Physics, 09.07.2020 01:01 gabmilex1

There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 nm to 400 nm. It is necessary for the production of vitamin D. UVB, with a wavelength in vacuum between 280 nm and 320 nm, is more dangerous because it is much more likely to cause skin cancer. (a) find the frequency ranges of UVA and UVB.

(b) what are the ranges of the wave numbers for UVA and UVB?

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Answer from: clietmaster112

(a) UVA - frequency range: 7.5*10^14 Hz to 9.37*10^14 Hz

UVB - frequency range: 9.37*10^14Hz to 1.07*10^15Hz

(b) UVA - wavenumber range: 1.57*10^7 m^-1 to 1.96*10^7m^-1

UVB - wavenumber range: 1.97*10^7 m^-1 to 2.24*10^7m^-1

Explanation:

In order to calculate the frequency ranges fro the two categories of ultraviolet light, you use the following formula:

$f=\frac{c}{\lambda}$ (1)

f: frequency

c: speed of light in vacuum = 3*10^8 m/s

λ: wavelength

(a) For UVA the wavelength range is λ1 = 320nm to λ2 = 400nm.

The frequency range is:

$f_1=\frac{c}{\lambda_1}=\frac{3*10^8m/s}{320*10^{-9}m}=9.37*10^{14}Hz\\\\f_2=\frac{c}{\lambda_2}=\frac{3*10^8m/s}{400*10^{-9}m}=7.5*10^{14}Hz$

frequency range: 7.5*10^14 Hz to 9.37*10^14 Hz

For UVB the wavelength range is λ1 = 280nm to λ2 = 320nm.

The frequency range is:

$f_1=\frac{3*10^8m/s}{280*10^{-9}m}=1.07*10^{15}Hz\\\\f_2=\frac{3*10^8m/s}{320*10^{-9}m}=9.37*10^{14}Hz\\\\$

frequency range: 9.37*10^14Hz to 1.07*10^15Hz

(b) The ranges for the wavenumbers are:

UVA:

$k_1=\frac{2\pi}{\lambda_1}=\frac{2\pi}{320*10^{-9}m}=1.96*10^7m^{-1}\\\\k_2=\frac{2\pi}{\lambda_2}=\frac{2\pi}{400*10^{-9}m}=1.57*10^7m^{-1}$

wavenumber range: 1.57*10^7 m^-1 to 1.96*10^7m^-1

UVB:

$k_1=\frac{2\pi}{280*10^{-9}m}=2.24*10^7m^{-1}\\\\k_2=\frac{2\pi}{320*10^{-9}m}=1.96*10^7m^{-1}$

wavenumber range: 1.97*10^7 m^-1 to 2.24*10^7m^-1

Answer from: Quest

Your answer is c. explanation: plasma is the fourth state of matter, derived from the gaseous state of matter. plasma can contain free electrons and/or ions. while plasma is never neutral, it can either be positively or negatively charged. gases, however, are electrically neutral.

Answer from: Quest

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.

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