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Return to Part One of Light Equations
Go to Part Two of Light Equations
These problems sometimes require metric conversions. If you wish to review metric conversions, click here.
Problem #1: Light with a frequency of 7.26 x 1014 Hz lies in the violet region of the visible spectrum. What is the wavelength of this frequency of light? Answer in units of nm.
Solution:
1) Substitute into λν = c:
(x) (7.26 x 1014 s¯1) = 3.00 x 108 m s¯1x = 4.13 x 10¯7 m
2) Convert from m to nm:
4.13 x 10¯7 m x (109 nm / 1 m) = 413 nm
Problem #2: When an electron beam strikes a block of copper, x-rays of frequency 1.07 x 1019 Hz are emitted. What is the wavelength of these x-rays? Answer in units of pm.
Solution:
1) Substitute into λν = c:
(x) (1.07 x 1019 s¯1) = 3.00 x 108 m s¯1x = 2.80 x 10¯11 m
2) Convert from m to pm:
2.80 x 10¯11 m x (1012 pm / 1 m) = 28 pm
By the way, if the question had asked for the answer in nm, it would have been 0.0280 nm. Notice that the wavelength unit in the above questions was deliberately picked to give a whole number. Generally speaking, the wavelength unit is picked to give a whole number, be it tens, hundreds or thousands.
Problem #3: Calculate the wavelength (in meters) of radiation a frequency of 5.00 x 1014 s¯1.
Solution:
1) Substitute into λν = c:
(x) (5.00 x 1014 s¯1) = 3.00 x 108 m s¯1x = 6.00 x 10¯7 m
2) By the way, the unit typically used for wavelengths of visible light is nanometers:
6.00 x 10¯7 m times (109 nm / 1 m) = 600 nm.Visible light has a wavelegth somewhere between 400 and 700 nm.
Sometimes you might be asked what color a radiation might be. Look at problems below for examples of this. What color might 600 nm be?
Problem #4: The radio station KUSC (in Southern California) broadcasts at 91.5 MHz. Calculate its wavelength in meters.
Solution:
1) Convert MHz to s¯1:
91.5 MHz = 91.5 x 106 s¯1 = 9.15 x 107 s¯1
1) Use λν = c:
(x) (9.15 x 107 s¯1) = 3.00 x 108 m/sx = 3.28 m
Problem #5: Calculate the wavelength of radiation emitted from radioactive cobalt with a frequency of 2.80 x 1020 s¯1. What region of the eletromagnetic spectrum does this lie in?
Solution:
1) Calculate the wavelength:
λν = c(x) (2.80 x 1020 s¯1) = 3.00 x 108 m/s
x = 1.07 x 10¯12 m
2) Determine the region of the EM spectrum:
Consult a convenient reference source.Gamma rays.
Problem #6: 1.50 x 1013 Hz? Does this radiation have a longer or shorter wavelength than red light?
Comment: there are a number of ways to answer this question. I'll do two.
Solution #1:
Calculate the frequency of 7000 Å (the longest wavelength of red light):
7000 Å = 7000 x 10¯8 cm = 7.00 x 10¯5 cm (three sig figs is a reasonable assumption)λν = c
(7.00 x 10¯5 cm) (x) = 3.00 x 1010 cm/s
x = 4.28 x 10¯14 s¯1
A lower frequency (like the value given in the problem) means a longer wavelength. The radiation in the problem has a longer wavelength than the red light I used.
Beyond red on the EM spectrum is infrared.
Solution #2:
Consult a convenient reference source.Compare 1.50 x 1013 Hz to the various values in the frequency column of the above web site.
Determine that the frequency given in the problem lies in the infrared, a region that has a longer wavelength than red light does.
Problem #7: What is the wavelength of sound waves having a frequency of 256.0 sec¯1 at 20 °C? Speed of sound = 340.0 m/sec. (The problem is about sound, but this does not change the basic idea of the equation "wavelength times frequency = speed."
Solution:
a) Use λν = speed:
(x) (256.0 sec¯1) = 340.0 m/secThe answer, to four sig figs, is 1.328 m.
Just for kicks: in centimeters, 132.8 cm, and in Ångströms, 1.328 x 1010 Å
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