Sample Solution from
Quantum Chemistry
2nd Edition
ISBN: 9781891389504
Chapter 1
Problem 1.1P
Try another sample solution
Interpretation Introduction

Interpretation:

The values of ν, ν¯ and E should be calculated for ultraviolet radiation having λ = 200 nm and also, the result should be compared with figure 1.19.

Concept Introduction:

The mathematical expression which gives the relation between velocity of light, frequency and wavelength is:

  ν=cλ

Where, ν = frequency

  c = velocity of light

  λ = wavelength

The mathematical expression for wave number is expressed as:

  ν¯=1λ

Where, ν¯ = wave number

  λ = wavelength

The mathematical expression for energy is expressed as:

  E=hcλ

Where, E = energy

  h = Plank’s constant ( 6.626×10−34 J⋅s )

  c = velocity of light

  λ = wavelength

Expert Solution

Answer

  ν=1.50×1015 Hz

  ν¯=5.00×104 cm−1

  E=9.93×10−19 J

In figure 1.19, the values of wavelength and frequency are close to the calculated values.

Explanation of Solution

Given information:

  λ =200 nm

The frequency is calculated as:

  ν=cλ

Convert the given value of wavelength in nm to m.

Since, 1 nm = 1×10−9 m

Thus,

Wavelength in m = 200 nm ×10−9 mnm

= 200×10−9 m

Put the values,

  ν=cλ

  ν=3×108 m/s200×10−9 m

  ν=1.50×1015 s−1

  ν=1.50×1015 s−1×(1 Hz1 s−1)

  ν=1.50×1015 Hz

Wave number is calculated as:

  ν¯=1λ

Convert the given value of wavelength in nm to cm.

Since, 1 nm = 1×10−7 cm

Thus,

Wavelength in m = 200 nm ×10−7 cmnm

= 200×10−7 cm

Put the values,

  ν¯=1λ

  ν¯=1200×10−7 cm

  ν¯=5.00×104 cm−1

Energy is calculated as:

  E=hcλ

Put the values,

  E=(6.626×10−34 J⋅s)(3×108 m/s)200×10−9 m

  E=9.93×10−19 J

Figure 1.19 is shown as:

  

In the above figure, the value of wavelength and frequency is 10−8 and 1016 which are very close to the calculated values that are ν=1.50×1015 Hz and λ=200×10−9 m .

Not sold yet?Try another sample solution