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?

Explanation:

, 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 108 m/s

x = 6.00 x 1015 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/s

The 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

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 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 #17: 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 #18: What color is light whose frequency is 7.39 x 1014 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 1014 s¯1) = 3.00 x 108 m/s

x = 4.065 x 10¯7 m

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

4.065 x 10¯7 m times (109 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 1014 Hz times ( 1 THz / 1012 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 1010 cm s¯1 and not have to do any conversions at all.

Solution:

(4.92 cm) (x) = 3.00 x 1010 cm s¯1

x = 6.10 x 109 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 1010 cm s¯1

x = 3.34 x 1014 s¯1

look at the !

pleaze mark brainliest!

- aluminum and copper chloride.

- the chloride ion to separate the aluminum from the oxygen so that the aluminum can react with the copper ions (and the water molecules).