A) 0.136 - 0.158 m
B) Microwaves
Explanation:
A)
For an ectromagnetic wave, there is a relationship between its frequency and its wavelength, contained in the wave equation:
[tex]\lambda=\frac{c}{f}[/tex]
where
[tex]\lambda[/tex] is the wavelength of the wave
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light in a vacuum
f is the frequency of the wave
For the waves in this problem, we have:
[tex]f_1=1.9\cdot 10^9 Hz[/tex] is the minimum frequency
[tex]f_2=2.2\cdot 10^9 Hz[/tex] is the maximum frequency
Therefore, the range of corresponding wavelengths is:
[tex]\lambda_1=\frac{3\cdot 10^8}{1.9\cdot 10^9}=0.158 m[/tex]
[tex]\lambda_2=\frac{3\cdot 10^8}{2.2\cdot 10^9}=0.136 m[/tex]
B)
Electromagnetic waves are oscillations of the electric and the magnetic field occurring in a plane perpendicular to the direction of motion the wave.
All electromagnetic waves travel in a vacuum always at the same speed, the speed of light ([tex]c=3.0\cdot 10^8 m/s[/tex]
).
Electromagnetic waves are classified into 7 different types, according to their wavelength and frequency. From the longest to the shortest wavelength, we have:
Radio waves (>1 m)
Microwaves (1 mm - 1 m)
Infrared (750 nm - 1 mm)
Visible light (380 nm - 750 nm)
Ultraviolet (10 nm - 380 nm)
X-rays (0.01 nm - 10 nm)
Gamma rays (<0.01 nm)
Therefore, we see that the waves in this problem (of wavelength between 0.136 and 0.158 m) are classified as microwaves.
