Chapter 22, Problem 7P

To determine

The wavelength of electromagnetic wave and its classification.

Answer to Problem 7P

  $311nm$ , Ultraviolet

Explanation of Solution

Given info:

Frequency, $f=9.66\times {10}^{14}Hz$

Speed of light, $c=3.00\times {10}^{8}m/s$

Formula used:

The relation between frequency and wavelength: $f=\frac{c}{\lambda }$

Where,

  $f$ is frequency

  $\lambda$ is wavelength

  $c$ is speed of light

Calculation:

Rearrange the relation between frequency and wavelength: $f=\frac{c}{\lambda }$ . So, $\lambda =\frac{c}{f}$

Substitute the known values in formula, $\lambda =\frac{c}{f}$

  $\lambda =\frac{3\times {10}^{8}m/s}{9.66\times {10}^{14}Hz}=3.11\times {10}^{-7}m=311nm$

This wavelength corresponds to ultraviolet region of the visible spectrum.

Conclusion: The required wavelength is $311nm$ and it lies in ultraviolet region of the visible spectrum.