Nature of radiation Questions in English

(A) The photoelectric effect is the emission of photoelectrons from a clean metal surface due to incident light whose frequency is greater than a threshold frequency.
The photoelectric effect supports the particle theory of light because it shows that the energy required to release electrons from a metal is dependent upon the frequency of the light,and not the intensity.
Therefore,light behaves as a stream of particles (photons) in this phenomenon,confirming the particle-like behavior of light.

(D) The photoelectric effect is the emission of electrons from a metal surface when light of a suitable frequency strikes it.
The energy of the incident photon is given by $E = h\nu$.
According to Einstein's photoelectric equation: $h\nu = \Phi + KE_{max}$,where $\Phi$ is the work function and $KE_{max}$ is the maximum kinetic energy of the emitted electrons.
$KE_{max} = h\nu - \Phi$.
Since $KE_{max} = \frac{1}{2}mv_{max}^2$,the velocity of emitted electrons ranges from $0$ to $v_{max}$.
This $v_{max}$ depends only on the frequency of the incident light $(\nu)$ and the nature of the metal $(\Phi)$,not on the intensity of the light.
Therefore,electrons are emitted with different velocities,none of which exceed the value corresponding to $KE_{max}$.

(C) In the photoelectric effect,the number of photoelectrons emitted per unit time is directly proportional to the intensity of the incident radiation,provided the frequency is above the threshold frequency.
Since the photocurrent is directly proportional to the number of emitted photoelectrons,it depends on the intensity of the incident light.
The frequency of the incident photon determines the maximum kinetic energy of the emitted photoelectrons,but it does not affect the magnitude of the photocurrent.
Therefore,the photocurrent is independent of the frequency of the incident photon.

(D) The photoelectric effect occurs when radiation with sufficient energy (frequency greater than the threshold frequency) strikes a metal surface.
$E = h\nu$
Since $X-rays$,$Gamma$ rays,$U.V.$ light,and even visible light (for certain metals like alkali metals) have frequencies that can exceed the threshold frequency of various materials,all of these can cause the photoelectric effect.
Therefore,the correct option is $D$.

(A) Phenomena such as interference,diffraction,and polarization are characteristic of wave motion,which light exhibits. However,the photoelectric effect cannot be explained by wave theory. It can only be explained by considering light as a stream of discrete energy packets called photons. Therefore,the photoelectric effect provides evidence for the particle nature of light.

(B) $1$. According to the de Broglie relation,$\lambda = \frac{h}{mv}$. Since the mass of an electron is much smaller than that of a proton,for the same speed,the electron will have a longer wavelength. This statement is true.
$2$. Violet light has a shorter wavelength (approx. $380-450 \ nm$) compared to red light (approx. $620-750 \ nm$). Therefore,the statement that violet radiations have a longer wavelength than red radiations is false.
$3$. Energy $E = \frac{hc}{\lambda}$. Since $E \propto \frac{1}{\lambda}$,light with a longer wavelength $(600 \ nm)$ has lower energy than light with a shorter wavelength $(500 \ nm)$. This statement is true.
$4$. Atoms emit radiation at specific discrete frequencies,resulting in a line spectrum. This statement is true.

(B) The splitting of spectral lines in the presence of an external electric field is known as the $Stark \ effect$.
In contrast,the splitting of spectral lines in the presence of an external magnetic field is known as the $Zeeman \ effect$.
Therefore,the correct option is $B$.

(B) The threshold wavelength $\lambda_0$ is $8000 \ \mathring{A} = 8000 \times 10^{-10} \ m$. The work function $\Phi = \frac{hc}{\lambda_0} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{8000 \times 10^{-10}} = 2.475 \times 10^{-19} \ J$.
When exposed to $\lambda = 5000 \ \mathring{A} = 5000 \times 10^{-10} \ m$,the energy of incident radiation is $E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}} = 3.96 \times 10^{-19} \ J$.
The kinetic energy $K.E. = E - \Phi = 3.96 \times 10^{-19} - 2.475 \times 10^{-19} = 1.485 \times 10^{-19} \ J$.
Using $K.E. = \frac{1}{2} m_e v^2$,we get $v = \sqrt{\frac{2 \times K.E.}{m_e}} = \sqrt{\frac{2 \times 1.485 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{0.326 \times 10^{12}} \approx 5.71 \times 10^5 \ ms^{-1}$.
Rounding to the nearest option,the answer is $5.74 \times 10^5 \ ms^{-1}$.

(C) Electromagnetic radiations consist of oscillating electric and magnetic fields and travel as photons.
Cathode rays are a stream of fast-moving electrons,which possess mass and charge.
Therefore,cathode rays are particle radiations,not electromagnetic radiations.

(B) The energy of a radiation is given by the formula $E = \frac{hc}{\lambda}$.
For the first radiation: $E_1 = \frac{hc}{\lambda_1} = 15 \ eV$ $(I)$
For the second radiation: $E_2 = \frac{hc}{\lambda_2} = 45 \ eV$ $(II)$
Dividing equation $(I)$ by equation $(II)$:
$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} = \frac{15}{45} = \frac{1}{3}$
Therefore,$\frac{\lambda_2}{\lambda_1} = \frac{1}{3}$,which implies $\lambda_1 = 3\lambda_2$.

(A) According to the law of conservation of energy,the energy of the absorbed photon is equal to the sum of the energies of the two re-emitted photons.
$E = E_{1} + E_{2}$
Since $E = \frac{hc}{\lambda}$,we have:
$\frac{hc}{\lambda} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}$
Dividing by $hc$:
$\frac{1}{\lambda} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}$
Given $\lambda = 300 \, nm$ and $\lambda_{1} = 496 \, nm$:
$\frac{1}{300} = \frac{1}{496} + \frac{1}{\lambda_{2}}$
$\frac{1}{\lambda_{2}} = \frac{1}{300} - \frac{1}{496} = \frac{496 - 300}{300 \times 496} = \frac{196}{148800}$
$\lambda_{2} = \frac{148800}{196} \approx 759.18 \, nm$
Rounding to the nearest integer,we get $759 \, nm$.

(B) The electromagnetic spectrum is arranged in order of increasing frequency and decreasing wavelength as follows: $Radio \ waves > Microwaves > Infrared > Visible > Ultraviolet > X-rays > Gamma \ rays$.
Since radio waves have the lowest frequency,they possess the maximum wavelength among the given options.

(A) The energy of the photon is given as $E = 6 \, MeV = 6 \times 10^6 \, eV$.
Converting energy to Joules: $E = 6 \times 10^6 \times 1.6 \times 10^{-19} \, J = 9.6 \times 10^{-13} \, J$.
For a photon,the relationship between energy $(E)$ and momentum $(p)$ is $E = pc$,where $c$ is the speed of light $(3 \times 10^8 \, m/s)$.
Therefore,$p = \frac{E}{c} = \frac{9.6 \times 10^{-13}}{3 \times 10^8} \, kg \cdot m/s$.
$p = 3.2 \times 10^{-21} \, kg \cdot m/s$.

(C) The kinetic energy $(K.E.)$ of the photoelectron is given by $K.E. = \frac{p^2}{2m}$.
According to the photoelectric equation,$\frac{hc}{\lambda} = W_o + K.E.$
Given that $K.E. \gg W_o$,we can approximate $\frac{hc}{\lambda} \approx K.E. = \frac{p^2}{2m}$.
For the first case,$\frac{hc}{\lambda} = \frac{p^2}{2m}$.
For the second case,let the new wavelength be $\lambda'$ and momentum be $p' = 1.5 \ p = \frac{3}{2} \ p$.
Then,$\frac{hc}{\lambda'} = \frac{(1.5 \ p)^2}{2m} = \frac{2.25 \ p^2}{2m}$.
Dividing the two equations: $\frac{\lambda}{\lambda'} = \frac{2.25 \ p^2 / 2m}{p^2 / 2m} = 2.25 = \frac{9}{4}$.
Therefore,$\lambda' = \frac{4}{9} \lambda$.

(B) Frequency $(v) = \frac{750 \text{ waves}}{60 \text{ seconds}} = 12.5 \, Hz$.
The speed of light $(c)$ is $3 \times 10^8 \, m/s$.
Using the relation $\lambda = \frac{c}{v}$,we get:
$\lambda = \frac{3 \times 10^8}{12.5} = 2.4 \times 10^7 \, m$.

(A) According to the theory of relativity,the mass of a particle moving with velocity $v$ is given by $m = \frac{m_{rest}}{\sqrt{1 - (\frac{v}{c})^2}}$,where $m$ is the relativistic mass and $m_{rest}$ is the rest mass.
Rearranging this formula to solve for the rest mass gives $m_{rest} = m \sqrt{1 - (\frac{v}{c})^2}$.

(C) $X$-rays are electromagnetic radiations with high energy and no charge.
Because they are electrically neutral,they are $NOT$ deflected by electric or magnetic fields.
Therefore,the statement that they are deflected by electric and magnetic fields is incorrect.

(A) From the given sketch,the distance shown represents one-fourth of the wavelength $(\lambda)$.
Therefore,the total wavelength is $\lambda = 4 \times 175 \, nm = 700 \, nm$.
The wavelength of $700 \, nm$ corresponds to the red color in the visible spectrum.

(B) Given: Wavelength $\lambda = 436 \, nm = 436 \times 10^{-9} \, m = 4.36 \times 10^{-7} \, m$.
Speed of light $c = 3 \times 10^8 \, m/s$.
The formula for frequency is $\nu = \frac{c}{\lambda}$.
Substituting the values: $\nu = \frac{3 \times 10^8 \, m/s}{4.36 \times 10^{-7} \, m} = 0.688 \times 10^{15} \, Hz = 6.88 \times 10^{14} \, Hz$.

(B) According to Einstein's photoelectric equation,$h\nu = h\nu_0 + K.E._{max}$,where $K.E._{max} = \frac{1}{2}mv^2$.
Since $h\nu_0$ (work function) is constant for a given metal,the kinetic energy $(K.E._{max})$ depends directly on the frequency $(\nu)$ of the incident light.
As $\nu = \frac{c}{\lambda}$,it also depends on the wavelength $(\lambda)$.

(C) The relationship between frequency $\nu$,speed of light $c$,and wavelength $\lambda$ is given by $\nu = \frac{c}{\lambda}$.
Given $\lambda = 5800 \ \mathring{A} = 5800 \times 10^{-10} \ m = 5.8 \times 10^{-7} \ m$.
The speed of light $c = 3 \times 10^8 \ m/s$.
Substituting the values: $\nu = \frac{3 \times 10^8 \ m/s}{5.8 \times 10^{-7} \ m} \approx 5.17 \times 10^{14} \ s^{-1}$.

(C) The energy of one Einstein is defined as the energy of one mole of photons.
$E = \frac{N_A \times h \times c}{\lambda}$
Given: $\lambda = 9000 \ \mathring{A} = 9000 \times 10^{-10} \ m = 9 \times 10^{-7} \ m$,$N_A = 6.022 \times 10^{23} \ mol^{-1}$,$h = 6.626 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$.
$E = \frac{6.022 \times 10^{23} \times 6.626 \times 10^{-34} \times 3 \times 10^8}{9 \times 10^{-7}}$
$E = \frac{11.97 \times 10^{-3}}{9 \times 10^{-7}} \approx 1.33 \times 10^5 \ J \ mol^{-1}$.

(A) The energy of radiation is given by $E = h\nu = \frac{hc}{\lambda}$.
For option $A$: $\lambda = 300 \, pm = 3 \times 10^{-10} \, m$. Thus,$E_A = \frac{hc}{3 \times 10^{-10}} = h \times (3 \times 10^8) / (3 \times 10^{-10}) = h \times 10^{18} \, J$.
For option $B$: $\lambda = 30 \, nm = 3 \times 10^{-8} \, m$. Thus,$E_B = \frac{hc}{3 \times 10^{-8}} = h \times (3 \times 10^8) / (3 \times 10^{-8}) = h \times 10^{16} \, J$.
For option $C$: $\nu = 3 \times 10^{12} \, s^{-1}$. Thus,$E_C = h \times 3 \times 10^{12} \, J$.
For option $D$: $\nu = 3 \times 10^{10} \, s^{-1}$. Thus,$E_D = h \times 3 \times 10^{10} \, J$.
Comparing the values,$E_A = h \times 10^{18}$ is the highest energy.

(B) line spectrum is a spectrum that contains radiation of only specific wavelengths. It is produced by excited atoms in the gaseous state. When atoms are excited,they emit light of specific frequencies,which appear as distinct lines in the spectrum. Therefore,a line spectrum is a characteristic of $Atoms$.

(C) The hydrogen spectrum consists of various series of spectral lines.
The Lyman series lies in the ultraviolet region.
The Balmer series lies in the visible region.
The Paschen,Brackett,and Pfund series lie in the infrared region.
Therefore,the correct answer is the Balmer series.

(B) The splitting of spectral lines in the presence of an external magnetic field is called the $Zeeman \ effect$.
In contrast,the splitting of spectral lines in the presence of an external electric field is known as the $Stark \ effect$.

(B) $1$. Frequency $(\nu)$ is the number of waves passing a point per second.
Given: $750$ waves in $1$ minute ($60$ seconds).
$\nu = \frac{750}{60} = 12.5 \ Hz$.
$2$. The relationship between speed of light $(c)$,frequency $(\nu)$,and wavelength $(\lambda)$ is $c = \nu \times \lambda$.
$3$. Using $c = 3 \times 10^8 \ m/s$:
$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{12.5} = 0.24 \times 10^8 \ m = 2.4 \times 10^7 \ m$.
Therefore,the correct option is $B$.

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Since $h$ and $c$ are constants,the energy $E$ is inversely proportional to the wavelength $\lambda$,i.e.,$E \propto \frac{1}{\lambda}$.
Therefore,the ratio of energies is $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Given $\lambda_1 = 3000 \ \mathring{A}$ and $\lambda_2 = 6000 \ \mathring{A}$.
Substituting the values,we get $\frac{E_1}{E_2} = \frac{6000}{3000} = \frac{2}{1}$.
Thus,the ratio is $2 : 1$.

(D) The energy of a photon is given by the formula $E = h\nu = \frac{hc}{\lambda}$.
From this relation,it is clear that the energy of a photon is inversely proportional to its wavelength $(\lambda)$.
Therefore,photons of different wavelengths possess different amounts of energy.
Since both the Assertion and the Reason are false,the correct option is $D$.

(D) The Assertion is incorrect because an absorption spectrum consists of dark lines separated by bright spaces,which occur when specific wavelengths are absorbed by a substance.
The Reason is also incorrect because an emission spectrum consists of bright lines on a dark background,representing the light emitted by excited atoms or molecules.
Therefore,both the Assertion and the Reason are incorrect.

(A) The wavelength,$\lambda$,is calculated using the formula $\lambda = c / \nu$,where $c$ is the speed of light $(3.00 \times 10^{8} \, m \, s^{-1})$ and $\nu$ is the frequency.
Given frequency $\nu = 1368 \, kHz = 1368 \times 10^{3} \, s^{-1}$.
Substituting the values:
$\lambda = \frac{3.00 \times 10^{8} \, m \, s^{-1}}{1368 \times 10^{3} \, s^{-1}}$
$\lambda = 219.3 \, m$
Since the wavelength is in the range of meters,this radiation belongs to the radio wave region of the electromagnetic spectrum.

Using the equation $v = \frac{c}{\lambda}$,where $c = 3.00 \times 10^{8} \, m \, s^{-1}$ is the speed of light.
For violet light $(\lambda = 400 \, nm = 400 \times 10^{-9} \, m)$:
$v = \frac{3.00 \times 10^{8} \, m \, s^{-1}}{400 \times 10^{-9} \, m} = 7.50 \times 10^{14} \, Hz$
For red light $(\lambda = 750 \, nm = 750 \times 10^{-9} \, m)$:
$v = \frac{3.00 \times 10^{8} \, m \, s^{-1}}{750 \times 10^{-9} \, m} = 4.00 \times 10^{14} \, Hz$
Thus,the range of the visible spectrum in terms of frequency is from $4.00 \times 10^{14} \, Hz$ to $7.50 \times 10^{14} \, Hz$.

$(a)$ Calculation of wavenumber $(\bar{\nu})$:
$\lambda = 5800 \, \mathring{A} = 5800 \times 10^{-10} \, m = 5.8 \times 10^{-7} \, m$.
$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{5.8 \times 10^{-7} \, m} \approx 1.724 \times 10^{6} \, m^{-1} = 1.724 \times 10^{4} \, cm^{-1}$.
$(b)$ Calculation of frequency $(\nu)$:
$\nu = \frac{c}{\lambda} = \frac{3 \times 10^{8} \, m \, s^{-1}}{5.8 \times 10^{-7} \, m} \approx 5.172 \times 10^{14} \, s^{-1}$.

The energy $(E)$ of one photon is given by the expression: $E = h \nu$.
Given:
$h = 6.626 \times 10^{-34} \ J \ s$
$\nu = 5 \times 10^{14} \ s^{-1}$
Energy of one photon:
$E = (6.626 \times 10^{-34} \ J \ s) \times (5 \times 10^{14} \ s^{-1}) = 3.313 \times 10^{-19} \ J$.
Energy of one mole of photons:
$E_{mole} = (3.313 \times 10^{-19} \ J) \times (6.022 \times 10^{23} \ mol^{-1}) = 199508.26 \ J \ mol^{-1} \approx 199.51 \ kJ \ mol^{-1}$.

(N/A) Given:
$\lambda = 580 \,nm = 580 \times 10^{-9} \,m$
$c = 3 \times 10^{8} \,m/s$
$1$. Frequency $(v)$:
Using the relation $v = \frac{c}{\lambda}$,
$v = \frac{3 \times 10^{8} \,m/s}{580 \times 10^{-9} \,m} = 5.17 \times 10^{14} \,s^{-1}$
$2$. Wave number $(\bar{v})$:
Using the relation $\bar{v} = \frac{1}{\lambda}$,
$\bar{v} = \frac{1}{580 \times 10^{-9} \,m} = 1.72 \times 10^{6} \,m^{-1}$
Thus,the frequency is $5.17 \times 10^{14} \,s^{-1}$ and the wave number is $1.72 \times 10^{6} \,m^{-1}$.

Frequency $(\nu)$ of light $= \frac{1}{\text{Period}}$
$= \frac{1}{2.0 \times 10^{-10} \,s} = 5.0 \times 10^{9} \,Hz$
Wavelength $(\lambda)$ of light $= \frac{c}{\nu}$
Where $c$ (velocity of light in vacuum) $= 3 \times 10^{8} \,m/s$
$\lambda = \frac{3 \times 10^{8} \,m/s}{5.0 \times 10^{9} \,s^{-1}} = 0.06 \,m = 6.0 \times 10^{-2} \,m$
Wave number $(\bar{\nu}) = \frac{1}{\lambda} = \frac{1}{6.0 \times 10^{-2} \,m} = 16.66 \,m^{-1}$

(N/A) The energy required to ionise one mole of sodium atoms is given by the formula $E = \frac{N_A h c}{\lambda}$.
Substituting the values:
$N_A = 6.022 \times 10^{23} \, mol^{-1}$
$h = 6.626 \times 10^{-34} \, J \cdot s$
$c = 3 \times 10^8 \, m \cdot s^{-1}$
$\lambda = 242 \times 10^{-9} \, m$
$E = \frac{(6.022 \times 10^{23} \, mol^{-1}) \times (6.626 \times 10^{-34} \, J \cdot s) \times (3 \times 10^8 \, m \cdot s^{-1})}{242 \times 10^{-9} \, m}$
$E \approx 494700 \, J \cdot mol^{-1}$
Converting to $kJ \cdot mol^{-1}$:
$E = 494.7 \, kJ \cdot mol^{-1}$

(N/A) Threshold wavelength of radiation $(\lambda_{0}) = 6800 \, \mathring{A} = 6800 \times 10^{-10} \, m$.
Threshold frequency $(v_{0})$ is given by the formula $v_{0} = \frac{c}{\lambda_{0}}$.
Substituting the values: $v_{0} = \frac{3 \times 10^{8} \, m s^{-1}}{6.8 \times 10^{-7} \, m} = 4.41 \times 10^{14} \, s^{-1}$.
Thus,the threshold frequency $(v_{0})$ of the metal is $4.41 \times 10^{14} \, s^{-1}$.
Work function $(W_{0})$ is given by $W_{0} = h v_{0}$.
Substituting the values: $W_{0} = (6.626 \times 10^{-34} \, J s)(4.41 \times 10^{14} \, s^{-1}) = 2.922 \times 10^{-19} \, J$.

(A) The electromagnetic spectrum follows the order of increasing frequency as: $FM$ radio waves $ < $ microwave radiation $<$ visible light (amber light) $ < $ $X$-rays $ < $ cosmic rays.
Therefore,the increasing order of frequency is: $(c) < (a) < (b) < (e) < (d)$.

The power of the laser is defined as the total energy emitted per unit time. Assuming the emission occurs over $1 \ s$ for the given number of photons:
Power $(P) = \frac{E}{t} = \frac{N h c}{\lambda \times t}$
Where:
$N = 5.6 \times 10^{24}$ (number of photons)
$h = 6.626 \times 10^{-34} \ J \ s$ (Planck's constant)
$c = 3 \times 10^{8} \ m \ s^{-1}$ (speed of light)
$\lambda = 337.1 \times 10^{-9} \ m$ (wavelength)
$t = 1 \ s$
Substituting the values:
$P = \frac{(5.6 \times 10^{24}) \times (6.626 \times 10^{-34}) \times (3 \times 10^{8})}{337.1 \times 10^{-9}}$
$P = \frac{11.13168 \times 10^{-1}}{337.1 \times 10^{-9}}$
$P \approx 3.302 \times 10^{6} \ W$

(N/A) According to the photoelectric effect equation:
$E = W_{0} + K.E_{max}$
where $E$ is the energy of the incident photon,$W_{0}$ is the work function,and $K.E_{max}$ is the maximum kinetic energy of the emitted photoelectron.
The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $h = 6.626 \times 10^{-34} \ J \ s$,$c = 3.0 \times 10^{8} \ m \ s^{-1}$,and $\lambda = 256.7 \times 10^{-9} \ m$:
$E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{256.7 \times 10^{-9}} \ J = 7.744 \times 10^{-19} \ J$.
Converting to electron-volts $(eV)$:
$E = \frac{7.744 \times 10^{-19} \ J}{1.602 \times 10^{-19} \ J/eV} \approx 4.83 \ eV$.
The stopping potential is $0.35 \ V$,so the maximum kinetic energy is $K.E_{max} = 0.35 \ eV$.
Substituting these values into the equation:
$W_{0} = E - K.E_{max} = 4.83 \ eV - 0.35 \ eV = 4.48 \ eV$.
The work function for silver metal is $4.48 \ eV$.

(N/A) Wilhelm Roentgen in $1895$ showed that when high-speed electrons strike a dense metal target in a cathode ray tube,they produce rays that cause fluorescence in materials placed outside the tube. He named them $X$-rays.
Production: $X$-rays are produced when fast-moving electrons are suddenly decelerated upon striking a dense metal anode (target).
Characteristics: $(i)$ $X$-rays are not deflected by electric or magnetic fields,indicating they are neutral.
$(ii)$ They possess very high penetrating power through matter.
$(iii)$ They have very short wavelengths,typically in the range of $\sim 0.1 \ nm$.
$(iv)$ They exhibit electromagnetic wave character.
Uses: Due to their high penetrating power,$X$-rays are extensively used in medical imaging to study the interior of the human body and in industrial applications to detect structural defects in objects.

(A) According to Maxwell's electromagnetic theory,when an electrically charged particle moves under acceleration,it produces alternating electrical and magnetic fields.
These fields are transmitted in the form of waves,which are known as electromagnetic waves or electromagnetic radiation.

(N/A) James Maxwell was the first to provide a comprehensive explanation regarding the interaction between charged bodies and the behavior of electrical and magnetic fields.
- When an electrically charged particle moves under acceleration,alternating electrical and magnetic fields are produced and transmitted.
- These fields are transmitted in the form of waves,which are known as electromagnetic waves or electromagnetic radiation.

(C) In earlier days,Newton proposed that light was made of particles.
In the $19^{th}$ century,it was established that light is a form of electromagnetic radiation.
Maxwell was the first to reveal that light waves are associated with oscillating electric and magnetic fields.

(N/A) Electromagnetic wave motion is complex in nature. Its key properties are as follows:
$(i)$ The oscillating electric and magnetic fields produced by oscillating charged particles are perpendicular to each other and both are perpendicular to the direction of propagation of the wave.
$(ii)$ Unlike sound waves or water waves,electromagnetic waves do not require a medium and can travel through a vacuum.
$(iii)$ There are many types of electromagnetic radiations,which differ from one another in wavelength $(\lambda)$ or frequency $(\nu)$. These constitute what is called the electromagnetic spectrum:
Radio frequency: $10^{6} \ Hz$ (broadcasting)
Microwave region: $10^{10} \ Hz$ (radar)
Infrared region: $10^{13} \ Hz$ (heating)
Visible light: $10^{15} \ Hz$ (component of sun's radiation)
Ultraviolet region: $10^{16} \ Hz$ (component of sun's radiation)
$(iv)$ These radiations are characterized by properties such as frequency $(\nu)$,wavelength $(\lambda)$,and wavenumber $(\bar{\nu})$.

There are many types of electromagnetic radiations,which differ from one another in wavelength (or frequency). These constitute what is called the electromagnetic spectrum. The electromagnetic spectrum is shown in the figure below:

(N/A) The electromagnetic radiations are characterized by frequency $(\nu)$,wavelength $(\lambda)$,and wave number $(\bar{\nu})$.
$1$. Frequency $(\nu)$: It is defined as the number of waves that pass a given point in one second. The $SI$ unit for frequency is hertz ($Hz$ or $s^{-1}$).
$2$. Wavelength $(\lambda)$: It is the distance between two successive crests or troughs of a wave. Its $SI$ unit is meter $(m)$,though smaller units like $\mathring{A}$,$nm$,or $cm$ are commonly used.
$3$. Wave number $(\bar{\nu})$: It is defined as the number of wavelengths per unit length. Its $SI$ unit is $m^{-1}$,and the commonly used unit is $cm^{-1}$.
In vacuum,all electromagnetic radiations travel at the same speed,known as the speed of light $(c)$,where $c = 3.0 \times 10^{8} \ m \ s^{-1}$.
The relationship between frequency $(\nu)$,wavelength $(\lambda)$,and the speed of light $(c)$ is given by the equation: $c = \nu \lambda$ or $\nu = \frac{c}{\lambda}$.
Additionally,the wave number is related to wavelength as: $\bar{\nu} = \frac{1}{\lambda}$.

(C-A-B-E-D) The electromagnetic spectrum in increasing order of frequency is: $FM$ radio waves < microwave radiation < visible light (amber) < $X$-rays < cosmic rays.
The order is: $(c) < (a) < (b) < (e) < (d)$.

Radiation Type	Frequency Range $(Hz)$
$(c)$ $FM$ radio	$\sim 10^{6}$
$(a)$ Microwave	$\sim 10^{10}$
$(b)$ Amber light	$\sim 10^{14}$
$(e)$ $X$-rays	$10^{17} - 10^{19}$
$(d)$ Cosmic rays	$> 10^{20}$

Given: Wavelength $(\lambda)$ = $3.6 \ \mathring{A} = 3.6 \times 10^{-10} \ m$.
Speed of light $(c)$ = $3 \times 10^{8} \ m/s$.
Frequency $(v)$ = $c / \lambda = (3 \times 10^{8} \ m/s) / (3.6 \times 10^{-10} \ m) = 8.333 \times 10^{17} \ s^{-1}$.
Wave number $(\bar{v})$ = $1 / \lambda = 1 / (3.6 \times 10^{-10} \ m) = 2.778 \times 10^{9} \ m^{-1}$.