Respuesta :
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
The frequency of the radiation is 6.0 × 10¹⁵ s⁻¹
The radiation is in the Ultra-violet region of the electromagnetic spectrum
From the question,
We are to determine the frequency of radiation with a wavelength of 5.00 × 10⁻⁸ m
From the formula
[tex]c = \nu \lambda[/tex]
We can write that
[tex]\nu = \frac{c}{\lambda}[/tex]
Where [tex]\nu[/tex] is the frequency
[tex]c[/tex] is the speed of light ([tex]c = 3.0 \times 10^{8} \ m/s[/tex])
and [tex]\lambda[/tex] is wavelength
From the given information
[tex]\lambda = 5.00 \times 10^{-8} \ m[/tex]
∴ [tex]\nu = \frac{3.0 \times 10^{8} }{5.00 \times 10^{-8} }[/tex]
[tex]\nu = 6.0 \times 10^{15} \ s^{-1}[/tex]
Hence, the frequency of the radiation is 6.0 × 10¹⁵ s⁻¹
The radiation is in the Ultra-violet region of the electromagnetic spectrum
Learn more here: https://brainly.com/question/24290571
