BIO Ultraviolet Radiation. There are two categories of ultraviolet light. Ultraviolet A(UVA) has a wavelength ranging from to . 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?

The number of waves that moves a fixed point in unit time is known as frequency.

$f=\frac{1}{T}$

Where,T =Time period in

The frequency, wavelength, and speed of light of any wave are related by the wavelength-frequency relationship.

The wavelength-frequency relationship is:

$c=\lambda f$

Where, c is the speed of light which is equal to $3\times {10}^{8}m/s$,$\lambda$ is the wavelength and f is the frequency.

The wavelength is inversely proportional to the frequency.

$\lambda =\frac{c}{f}$or$\frac{v}{f}$

Where, v is velocity of wave in m/s.

The relation between wave number and wavelength is:

$k=\frac{2\mathrm{\pi }}{\lambda }$

Where, k is wavenumber in rad/m.

Given that,

Range for $UVA:\lambda =320nmto400nm$ to

Range for $UVB:\lambda =280nmto320nm$

First determine for UVA:

Take$\lambda =320$

The wavelength-frequency relationship is:

$\begin{array}{r}f=\frac{c}{\lambda }\\ =\frac{3\times {10}^8}{320\times {10}^{-9}}\\ =9.375\times {10}^{14}\mathrm{Hz}\end{array}$

Take$\lambda =400nm$

$\begin{array}{r}f=\frac{c}{\lambda }\\ =\frac{3\times {10}^8}{400\times {10}^{-9}}\\ =7.5\times {10}^{14}\mathrm{Hz}\end{array}$

Now, determine for UVB:

Take$\lambda =280nm$

The wavelength-frequency relationship is:

$\begin{array}{r}f=\frac{c}{\lambda }\\ =\frac{3\times {10}^8}{280\times {10}^{-9}}\\ =1.071\times {10}^{15}\mathrm{Hz}\end{array}$

Take$\lambda =320nm$

$\begin{array}{r}f=\frac{c}{\lambda }\\ =\frac{3\times {10}^8}{320\times {10}^{-9}}\\ =9.375\times {10}^{14}\mathrm{Hz}\end{array}$

Hence, the frequency ranges of UVA and UBA are$7.5\times {10}^{14}Hz$ to$9.375\times {10}^{14}Hz$ and$9.375\times {10}^{14}Hz$ to$1.071\times {10}^{15}Hz$ respectively.

Given that,

Range for $UVA:\lambda =320nmto400nm$

Range for role="math" localid="1668324761924" $UVB:\lambda =280nmto320nm$

First determine for UVA:

Take$\lambda =320nm$

The relation between wave number and wavelength is:

$\begin{array}{r}k=\frac{2\pi }{\lambda }\\ =\frac{2\pi }{320\times {10}^{-9}}\\ =19.63\times {10}^6{\mathrm{m}}^{-1}\end{array}$

Take$\lambda =400nm$

The relation between wave number and wavelength is:

$\begin{array}{r}k=\frac{2\pi }{\lambda }\\ =\frac{2\pi }{400\times {10}^{-9}}\\ =15.71\times {10}^6{\mathrm{m}}^{-1}\end{array}$

Now, determine for :

Take$\lambda =280nm$

The relation between wave number and wavelength is:

$\begin{array}{r}k=\frac{2\pi }{\lambda }\\ =\frac{2\pi }{280\times {10}^{-9}}\\ =22.44\times {10}^6{\mathrm{m}}^{-1}\end{array}$

Take$\lambda =320nm$

The relation between wave number and wavelength is:

$\begin{array}{r}k=\frac{2\pi }{\lambda }\\ =\frac{2\pi }{320\times {10}^{-9}}\\ =19.63\times {10}^6{\mathrm{m}}^{-1}\end{array}$

The frequency of wave is:

$f=\frac{1}{T}\Rightarrow T=\frac{1}{f}$

Substitute the values of variables

$\begin{array}{r}T=\frac{1}{f}\\ =\frac{1}{3\times {10}^{19}}\\ =3.3\times {10}^{-19}\sec \end{array}$

Hence the range of wave number for UVA and UBA are$15.71\times {10}^6{m}^{-1}$ to$19.63\times {10}^6{m}^{-1}$ and$19.63\times {10}^6{m}^{-1}$ to$22.44\times {10}^6{m}^{-1}$ respectively.