

The frequency per mole of photons is,

ν=cλ                                    (b)

Combining equation (a) and (b),

The energy per photon is,

E=hcλ

Therefore,

λ=hcE=6

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