Sample Solution from
Physical Chemistry: A Molecular Approach
97th Edition
ISBN: 9780935702996
Chapter 1
Problem 1.1P
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Interpretation Introduction

Interpretation:The value of ν ,  ν¯  and E needs to be determined for ultraviolet radiation with λ = 200 nm and the result needs to be compared with given figure of electromagnetic spectrum.

Concept Introduction:Electromagnetic spectrum involves different radiations from the lowest to the highest frequency values. Some common radiations which are part of electromagnetic radiations are infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

The relation between energy and wavelength is:

  E = h × cλ  

Here:

h= Plank’s constant = 6.626 x 10-34 J s

c= speed of light = 2.99 x 108 m/s

λ= wavelength; E = energy

Expert Solution

Answer

  E = 9.90× 10-19J

  ν¯  =  5×106 m-1

  ν  =  1.5×1015 sec−1

Explanation of Solution

The wavelength is given as follows:

  λ = 200 nm

Convert wavelength from nm to m:

1 nm = 10-9m

   200 nm × 10-9m1 nm=200 ×10-9m

Calculate frequency ν :

  ν  =  cλν  =  2 .99×10 8  m/sec 200×10 -9  mν  =  1.5×1015 sec−1

Calculate wavenumber ν¯ :

  ν¯ =  1λν¯  =  1 200×10 -9  mν¯  =  5×106 m-1

Calculate energy E :

  E = h × cλ  E = 6 .626 × 10 -34 J s × 2 .99 × 108 m/s 200×10 -9 m  E = 9.90× 10-19J

Conclusion

Thus, E = 9.90× 10-19J

  ν¯  =  5×106 m-1

  ν  =  1.5×1015 sec−1

The range of frequency and wavelength confirm the Ultraviolet region as it lies from 10-8 to 10-10 m of wavelength.

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