Question

The speed of electromagnetic waves (which include visible light, radio, and x rays) in vacuum is 3.0 x 108 m/s. Wavelengths of visible light waves range from about 400 nm in the violet to about 700 nm in the red. What are the (a) minimum and (b) maximum frequencies of these waves? The range of frequencies for shortwave radio (for example, FM radio and VHF television) is 1.5 to 300 MHz. What are the (c) minimum and (d) maximum wavelengths of these waves? X ray wavelengths range from about 5.0 nm to about 1.0 x 10-2 nm. What are the (e) minimum and (f) maximum frequencies of x rays?

Answer 1

(a)
`4.29\cdot 10^(14)Hz`

The frequency of an electromagnetic wave is given by

`f=(c)/(\lambda)`

where

c is the speed of light

`\lambda` is the wavelength

We notice from the formula that the frequency is inversely proportional to the wavelength, so the minimum frequency corresponds to the maximum wavelength, and viceversa

The maximum value of the wavelength of the visible light waves is

`\lambda_(max) = 700 nm = 7.0\cdot 10^(-7) m` (red light)

so the minimum frequency of visible light is

`f_(min) = (c)/(\lambda_(max))=(3.0\cdot 10^8 m/s)/(7.00\cdot 10^(-7)m)=4.29\cdot 10^(14)Hz`

(b)
`7.50\cdot 10^(14)Hz`

The maximum frequency corresponds to the minimum wavelength;

The minimum wavelength is

`\lambda_(min) = 400 nm = 4.0\cdot 10^(-7) m` (violet light)

so the maximum frequency of visible light is

`f_(max) = (c)/(\lambda_(min))=(3.0\cdot 10^8 m/s)/(4.00\cdot 10^(-7)m)=7.50\cdot 10^(14)Hz`

(c) 1 m

The wavelength of an electromagnetic wave is given by

`\lambda=(c)/(f)`

as before, we notice that the minimum wavelength corresponds to the maximum frequency, and viceversa.

The maximum frequency of shortwave radio waves is

`f_(max)=300 MHz = 3.0\cdot 10^8 Hz`

so the minimum wavelength of these waves is

`\lambda_(min) = (c)/(f_(max))=(3.0\cdot 10^8 m/s)/(3.0\cdot 10^8 Hz)=1 m`

(d) 200 m

The minimum frequency of shortwave radio waves is

`f_(min)=1.5 MHz = 1.5\cdot 10^6 Hz`

so the maximum wavelength of these waves is

`\lambda_(max) = (c)/(f_(min))=(3.0\cdot 10^8 m/s)/(1.5\cdot 10^6 Hz)=200 m`

(e)
`6.0\cdot 10^(16)Hz`

As in part (a) and (b), we can find the frequency of the X-rays by using the formula

`f=(c)/(\lambda)`

The maximum wavelength of the x-rays is

`\lambda_(max) = 5.0 nm = 5.0\cdot 10^(-9) m`

so the minimum frequency is

`f_(min) = (c)/(\lambda_(max))=(3.0\cdot 10^8 m/s)/(5.0\cdot 10^(-9)m)=6.0\cdot 10^(16)Hz`

(f)
`3.0\cdot 10^(19)Hz`

The maximum frequency corresponds to the minimum wavelength;

The minimum wavelength is

`\lambda_(min) = 1.0\cdot 10^(-2) nm = 1.0\cdot 10^(-11) m`

so the maximum frequency of the x-rays is

`f_(max) = (c)/(\lambda_(min))=(3.0\cdot 10^8 m/s)/(1.0\cdot 10^(-11)m)=3.0\cdot 10^(19)Hz`