The speed of electromagnetic waves (which include visible light, radio, and x rays) in vacuum is 3.0 x 108 m/s. wavelengths of visible light waves range from about 400 nm in the violet to about 700 nm in the red. what are the (a) minimum and (b) maximum frequencies of these waves? the range of frequencies for shortwave radio (for example, fm radio and vhf television) is 1.5 to 300 mhz. what are the (c) minimum and (d) maximum wavelengths of these waves? x ray wavelengths range from about 5.0 nm to about 1.0 x 10-2 nm. what are the (e) minimum and (f) maximum frequencies of x rays?

(a)

The frequency of an electromagnetic wave is given by

where

c is the speed of light

is the wavelength

We notice from the formula that the frequency is inversely proportional to the wavelength, so the minimum frequency corresponds to the maximum wavelength, and viceversa

The maximum value of the wavelength of the visible light waves is

(red light)

so the minimum frequency of visible light is

(b)

The maximum frequency corresponds to the minimum wavelength;

The minimum wavelength is

(violet light)

so the maximum frequency of visible light is

(c) 1 m

The wavelength of an electromagnetic wave is given by

as before, we notice that the minimum wavelength corresponds to the maximum frequency, and viceversa.

The maximum frequency of shortwave radio waves is

so the minimum wavelength of these waves is

(d) 200 m

The minimum frequency of shortwave radio waves is

so the maximum wavelength of these waves is

(e)

As in part (a) and (b), we can find the frequency of the X-rays by using the formula

The maximum wavelength of the x-rays is

so the minimum frequency is

(f)

The maximum frequency corresponds to the minimum wavelength;

The minimum wavelength is

so the maximum frequency of the x-rays is

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

k2s for potassium sulphide