Electromagnetic Spectrum

Assessment Problems

  • 1. The speed of light is about 3.0 x 108 m/s. How far will light travel in 1 year?

    Answer: We need to determine the number of seconds in a year.


    (1 year)(365 days/year)(24 hr/day)(60 minutes/hr)(60 sec/min) = 3.15 x 107 seconds.

    Then, (3.0 x 108 m/s)(3.15 x 107 s) = 9.5 x 1015 m.


  • 2. A portion of the electromagnetic spectrum emits radiation with a frequency ν and has a wavelength, λ. What would be the wavelength of radiation with frequency 2ν?

    Answer: Since c = νλ and c is the speed of light, a constant, then doubling the frequency would require that the wavelength be halved or be expressed as λ/2.

  • 3. An electromagnetic wave has a frequency of 9.6 x 1014 Hz.  What is its wavelength?  (One Hz equals one cycle per second or 1/s.)

    Answer: c = νλ  Then, λ = c/ν and λ = 3.0 x 108 m/s/9.6 x 1014 1/s = 3.1 x 10-7 m.

  • 4. Is the wave in problem 3 in the visible portion of the electromagnetic spectrum, the infrared or the ultraviolet?  (The range of wavelengths in the visible portion of the electromagnetic spectrum range from about 3.8 x 10-7 m to 7.5 x 10-7 m.)

    Answer: Since the wavelength in problem 3 is 3.1 x 10-7 m and therefore shorter than the shortest wavelength in the visible portion of the visible spectrum, the wave would be in the ultraviolet region

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