Electromagnetic Spectrum Questions in English

(N/A) Radio waves are produced by the acceleration of charges in conducting wires. These waves typically fall in the frequency range of $500 \text{ kHz}$ to $1000 \text{ MHz}$.
Types of radio waves:
$(i)$ $AM$ (Amplitude Modulation) band: $530 \text{ kHz}$ to $1710 \text{ kHz}$.
$(ii)$ $SW$ (Short Wave) band: Used for long-distance communication via ionospheric reflection.
$(iii)$ $TV$ transmission: Frequencies range from $54 \text{ MHz}$ to $890 \text{ MHz}$.
$(iv)$ $FM$ (Frequency Modulation) band: $88 \text{ MHz}$ to $108 \text{ MHz}$.
$(v)$ Cellular phones: Use $UHF$ (Ultra High Frequency) bands for voice communication.
Uses: Radio waves are primarily used in radio,television,and cellular communication systems.
Summary:
- Wavelength range: $0.1 \text{ m}$ to $600 \text{ m}$.
- Frequency range: $500 \text{ kHz}$ to $1000 \text{ MHz}$.
- Source: Accelerated charges in oscillating circuits.
- Inventor: Guglielmo Marconi in $1895$.
- Properties: They exhibit reflection and refraction.

(A) Microwaves are produced by special types of vacuum tubes like Klystron,magnetron,or Gunn diodes.
They are radio waves of very short wavelength,and their frequency range varies from $0.3 \text{ GHz}$ to $300 \text{ GHz}$.
Uses of Microwaves:
$(i)$ They are used in radar systems for aircraft navigation.
$(ii)$ They are used for the control of fighter planes,naval operations,and satellite communication.
$(iii)$ Radar provides the basis for speed guns used to measure the speed of balls (tennis and cricket) and vehicles.
$(iv)$ They are used in microwave ovens. In such ovens,the frequency of the microwave resonates with the frequency of water molecules. Hence,the energy of the wave is transferred as kinetic energy to water molecules,heating the food. Domestic microwave ovens typically use $0.915 \text{ GHz}$ or $2.45 \text{ GHz}$.
Infrared waves are produced by hot bodies and their molecules. They lie between the microwave and visible regions of the electromagnetic spectrum. $LED$s can also produce infrared waves. These are often called heat waves because they are readily absorbed by water molecules,$CO_2$,and $NH_3$ present in matter. By absorbing these waves,the kinetic energy of molecules increases,causing them to heat up.
Uses of Infrared Waves:
$(i)$ Infrared bulbs are used in physical therapy.
$(ii)$ They maintain the average temperature of the Earth through the greenhouse effect.
$(iii)$ Infrared sensors are used for military purposes and monitoring crop growth in agriculture.
$(iv)$ $LED$s are used in remote controls for $TV$s,video recorders,and $Hi-Fi$ systems.

(N/A) Visible light is a part of the electromagnetic radiation emitted by the Sun. These rays can also be produced by flames,bulbs,and incandescent lamps.
This is the part of the electromagnetic spectrum detected by the human eye.
Its frequency varies from $4 \times 10^{14} \text{ Hz}$ to $7 \times 10^{14} \text{ Hz}$. Its wavelength is in the range of $700 \text{ nm}$ to $400 \text{ nm}$.
Uses of Visible Rays:
$(i)$ The human eye is sensitive to visible radiation.
$(ii)$ Due to visible light,we can see objects around us.
$(iii)$ Different animals are sensitive to different ranges of wavelengths. For example,snakes can detect infrared waves.
$(iv)$ For some insects,the visible range extends into the ultraviolet range.
Ultraviolet $(UV)$ radiation is produced by special types of lamps and very hot bodies. The Sun is also an important source of $UV$ radiation.
Most of the ultraviolet rays coming from the Sun are absorbed by the ozone layer,which is at a height of $40-50 \text{ km}$.
Excessive ultraviolet light is harmful to the human body. When the body is exposed to $UV$ light for a long period,melanin is produced in the body,causing the skin to darken.
Sparks produced during welding contain a lot of $UV$ light; hence,to protect the eyes,special face masks containing black-colored glass are used.
$UV$ radiation is absorbed by ordinary glass. Hence,one cannot get tanned or sunburned through glass windows.
Uses of Ultraviolet Rays:
$(i)$ Ultraviolet waves have a very short wavelength ($400 \text{ nm}$ to $0.6 \text{ nm}$),so they can be focused into a very narrow beam of high precision. It is used in $LASIK$ (Laser-Assisted in Situ Keratomileusis).
$(ii)$ In some water purifiers,$UV$ bulbs are used to kill germs.
$(iii)$ In the atmosphere,the ozone layer acts as a protective layer. The depletion of the ozone layer (due to the use of $CFCs$) is a matter of international concern.

(A) The electromagnetic spectrum is the entire range of all possible frequencies of electromagnetic radiation. It encompasses all types of electromagnetic radiation,including radio waves,microwaves,infrared,visible light,ultraviolet,$X$-rays,and gamma rays. These waves are ordered by frequency or wavelength.

(N/A) The frequency range for radio waves is generally considered to be from $3 \ kHz$ to $300 \ GHz$.
Specifically,the $FM$ (Frequency Modulation) radio band operates in the range of $88 \ MHz$ to $108 \ MHz$.

(B) microwave oven uses electromagnetic waves known as microwaves to heat food.
These waves have a frequency typically around $2.45 \ GHz$.
When these waves are absorbed by water molecules in the food,they cause the molecules to rotate rapidly,generating heat through molecular friction.
Therefore,the correct answer is microwaves.

(A) Infrared waves are produced by the vibration of atoms and molecules in a substance. When these particles vibrate,they emit electromagnetic radiation in the infrared region of the spectrum. This is why infrared radiation is often referred to as 'heat radiation',as it is emitted by all objects at temperatures above absolute zero.

(A) Infrared rays are commonly known as heat waves.
These waves are produced by hot bodies and molecules.
When these waves are incident on a substance,they cause the molecules to vibrate,thereby increasing the internal energy and temperature of the substance.

(B) Remote control switches for household appliances like televisions,air conditioners,and music systems typically operate using $Infrared$ $(IR)$ waves. These waves have wavelengths ranging from approximately $1 \ mm$ to $700 \ nm$,which are ideal for short-range wireless communication within a room.

(N/A) The visible light spectrum is the portion of the electromagnetic spectrum that is visible to the human eye.
Visible light typically covers a frequency range from approximately $4 \times 10^{14} Hz$ to $7.5 \times 10^{14} Hz$ (or $400 THz$ to $750 THz$).

(B) Snakes,particularly pit vipers,possess specialized sensory organs known as pit organs located between their eyes and nostrils. These organs are highly sensitive to thermal radiation,which corresponds to the infrared region of the electromagnetic spectrum. This allows them to detect the body heat of their prey even in total darkness. Therefore,snakes can detect infrared waves.

(N/A) Ultraviolet $(UV)$ waves are produced by special lamps and very hot bodies. The most significant natural source of ultraviolet radiation is the Sun. Other sources include mercury vapor lamps,arc lamps,and welding arcs.

(N/A) Ultraviolet $(UV)$ waves are electromagnetic waves that occupy the spectral region between visible light and $X$-rays.
The wavelength range for ultraviolet waves is approximately $1 \times 10^{-7} \, m$ to $4 \times 10^{-7} \, m$ (or $100 \, nm$ to $400 \, nm$).

(B) When human skin is exposed to sunlight for a long period,it absorbs ultraviolet $(UV)$ radiation.
To protect the deeper layers of the skin from the harmful effects of these $UV$ rays,the body produces more melanin,a pigment that darkens the skin.
This process is the body's natural defense mechanism against $UV$-induced damage.

(A) $LASIK$ stands for Laser-Assisted In Situ Keratomileusis.
In this surgery,an excimer laser is used to reshape the cornea of the eye.
The excimer laser emits light in the ultraviolet range,specifically at a wavelength of $193 \ nm$.
Therefore,ultraviolet rays are used for $LASIK$ surgery.

(A) The $UV$ (ultraviolet) radiation emitted by the Sun is harmful to living organisms as it can cause skin cancer and damage $DNA$.
The Earth's atmosphere contains a layer of ozone $(O_3)$ in the stratosphere,which is known as the ozone layer.
This ozone layer absorbs most of the harmful $UV$ radiation from the Sun,preventing it from reaching the Earth's surface and thus protecting us.

(N/A) The wavelength range of $X$-rays typically extends from approximately $0.01 \ nm$ to $10 \ nm$ ($10^{-11} \ m$ to $10^{-8} \ m$).
$X$-rays are electromagnetic waves with high energy and high frequency,positioned between ultraviolet rays and gamma rays in the electromagnetic spectrum.

(B) Gamma rays are high-energy electromagnetic waves with very short wavelengths.
Due to their high energy and penetrating power,they are used in medical treatments,specifically in radiotherapy,to kill or destroy cancer cells by damaging their $DNA$.

(A) $(i)$ In satellite communication,microwaves are used,so ${\lambda _1}$ is the wavelength of microwaves.
$(ii)$ Ultraviolet radiations are used to kill germs in water purifiers,so ${\lambda _2}$ is the wavelength of $UV$ rays.
$(iii)$ $X$-rays are used to detect leakage of oil in underground pipelines,so ${\lambda _3}$ is the wavelength of $X$-rays.
$(iv)$ Infrared rays are used to improve visibility on runways during fog and mist conditions,so ${\lambda _4}$ is the wavelength of infrared rays.
$(b)$ The order of wavelengths in the electromagnetic spectrum is: ${\lambda _{X-rays}} < {\lambda _{UV}} < {\lambda _{IR}} < {\lambda _{Micro}}$.
Therefore,the ascending order is: ${\lambda _3} < {\lambda _2} < {\lambda _4} < {\lambda _1}$.
$(c)$ $(i)$ Microwaves are used in $RADAR$.
$(ii)$ $UV$ rays are used in $LASIK$ eye surgery.
$(iii)$ $X$-rays are used to detect fractures in bones.
$(iv)$ Infrared rays are used in optical communication.

(C) The electromagnetic spectrum is ordered by wavelength.
Radio waves have the longest wavelengths,followed by microwaves,infrared,visible light,ultraviolet,$X$-rays,and gamma rays,which have the shortest wavelengths.
Therefore,the correct order of decreasing wavelength is:
$\lambda_{\text{radio waves}} > \lambda_{\text{microwaves}} > \lambda_{\text{visible}} > \lambda_{\text{x-rays}}$.

(D) The wavelength ranges for the given electromagnetic radiations are as follows:
$(a)$ Microwave: The wavelength is approximately $10^{-3} \, m$ to $0.3 \, m$. Thus, $(a) \rightarrow (iv)$.
$(b)$ Gamma rays: These have the shortest wavelengths, typically less than $10^{-13} \, m$ (often cited as $10^{-15} \, m$ in this context). Thus, $(b) \rightarrow (ii)$.
$(c)$ $A.M.$ radio waves: These have very long wavelengths, typically in the range of $100 \, m$ to $1000 \, m$. Thus, $(c) \rightarrow (i)$.
$(d)$ $X$-rays: These have wavelengths in the range of $10^{-8} \, m$ to $10^{-12} \, m$ (often cited as $10^{-10} \, m$). Thus, $(d) \rightarrow (iii)$.
Therefore, the correct matching is $(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)$.

(D) The electromagnetic spectrum is ordered by wavelength.
Gamma rays have the shortest wavelength,typically ranging from $10^{-10} \ m$ to $10^{-14} \ m$.
Comparing the given options:
$1$. Microwaves: $10^{-3} \ m$ to $10^{-1} \ m$
$2$. Ultraviolet rays: $10^{-7} \ m$ to $10^{-8} \ m$
$3$. $X$-rays: $10^{-8} \ m$ to $10^{-10} \ m$
$4$. Gamma-rays: $< 10^{-10} \ m$
Therefore,Gamma-rays have the shortest wavelength.

(A) The atmosphere is stratified based on altitude as follows:
$1$. Troposphere: $0$ to $12\, km$. Thus,$(a) \rightarrow (iv)$.
$2$. Stratosphere: $12$ to $50\, km$. Thus,$(b) \rightarrow (iii)$.
$3$. Mesosphere: $50$ to $85\, km$. Thus,$(c) \rightarrow (ii)$.
$4$. Thermosphere: Above $85\, km$. Thus,$(d) \rightarrow (i)$.
Therefore,the correct matching is $(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)$.

(D) The source of microwave frequency is a magnetron.
$(b)$ The source of infrared frequency is the vibration of atoms and molecules.
$(c)$ The source of Gamma rays is the radioactive decay of the nucleus.
$(d)$ The source of $X$-rays is the transition of inner shell electrons.
Therefore,the correct matching is $(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)$.

(C) The electromagnetic spectrum classifies waves based on their wavelength range:
$(a)$ $AM$ radio waves have long wavelengths, typically around $10^{2} \, m$ (Matches with $ii$).
$(b)$ Microwaves have wavelengths in the range of $10^{-2} \, m$ (Matches with $iii$).
$(c)$ Infrared radiations have wavelengths around $10^{-4} \, m$ (Matches with $iv$).
$(d)$ $X$-rays have very short wavelengths, typically around $10^{-10} \, m$ (Matches with $i$).
Therefore, the correct matching is $(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)$.

(B) According to the electromagnetic spectrum, the wavelength $(\lambda)$ increases in the following order:

$\gamma$-ray

$X$-ray

Ultraviolet

Visible

Infrared

Microwave

Radio wave

Therefore, the correct ascending order of wavelengths is: $\lambda_{\text{gamma-ray}} < \lambda_{\text{X-ray}} < \lambda_{\text{visible}} < \lambda_{\text{microwave}}$.

(A) The correct matches are as follows:
$(a)$ Ultraviolet rays are used for sterilizing surgical instruments $(iii)$.
$(b)$ Microwaves are used in Radar systems for aircraft navigation $(iv)$.
$(c)$ Infrared waves are responsible for the Greenhouse effect $(ii)$.
$(d)$ $X$-rays are used to study crystal structures due to their short wavelengths $(i)$.
Therefore,the correct sequence is $(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)$.

(A) $UV$ rays are used for water purification as they kill bacteria.
$X-$rays are used as a diagnostic tool in medicine to detect fractures.
Microwaves are used for communication and radar systems.
Infrared waves have longer wavelengths and show less scattering,therefore they are used for improving visibility in foggy conditions.

(C) The general equation for a plane electromagnetic wave is $B = B_0 \sin(kx + \omega t)$.
Comparing this with the given equation $B = 2 \times 10^{-7} \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)$,we get:
Angular frequency $\omega = 1.5 \times 10^{11} \, \text{rad/s}$ and wave number $k = 0.5 \times 10^3 \, \text{m}^{-1}$.
The frequency $f$ is given by $f = \frac{\omega}{2\pi} = \frac{1.5 \times 10^{11}}{2 \times 3.14} \approx 2.39 \times 10^{10} \, \text{Hz}$.
The wavelength $\lambda$ is given by $\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{0.5 \times 10^3} \approx 1.25 \times 10^{-2} \, \text{m} = 1.25 \, \text{cm}$.
Electromagnetic waves with wavelengths in the range of $1 \, \text{mm}$ to $1 \, \text{m}$ (or frequencies between $300 \, \text{MHz}$ and $300 \, \text{GHz}$) are classified as microwaves.
Therefore,the correct option is $C$.

(A) The ozone layer in the Earth's atmosphere is primarily responsible for absorbing ultraviolet $(UV)$ radiation from the Sun.
Specifically,it effectively absorbs radiation with wavelengths shorter than approximately $3 \times 10^{-7} \,m$ (or $300 \,nm$),which corresponds to the $UV$-$C$ and parts of the $UV$-$B$ spectrum.
Therefore,radiation with wavelengths less than $3 \times 10^{-7} \,m$ is blocked by the ozone layer.

(D) The correct option is $(d)$.
Both $X$-rays and $\gamma$-rays are high-energy electromagnetic radiations.
Due to their high penetrating power and ionizing nature,they are effectively used in radiotherapy to destroy cancerous cells by damaging their $DNA$.

(C) The power $P$ of the source is $15 \text{ kW} = 15 \times 10^3 \text{ W}$.
The number of photons produced per second $n = 10^{16} \text{ s}^{-1}$.
The energy of one photon $E$ is given by $E = \frac{P}{n} = \frac{15 \times 10^3}{10^{16}} = 15 \times 10^{-13} \text{ J}$.
Using the relation $E = h\nu$,we find the frequency $\nu$:
$\nu = \frac{E}{h} = \frac{15 \times 10^{-13}}{6 \times 10^{-34}} = 2.5 \times 10^{21} \text{ Hz}$.
Electromagnetic radiation with a frequency of $2.5 \times 10^{21} \text{ Hz}$ falls in the range of Gamma rays (which typically have frequencies greater than $10^{19} \text{ Hz}$).

(C) The correct matches are as follows:
$1$. Microwaves $(A)$ are used in radar systems for aircraft navigation $(IV)$.
$2$. $UV$ rays $(B)$ are used in Lasik eye surgery $(III)$.
$3$. Infra-red rays $(C)$ are used in physiotherapy $(I)$ to provide heat treatment.
$4$. $X$-rays $(D)$ are used in the treatment of cancer $(II)$ (radiotherapy).
Thus,the correct matching is $A-IV, B-III, C-I, D-II$.

(B) The correct matches are as follows:
$1$. Microwaves are produced by Klystron valves or Magnetron valves $(A-IV)$.
$2$. Gamma rays are produced by the radioactive decay of the nucleus $(B-I)$.
$3$. Radio waves are produced by the rapid acceleration and deceleration of electrons in aerials $(C-II)$.
$4$. $X$-rays are produced when high-energy electrons strike a metal target,causing transitions of inner shell electrons $(D-III)$.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(A) Optical communication typically uses infrared or visible light,which falls in the frequency range of $1 \, THz$ to $1000 \, THz$.
Microwaves used in $RADAR$ have frequencies in the range of $1 \, GHz$ to $300 \, GHz$.
Since frequency $f$ and wavelength $\lambda$ are inversely related by $c = f\lambda$,higher frequency waves have shorter wavelengths.
Therefore,$EM$ waves used for optical communication have shorter wavelengths than microwaves,making Assertion $A$ false.
Regarding Reason $R$,the energy of a photon is given by $E = hf$. Since infrared waves have higher frequencies than microwaves,they are indeed more energetic. Thus,Reason $R$ is true.
Conclusion: $A$ is false but $R$ is true.

(B) The wavelength ranges for the given electromagnetic waves are as follows:
$(A)$ Microwave: The wavelength range is $0.1\,m$ to $1\,mm$ (matches $(IV)$).
$(B)$ Ultraviolet: The wavelength range is $400\,nm$ to $1\,nm$ (matches $(I)$).
$(C)$ $X$-Ray: The wavelength range is $1\,nm$ to $10^{-3}\,nm$ (matches $(II)$).
$(D)$ Infra-red: The wavelength range is $1\,mm$ to $700\,nm$ (matches $(III)$).
Therefore,the correct matching is $(A)-(IV), (B)-(I), (C)-(II), (D)-(III)$.

(C) The electromagnetic spectrum in increasing order of wavelength is as follows:
Gamma rays < $X$-rays < Ultraviolet rays < Visible light < Infrared waves < Microwaves < Radio waves.
Given the wavelengths:
Gamma rays: $\lambda_1$
$X$-rays: $\lambda_2$
Infrared waves: $\lambda_3$
Microwaves: $\lambda_4$
Comparing these,we find that $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$.
Therefore,the ascending order is $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$.

(A) The electromagnetic spectrum is categorized by wavelength ranges:
$1$. Infrared radiation has wavelengths ranging from $1 \, mm$ to $700 \, nm$ (or $10^{-3} \, m$ to $7 \times 10^{-7} \, m$). Thus,$(A)-(iii)$.
$2$. Ultraviolet radiation has wavelengths ranging from $400 \, nm$ to $1 \, nm$. Thus,$(B)-(ii)$.
$3$. $X$-rays have wavelengths ranging from $1 \, nm$ to $10^{-3} \, nm$. Thus,$(C)-(iv)$.
$4$. Gamma rays have the shortest wavelengths,typically $ < 10^{-3} \, nm$. Thus,$(D)-(i)$.
Therefore,the correct matching is $(A)-(iii), (B)-(ii), (C)-(iv), (D)-(i)$.

(A) The electromagnetic spectrum in increasing order of wavelength is as follows:
$X-$rays < Ultraviolet $(UV)$ rays < Infrared $(IR)$ rays < Microwaves.
Given the wavelengths:
$(A)$ Microwaves: $\lambda_1$
$(B)$ Ultraviolet rays: $\lambda_2$
$(C)$ Infrared rays: $\lambda_3$
$(D)$ $X-$rays: $\lambda_4$
Comparing these,we have $\lambda_4 < \lambda_2 < \lambda_3 < \lambda_1$.
Therefore,the correct ascending order is $\lambda_4 < \lambda_2 < \lambda_3 < \lambda_1$.

(C) The energy required to dissociate a carbon monoxide molecule is $E = 11 \ eV$.
Converting this energy into Joules: $E = 11 \times 1.6 \times 10^{-19} \ J = 1.76 \times 10^{-18} \ J$.
Using the relation $E = h\nu$,where $h = 6.626 \times 10^{-34} \ J \cdot s$ is Planck's constant and $\nu$ is the frequency:
$\nu = \frac{E}{h} = \frac{1.76 \times 10^{-18}}{6.626 \times 10^{-34}} \approx 2.66 \times 10^{15} \ Hz$.
The electromagnetic spectrum range for ultraviolet radiation is approximately $7.5 \times 10^{14} \ Hz$ to $3 \times 10^{16} \ Hz$.
Since the calculated frequency $2.66 \times 10^{15} \ Hz$ falls within this range,the radiation belongs to the ultraviolet region.

(B) $UV$ rays are used for water purification because they can kill bacteria and microorganisms.
$(b)$ $X-$rays are high-energy electromagnetic waves used as a diagnostic tool in medicine to detect fractures and internal injuries.
$(c)$ Microwaves are used in communication systems and Radar technology due to their ability to carry information and reflect off objects.
$(d)$ Infrared waves have longer wavelengths and show less scattering, making them useful for improving visibility in foggy conditions.
Therefore, the correct matching is: $(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)$.

(B) The electromagnetic spectrum is categorized based on wavelength ranges.
- The wavelength range of $100 \text{ Å}$ to $400 \text{ Å}$ falls within the ultraviolet $(UV)$ region.
- $X$-rays typically range from $0.01 \text{ Å}$ to $100 \text{ Å}$.
- The visible region ranges from approximately $4000 \text{ Å}$ to $7000 \text{ Å}$.
- The infrared region starts above $7000 \text{ Å}$.
Therefore,the correct classification for the range $100 \text{ Å}$ to $400 \text{ Å}$ is the $UV$ region.

(C) Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field. They do not require a medium to travel.
Light rays,$X$-rays,and $Gamma$ rays are all parts of the electromagnetic spectrum.
Alpha rays consist of high-energy helium nuclei ($He^{2+}$ particles),which are charged particles with mass. Therefore,alpha rays are particle radiation,not electromagnetic waves.

(D) The ozone layer in the Earth's stratosphere absorbs the majority of the Sun's harmful ultraviolet $(UV)$ radiation. This absorption is crucial as it protects living organisms on Earth from the damaging effects of high-energy $UV$ rays,such as skin cancer and cataracts.

(A) Cellular phones operate by transmitting and receiving signals within the Ultra High Frequency $(UHF)$ band,which typically ranges from $300 \ MHz$ to $3 \ GHz$. This frequency range is ideal for mobile communication because it allows for high data rates and supports the compact antenna sizes required for portable devices. Therefore,the correct option is $A$.

(A) The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye.
Visible light has a wavelength range of approximately $400 \text{ nm}$ to $700 \text{ nm}$.
Using the relation $c = f \lambda$,where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light,we can calculate the frequency $f = c / \lambda$.
For $\lambda = 700 \text{ nm} = 700 \times 10^{-9} \text{ m}$,$f = (3 \times 10^8) / (700 \times 10^{-9}) \approx 4.29 \times 10^{14} \text{ Hz} = 429 \text{ THz}$.
For $\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}$,$f = (3 \times 10^8) / (400 \times 10^{-9}) = 7.5 \times 10^{14} \text{ Hz} = 750 \text{ THz}$.
Thus,the frequency range is approximately $400 \text{ THz}$ to $750 \text{ THz}$,which aligns with option $A$.

(D) The standard frequency range allocated for $FM$ (Frequency Modulated) radio broadcasting is $88 \text{ MHz}$ to $108 \text{ MHz}$.
This band is part of the Very High Frequency $(VHF)$ spectrum.
Therefore,the correct option is $D$.

(D) The electromagnetic spectrum is categorized based on wavelength and frequency.
Ultraviolet $(UV)$ radiation lies between the visible light spectrum and $X$-rays.
The wavelength range for ultraviolet radiation is approximately $400 \ nm$ to $1.0 \ nm$ (or $10^{-7} \ m$ to $10^{-9} \ m$).
Therefore,the correct option is $D$.

(B) The correct order is $v_r < v_v < v_{uv}$.
In the electromagnetic spectrum,waves are arranged in increasing order of frequency as follows:
Radio Waves $\rightarrow$ Microwaves $\rightarrow$ Infrared rays $\rightarrow$ Visible light $\rightarrow$ Ultraviolet rays $\rightarrow$ $X$-rays $\rightarrow$ Gamma rays.
Since Radio waves have the lowest frequency and Ultraviolet waves have a higher frequency than Visible light,the correct relationship is $v_r < v_v < v_{uv}$.

(B) The electromagnetic spectrum is ordered by wavelength as follows: Gamma rays < $X$-rays < Ultraviolet < Visible < Infrared < Microwaves < Radio waves.
Given that $\lambda_r$ is the wavelength of Gamma rays,$\lambda_x$ is the wavelength of $X$-rays,and $\lambda_m$ is the wavelength of microwaves.
According to the electromagnetic spectrum,the wavelength of Gamma rays is the shortest and the wavelength of microwaves is much longer than that of $X$-rays.
Therefore,the correct order is $\lambda_r < \lambda_x < \lambda_m$.