(a) They are transverse in nature, and travel with the speed of light in vacuum.
(b) The selfless attitude of the scientist, service mindedness and modesty.

Energy of photon, \(\mathrm{E}=\mathrm{hv}\)
This implies,\(\mathrm{E}=\mathrm{h} \frac{\mathrm{c}}{\lambda}\)
Where, \(\mathrm{h}=6.62 \times 10^{-34} \mathrm{js}\)
\( \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}\)
If wave length \(\lambda\) is in meter and energy is in joule then, we will divide E by \(1.6 \times 10^{-19}\) to convert into eV (Electron volt).
\(\therefore \mathrm{E}=\frac{\mathrm{hc}}{\lambda \times 1.6 \times 10^{-19}} \mathrm{eV}\)
(1) For y - rays wave length ranges from  to less that \(10^{-14} \mathrm{~m}\)
Therefore,  \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \times 10^3 \mathrm{eV} \approx 10^4 \mathrm{eV}\)
Thus, \( \lambda=10^{-10} \mathrm{~m}, \text { energy }=10^4 \mathrm{eV} \text { and }\) \( \lambda=10^{-14} \mathrm{~m} \text {, energy }=10^8 \mathrm{eV}\)
Energy of y - rays ranges between 10⁴ to \(10^8 \mathrm{eV}\)
(2) For X - rays wave length ranges from \(10^{-8} \mathrm{~m}\) to \(10^{-7} \mathrm{~m}\) For \(\lambda=10^{-8}\)
Therefore, \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-8} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \approx 10^2 \mathrm{eV}\)
\( \lambda=10^{-13} \mathrm{~m} \text {, energy }=10^7 \mathrm{eV}\)
(3) For violet radiation \(\lambda\) ranges from \(4 \times 10^{-7}\) to \(6 \times 10^{-10}\)
Therefore, for \(\lambda=4 \times 10^{-7}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV} =3.1 \mathrm{eV} \approx 10^{10} \mathrm{eV}\)
\(\lambda=6 \times 10^{-10} \mathrm{~m} \text {, Energy }=10^3 \mathrm{eV}\)
Energy of ultraviolet radiation vary between \(10^{10}\) to \(10^3 \mathrm{eV}\).
(4) For visible radiations wave length range from \(4 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-7} \mathrm{~m}\)
Therefore,
For \(\lambda=4 \times 10^{-7} \mathrm{~m}\), and Energy \(=10^{10} \mathrm{eV}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{7 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{VV} \)
\(=1.77 \mathrm{eV} \approx 10^{\circ} \mathrm{eV}\)
(5) For infrared radiation $\lambda$ range from \(7 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-14} \mathrm{~m}\)
Therefore, \(\lambda=7 \times 10^{-7} \text {, energy }=10^{\circ} \mathrm{eV}\)
For \(\lambda=7 \times 10^{-4} \text {, energy }=\frac{1}{1000} \text { times }\)
the other order of  \(10^{-3}\)eV
(6) For micro waves $\lambda$ ranges from 1 mm to 0.3 m
For \(\lambda=1 \mathrm{~mm}\) or \(10^{-3}\)
energy is equal to \( \text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-3} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-3} \mathrm{eV} \approx 10^{-3} \mathrm{eV}\)
For \(\lambda=0.3 \mathrm{~m} \text {, Energy }=4.1 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV} \text {. }\)
(7) For Radio waves $\lambda$ ranges from 1 m to few km For $\lambda=1 \mathrm{~m}$
For λ=1m
Energy is equal to
\(=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^0 \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV}\)
Energy for λ of the order of few km≈10⁻⁶eV
The Energy of a photon that a source produces indicates the spacing of relevant energy levels of the source

Given, amplitude of an electromagnetic wave,
E₀ = 120 N/C
Frequency of wave, v = 50 MHz = 50 \(\times\) 10⁶ Hz
(i) Speed of light in vacuum, \(c=\frac{E_0}{B_0}\)
\(\begin{aligned} B_0=\frac{E_0}{c}= & \frac{120}{3 \times 10^8}=40 \times 10^{-8} \end{aligned}\)
= 400 \(\times\) 10^-9 T = 400 nT
Angular frequency of electromagnetic wave,
\(\omega=2 \pi \nu=2 \times 3.14 \times 50 \times 10^6\)
\(\omega=3.14 \times 10^8 \mathrm{rad} / \mathrm{s}\)
Wave number of electromagnetic wave,
\(k=\frac{\omega}{c}=\frac{3.14 \times 10^8}{3 \times 10^8}=1.05 \mathrm{rad} / \mathrm{m}\)
Wavelength of electromagnetic wave,
\(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{50 \times 10^6}=6.00 \mathrm{~m}\)
(ii) Expression of electric field, E = E₀ sin (kx - \(\omega\)t)
E = 120 sin (1.05x - 3.14 \(\times\)10⁸ t)
Expression of magnetic field B,
B = B₀ sin (kx - \(\omega\)t)
E = 120 sin (kx - \(\omega\)t)
B = 4 \(\times\)10^-7 sin (1.05x - 3.14 \(\times\) 10⁸ t)

Acharge oscillating with some frequency, produces an oscillating electric field in space, which in turn produces an oscillating magnetic field perpendicular to the electric field, this process goes on repeating, producing em waves in space perpendicular to both the fields.
    
Directions of \(\overrightarrow { E } \) and \(\overrightarrow { B } \) are perpendicular to each other and also perpendicular to direction of propagation of em waves.
Ampere's circuital law is given by as \(\phi \overrightarrow { B } .\overrightarrow { dl } ={ \mu }_{ 0 }{ i }_{ c }\)
But for a circuit containing capacitor, during its charging/discharging the current within the plates of the capacitor varies, (producing displacement current i _d)
Therefore, the above equation, as generalized by Maxwell, is given as \(\phi \overrightarrow { B } .\overrightarrow { dl } ={ \mu }_{ 0 }{ i }_{ c }+{ \mu }_{ 0 }{ i }_{ d }\) 
During the process of charging capacitor, electric flux (\(\phi _{ c }\)) between the plates of capacitor changes with time, which produces the current within the plates of capacitor. This current, being proportional to \(\frac { d\phi \epsilon }{ dt } \), we have \(i={ \epsilon }_{ 0 }\frac { d\phi \epsilon }{ dt } \) 

(i) Microwave
(ii) Infrared waves are used to treat muscular strain.
(iii) X-rays are used as a diagnostic tool in medicine. X-rays are produced by X-ray tubes or inner shell electrons.

Given, amplitude of the magnetic field part of harmonic electromagnetic wave,
B₀ = 510 nT = 510 \(\times\)10^-9 T
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB₀
= 3 × 10⁸ × 510 × 10⁻⁹ = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.

f₁=7.5×10⁶ Hz
f₂=12×10⁶ Hz
λ₁=c/f₁=40 m
λ₂=c/f₂=25 m
So the range is 40m to 25m

Using Eq, the magnitude of B is
\(B=\frac { E }{ c } \) 
\(=\frac { 6.3V/m }{ 3\times { 10 }^{ 8 }m/s } =2.1\times { 10 }^{ -8 }T\)
To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction.
Since, \((+\overrightarrow{\mathbf{j}}) \times(+\hat{\mathbf{k}})=\overrightarrow{\mathbf{i}}, \mathbf{B}\) is along the z-direction.
Thus, \(\mathbf{B}=2.1 \times 10^{-8} \hat{\mathbf{k}} \mathrm{T}\)

It is given that a plane electromagnetic wave travels in vacuum along z-direction and the frequency of the electromagnetic wave is 30MHz.
We can say that electric field and magnetic field will be in x-plane because the electromagnetic wave travels along the z-direction and both fields are mutually perpendicular to each other.
The formula of the wavelength of a wave is,
λ=c/ν
Substitute the values in the above expression,
λ=(3×10⁸)/(30×10⁶)=10m.
Thus, the value of wavelength is 10 m and the direction of electric and magnetic fields will be in x-y plane.

Given \(v=40\times { 10 }^{ 6 }Hz\) 
\(\therefore \)\(T=\frac { 1 }{ v } =\frac { 1 }{ 40\times 10^{ -6 } } =0.25\times 10^{ -6 }/s\) 
Magnetic field
\({ B }_{ 0 }=\frac { E_{ 0 } }{ c } =\frac { 750 }{ 3\times 10^{ 18 } } =2.5\times 10^{ -6 }T \ along \ (z \ direction)\) 
And angular frequency 
\(\omega =2\pi v=2\times \pi \times 40\times { 10 }^{ 6 }=8\pi \times { 10 }^{ 7 } \ Hz\)​​​​​

It is given that a charged particle oscillates about its mean equilibrium position and the frequency of charged particle is 10⁹  Hz.
The frequency of an electromagnetic wave produced by the oscillator will be same to the value of frequency of a charged particle that is oscillating about its mean position.
Thus, the value of the frequency of an electromagnetic wave produced by oscillator is 10⁹ Hz.

(a) Comparing the given equation with
\(B_{y}=B_{0} \sin \left[2 \pi\left(\frac{x}{\lambda}+\frac{t}{T}\right)\right]\)
We get, \(\lambda=\frac{2 \pi}{0.5 \times 10^{3}} \mathrm{~m}=1.26 \mathrm{~cm}\)
and \(\frac{1}{T}=v=\left(1.5 \times 10^{11}\right) / 2 \pi=23.9 \mathrm{GHz}\)
(b) E₀ = B₀c = 2 x 10^–7 T x 3 x 10₈ m/s = 6 x 10¹ V/m
The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along the z-axis is obtained as
E_z = 60 sin (0.5 x 10³x + 1.5 x 10¹¹ t) V/m.

X-rays because \(\lambda=0.825 \stackrel{\circ}{\AA}\)

The speed of light (3 × 10⁸ m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

A charge moving with a circular orbit can produce electromagnetic waves because a circular motion is an accelerated motion and accelerated charges produce e.m.waves

(i) (a)
(ii) (a): Greenhouse effect is due to infrared rays.
(iii) (a): Ozone layer absorbs the harmful ultraviolet radiations coming from the sun.
(iv) (a): Ozone layer lies in stratosphere.
(v) (b): The atmosphere of earth is richest in infrared radiation.

(i) (b): Infrared rays can be converted into electric energy as in solar cell.
(ii) (d): Radiowaves have longest wavelength.
(iii) (c) : Cathode rays are invisible fast moving streams of electrons emitted by the cathode of a discharge tube which is maintained at a pressure of about 0.01 mm of mercury.
(iv) (c): \(\Upsilon\)-rays have minimum wavelength
(v) (a): \(\lambda_{\text {micro }}>\lambda_{\text {infra }}>\lambda_{\text {ultra }}>\lambda_{\text {gamma }}\)