Electromagnetic waves - class-XII

The matter-wave picture of electromagnetic wave/radiation elegantly incorporated  the Heisenberg uncertainty principle.

$\displaystyle c=\nu \lambda $
$c=$speed
$\nu=$Frequency
$\displaystyle \lambda $ = wavelength

The wavelength range of $4000-8000 A^o$ is known as visible spectrum as waves within this wavelength range create a sensation of vision in our eyes.

Energy of light ray      $E = h\nu$                $\implies E \propto \nu$

The portion of spectrum just beyond the red end is called infrared spectrum, while the portion of the spectrum just before the voilet end is called the ultravoilet spectrum.

Electromagnetic radiations are present all around us in different forms such as microwaves, radio waves, gamma rays, and X-rays. These radiations can be defined as a form of energy produced by the movement of electrically charged particles that can be found in matter or vacuum or by oscillating magnetic or electric disturbance.

The refractive index of the ionosphere is given by n = sqrt(1 - (f_c/f)^2). For f > f_c, the refractive index is real and less than 1, but as f increases, the ionosphere becomes less effective at refracting the wave back to Earth, eventually allowing it to pass through.

Cosmic Rays             ${{10}^{-14}}m\,\,\,to\,\,{{10}^{-12}}m$

$\gamma $ Rays                        ${{10}^{-12}}m\,\,\,to\,\,{{10}^{-10}}m$

$x$ Rays                        ${{10}^{-10}}m\,\,\,to\,\,{{10}^{-09}}m$

Ultraviolet Rays         ${{10}^{-07}}m\,\,\,to\,\,\,4\times {{10}^{-07}}m$ 

Hence, Cosmic Rays have smallest wavelength.

$ E=\dfrac{hc}{\lambda } $

$ \lambda =\dfrac{hc}{E}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{14.4\times 1.6\times {{10}^{-19}}}=0.86\times {{10}^{-09}} $

Cosmic Rays                ${{10}^{-14}}m\,\,\,to\,\,{{10}^{-12}}m$

$\gamma$ Rays                        ${{10}^{-12}}m\,\,\,to\,\,{{10}^{-10}}m$

$x$ Rays                        ${{10}^{-10}}m\,\,\,to\,\,{{10}^{-09}}m$

Ultraviolet Rays         ${{10}^{-7}}m\,\,\,to\,\,\,4\times {{10}^{-7}}m$

Hence, it lies in $x$ Rays region.

The ozone layer prevents most harmful UV wavelengths of ultraviolet light (UV light) from passing through the Earth's atmosphere. These wavelengths cause skin cancer, sunburn and cataracts, which were projected to increase dramatically as a result of thinning ozone, as well as harming plants and animals.

The energy of the absorbed photon is E = hc/lambda = 12420/304 approx 40.8 eV. For a hydrogen-like atom, E = 13.6 * Z^2 * (1/n1^2 - 1/n2^2). Given the emission of 6 lines, the final state must be n=4 (since 4*3/2 = 6). The transition is from n_initial to n_final = 4. Solving for Z and n_initial requires specific energy level data.

For impure spectrum:

1. The order of colours is are straight, hence option a is false
2. Order of colours are regular
3. Colours are overlapped each other in impure spectrum
4. Colours are present in impure spectrum

Intensity or power per unit area of the radiations,
$P = pv$
$\Rightarrow p=\dfrac{p}{v} = \dfrac{0.5}{3\times 10^8}=0.166\times 10^{-8}N/m^2$

Radiation pressure is given by   $P _R = \dfrac{(1+\alpha) I}{c}$
where $\alpha$ is the coefficient of reflection of the surface.
For completely reflecting surface  $\alpha =1$
For completely absorbing surface  $\alpha = 0$
So,  radiation pressure depends on the nature of surface on which the light is falling but independent of wavelength of light falling.