Problems #11 - 20

Wavelength-Frequency Problems #1 - 10 | Go to Part Two of Light Equations |

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**Given Wavelength, Calculate Frequency**

**Example #11:** What is the frequency of radiation with a wavelength of 5.00 x 10¯^{8} m? In what region of the electromagnetic spectrum is this radiation?

**Solution:**

1) Use λν = c to determine the frequency:

(5.00 x 10¯^{8}m) (x) = 3.00 x 10^{8}m/sx = 6.00 x 10

^{15}s¯^{1}

2) Determine the electromagnetic spectrum region:

Consult a convenient reference source.This frequency is right in the middle of the ultraviolet region of the spectrum.

**Problem #12:** What is the wavelength of sound waves having a frequency of 256.0 sec¯^{1} at 20 °C? Speed of sound = 340.0 m/s. (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 s¯^{1}) = 340.0 m/sThe answer, to four sig figs, is 1.328 m.

Just for kicks: in centimeters, 132.8 cm, and in Ångströms, 1.328 x 10

^{10}Å

Probs 13, 14, 15 here soon

**Given Frequency, Calculate Wavelength**

**Problem #16:** Calculate the wavelength of radiation emitted from radioactive cobalt with a frequency of 2.80 x 10^{20} s¯^{1}. What region of the eletromagnetic spectrum does this lie in?

**Solution:**

1) Calculate the wavelength:

λν = c(x) (2.80 x 10

^{20}s¯^{1}) = 3.00 x 10^{8}m/sx = 1.07 x 10¯

^{12}m

2) Determine the region of the EM spectrum:

Consult a convenient reference source.Gamma rays.

**Problem #17:** 1.50 x 10^{13} 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 10^{10}cm/sx = 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 10

^{13}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 #18:** What color is light whose frequency is 7.39 x 10^{14} Hertz?

**Solution:**

The usual manner to solve this problem is to determine the wavelength, then compare the wavelength to a color chart of the electromagnetic spectrum. Here is an example of a color chart. You can find many more on the Internet.

1) Determine wavelength:

λν = c(x) (7.39 x 10

^{14}s¯^{1}) = 3.00 x 10^{8}m/sx = 4.065 x 10¯

^{7}m

2) EM color charts usually express the wavelength in nm:

4.065 x 10¯^{7}m times (10^{9}nm / 1 m)= 406.5 nm

3) When you examine a color chart, you see that the given wavelength is in the violet region.

An alternate method would be to find a color chart which associates frequency values with the colors of the visible spectrum. This type of chart is not as common as a wavelength-based chart, but they can be found. Here is an example.

Using that chart will require that you convert Hz to THz (terahertz).

7.39 x 10^{14} Hz times ( 1 THz / 10^{12} Hz) = 739 THz

**Problem #19:** Calculate the frequency of radiation with a wavelength of 4.92 cm.

Comment: since the wavelength is already in cm, we can use c = 3.00 x 10^{10} cm s¯^{1} and not have to do any conversions at all.

**Solution:**

(4.92 cm) (x) = 3.00 x 10^{10}cm s¯^{1}x = 6.10 x 10

^{9}s¯^{1}

**Problem #20:** Calculate the frequency of radiation with a wavelength of 8973 Å.

Comment: since 1 Å = 10¯^{8} cm, therefore 8973 Å = 8973 x 10¯^{8} cm. Converting to scientific notation gives 8.973 x 10¯^{5} cm. This is another place where the cm s¯^{1} value for c can be used, since Å converts to cm very easily.

**Solution:**

(8.973 x 10¯^{5}cm) (x) = 3.00 x 10^{10}cm s¯^{1}x = 3.34 x 10

^{14}s¯^{1}

Wavelength-Frequency Problems #1 - 10 | Go to Part Two of Light Equations |

Return to Part One of Light Equations | Return to Electrons in Atoms menu |