Electromagnetic Spectrum Questions in English

(B) Infrared radiation is a type of electromagnetic radiation with wavelengths longer than those of visible light.
It is primarily detected by measuring the heat it produces when absorbed by a material.
$A$ $Bolometer$ is a device used to measure the power of incident electromagnetic radiation via the heating of a material with a temperature-dependent electrical resistance.
Therefore,infrared radiation can be detected by a $Bolometer$.

(A) The electromagnetic spectrum is arranged in order of increasing frequency and decreasing wavelength.
Radio waves have the lowest frequency and the longest wavelength in the electromagnetic spectrum.
Therefore,among the given options,radio waves have the longest wavelength.

(A) bolometer is a device used for measuring the power of incident electromagnetic radiation via the heating of a material with a temperature-dependent electrical resistance. It is primarily used to detect infrared $(IR)$ radiation. When infrared radiation falls on the bolometer,it causes a change in temperature,which leads to a change in electrical resistance,allowing for the detection and measurement of the radiation.

(C) Gamma rays possess high energy and high penetrating power.
In the medical field,they are used in radiotherapy to target and destroy cancerous cells by damaging their $DNA$.
Care is taken to focus the beam precisely so that healthy cells are minimally affected.
Therefore,the correct option is $C$.

(B) The electromagnetic spectrum is divided into various frequency bands for communication purposes.
According to the standard classification:
$1$. Medium frequency $(MF)$ range is $0.3-3\, MHz$.
$2$. High frequency $(HF)$ range is $3-30\, MHz$.
$3$. Very high frequency $(VHF)$ range is $30-300\, MHz$.
Therefore,the frequency range from $3\, MHz$ to $30\, MHz$ is known as the High frequency $(HF)$ band.

(A) The greenhouse effect is primarily caused by the trapping of heat within the Earth's atmosphere.
Solar radiation reaches the Earth,and the surface absorbs this energy,re-emitting it as thermal radiation.
Greenhouse gases (such as $CO_2$,$CH_4$,and water vapor) absorb this outgoing thermal radiation,which corresponds to the infrared region of the electromagnetic spectrum.
Therefore,infrared rays are responsible for the greenhouse effect.

(C) The electromagnetic spectrum in order of increasing frequency (and decreasing wavelength) is: Radio waves,Microwaves,Infrared,Visible light,Ultraviolet,$X$-rays,and Gamma rays.
Comparing the given options:
$1$. Radio waves: Longest wavelength.
$2$. Microwaves: Shorter than radio waves.
$3$. Visible light: Shorter than microwaves.
$4$. Ultraviolet rays: Shorter than visible light.
Therefore,among the given options,Ultraviolet rays have the shortest wavelength.

(A) The electromagnetic spectrum classifies waves based on their wavelength or frequency.
Infrared rays have wavelengths ranging from approximately $1 \, mm$ to $700 \, nm$ ($10^{-3} \, m$ to $7 \times 10^{-7} \, m$).
Converting these values to centimeters $(1 \, m = 100 \, cm)$:
$10^{-3} \, m = 10^{-1} \, cm$
$7 \times 10^{-7} \, m = 7 \times 10^{-5} \, cm$.
Comparing the given options,$10^{-4} \, cm$ falls within the infrared wavelength range,while the other values are closer to or within the visible and ultraviolet regions.
Therefore,the correct option is $A$.

(A) The $AM$ (Amplitude Modulation) radio band typically operates in the frequency range of $535 \, kHz$ to $1605 \, kHz$.
Since $1 \, MHz = 1000 \, kHz$,this range is equivalent to $0.535 \, MHz$ to $1.605 \, MHz$.
Comparing this to the given options,this range is clearly less than $30 \, MHz$.

(C) The electromagnetic spectrum is arranged in order of increasing frequency or decreasing wavelength. The order is: Radio waves,Microwaves,Infrared rays,Visible light,Ultraviolet rays,$X$-rays,and Gamma rays.
Visible light lies between Infrared rays and Ultraviolet rays in the electromagnetic spectrum.
Therefore,the correct option is $C$.

(B) The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye.
It typically ranges from approximately $4000 \, \mathring{A}$ (violet) to $7000 \, \mathring{A}$ (red).
This corresponds to a frequency range of approximately $4 \times 10^{14} \, \text{Hz}$ to $7.5 \times 10^{14} \, \text{Hz}$.

(A) The electromagnetic spectrum is categorized based on frequency and wavelength. Microwaves are a part of the electromagnetic spectrum with frequencies typically ranging from $0.3 \, GHz$ to $300 \, GHz$ (or $3 \times 10^8 \, Hz$ to $3 \times 10^{11} \, Hz$). Among the given options,the range $0.3 \, GHz$ to $300 \, GHz$ is the standard definition for microwaves.

(A) The electromagnetic spectrum is arranged in order of increasing frequency as follows: Radio waves < Microwaves < Infrared < Visible light < Ultraviolet < $X$-rays < Gamma rays.
Since radio waves appear at the beginning of this sequence,they possess the lowest frequency among the given options.

(B) The ozone layer or ozone shield refers to a region of Earth's stratosphere that absorbs most of the Sun's ultraviolet $(UV)$ radiation.
It contains high concentrations of ozone $(O_3)$ relative to other parts of the atmosphere,although still very small relative to other gases in the stratosphere.

(A) The ozone layer is located in the stratosphere of the Earth's atmosphere. Its primary function is to absorb most of the Sun's harmful ultraviolet $(UV)$ radiation. By filtering out these high-energy rays,it protects living organisms on Earth from potential damage such as skin cancer,cataracts,and harm to ecosystems. Therefore,the correct option is $A$.

(B) The electromagnetic spectrum is categorized based on frequency ranges.
$3 \times 10^9 \, Hz$ is equivalent to $3 \, GHz$.
$3 \times 10^{10} \, Hz$ is equivalent to $30 \, GHz$.
The frequency range from $3 \, GHz$ to $30 \, GHz$ is classified as the Super High Frequency $(SHF)$ band.

(A) The electromagnetic spectrum is ordered by increasing wavelength and decreasing frequency.
The order of frequency for the given radiations is:
Frequency of $\gamma$-rays > Frequency of $X$-rays > Frequency of ultraviolet radiations.
Given that the frequencies are $B, A,$ and $C$ respectively:
$B$ (frequency of $\gamma$-rays) > $A$ (frequency of $X$-rays) > $C$ (frequency of ultraviolet radiations).
Therefore,the correct relation is $B > A > C$.

(B) The electromagnetic spectrum encompasses a vast range of wavelengths.
Gamma rays have the shortest wavelengths,starting from approximately $10^{-16} \ m$ to $10^{-10} \ m$.
Radio waves have the longest wavelengths,extending up to $10^{8} \ m$ or even longer.
Therefore,the broad range of the electromagnetic spectrum is approximately from $10^{-16} \ m$ to $10^{8} \ m$.
Among the given options,the range $10^{-15} \ m$ to $10^{8} \ m$ is the most accurate representation of the electromagnetic spectrum.

(A) The electromagnetic spectrum is categorized by wavelength ranges:
$1$. Radio waves have long wavelengths,typically $10^3 \ m$ to $10^{-1} \ m$ (or higher,up to $10^7 \ m$ for long-wave radio).
$2$. $X$-rays have very short wavelengths,typically in the range of $10^{-8} \ m$ to $10^{-12} \ m$. Thus,$10^{-10} \ m$ corresponds to $X$-rays.
$3$. Visible light has wavelengths in the range of approximately $4 \times 10^{-7} \ m$ to $7 \times 10^{-7} \ m$. Thus,$10^{-7} \ m$ (or $100 \ nm$) is close to the ultraviolet/visible boundary,but in the context of standard multiple-choice options,it is categorized as visible light.
Therefore,$10^7 \ m$ is a Radio wave,$10^{-10} \ m$ is an $X$-ray,and $10^{-7} \ m$ is Visible light.

(A) Infrared rays are electromagnetic waves with wavelengths longer than visible light.
They are commonly detected using a spectrometer,which can analyze the spectral distribution of radiation.
Thermopiles,bolometers,and infrared photographic film are also used for detection,but among the given options,a spectrometer is the correct instrument for analyzing and detecting these rays.

(B) The power of the source is given by $P = n \cdot E_{photon} = n \cdot \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.
From this,the wavelength $\lambda$ is given by $\lambda = \frac{nhc}{P}$.
Substituting the given values: $n = 10^{20}$,$h = 6.625 \times 10^{-34} \, J \cdot s$,$c = 3 \times 10^8 \, m/s$,and $P = 4 \times 10^3 \, W$.
$\lambda = \frac{10^{20} \times 6.625 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^3} \, m$.
$\lambda = \frac{19.875 \times 10^{-6}}{4 \times 10^3} \, m = 4.96875 \times 10^{-9} \, m \approx 5 \times 10^{-9} \, m$.
This wavelength range ($10^{-8} \, m$ to $10^{-11} \, m$) corresponds to $X -$ rays.

(B) Given: $\lambda_1 = 3 \times 10^{-3} \, m$,$\lambda_2 = 1 \, m$,and $c = 3 \times 10^8 \, m/s$.
The frequency $f$ is given by $f = \frac{c}{\lambda}$.
For $\lambda_1 = 3 \times 10^{-3} \, m$:
$f_1 = \frac{3 \times 10^8}{3 \times 10^{-3}} = 10^{11} \, Hz = 10^5 \, MHz$.
For $\lambda_2 = 1 \, m$:
$f_2 = \frac{3 \times 10^8}{1} = 3 \times 10^8 \, Hz = 3 \times 10^2 \, MHz$.
Thus,the frequency range is $3 \times 10^2 \, MHz$ to $10^5 \, MHz$.

(C) The energy of the photon is given by $E = 14.4 \, keV = 14.4 \times 10^3 \times 1.6 \times 10^{-19} \, J$.
Using the relation $\lambda = \frac{hc}{E}$,where $h = 6.63 \times 10^{-34} \, J \cdot s$ and $c = 3 \times 10^8 \, m/s$:
$\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{14.4 \times 10^3 \times 1.6 \times 10^{-19}} \approx 0.86 \times 10^{-10} \, m = 0.86 \, \mathring{A}$.
Since the wavelength is in the range of $0.01 \, \mathring{A}$ to $100 \, \mathring{A}$,this electromagnetic wave belongs to the $X$-ray region.

(A) The electromagnetic spectrum is ordered by frequency and wavelength. The order of increasing wavelength is: $\gamma-$ rays < $X-$ rays < $UV$ rays < Visible light < Infrared < Microwaves < Radio waves.
Since $\gamma-$ rays have the highest frequency,they possess the smallest wavelength among the given options.

(D) The electromagnetic spectrum in order of increasing frequency is: radio waves,microwaves,infrared radiation,visible light,ultraviolet radiation,$X$-rays,and gamma rays.
The frequency of gamma rays is the highest among these,typically being greater than $3 \times 10^{19} \text{ Hz}$.
Therefore,$\gamma$-rays have the maximum frequency in the electromagnetic spectrum.

(A) The electromagnetic spectrum in terms of decreasing wavelength is given by:
Microwaves > Infrared > Ultraviolet > Gamma rays.
Therefore,the correct decreasing order is:
$\lambda_{\text{Microwave}} > \lambda_{\text{Infrared}} > \lambda_{\text{Ultraviolet}} > \lambda_{\text{Gamma rays}}$.

(B) The energy $E$ of a photon is related to its wavelength $\lambda$ by the formula $\lambda = \frac{hc}{E}$.
Given,$E = 15\, keV = 15 \times 10^3\, eV$.
The value of $hc$ is approximately $1240\, eV\, nm$.
Substituting these values,we get $\lambda = \frac{1240\, eV\, nm}{15 \times 10^3\, eV} \approx 0.083\, nm$.
The wavelength range for $X$-rays typically spans from $1\, nm$ to $10^{-3}\, nm$.
Since $0.083\, nm$ falls within this range,the electromagnetic waves belong to the $X$-ray part of the spectrum.

(B) Given: Power $P = 4 \ kW = 4000 \ W$,Number of photons $n = 10^{20}$,Time $t = 1 \ s$.
The energy of one photon is given by $E = \frac{P \cdot t}{n}$.
$E = \frac{4000 \times 1}{10^{20}} = 4 \times 10^{-17} \ J$.
To find the frequency $\nu$,we use $E = h\nu$,where $h = 6.63 \times 10^{-34} \ J \cdot s$.
$\nu = \frac{E}{h} = \frac{4 \times 10^{-17}}{6.63 \times 10^{-34}} \approx 6.03 \times 10^{16} \ Hz$.
The frequency range for $X$-rays is typically $3 \times 10^{16} \ Hz$ to $3 \times 10^{19} \ Hz$.
Since the calculated frequency falls within this range,the radiation lies in the $X$-ray region.

(B) The power $P$ of the source is given by $P = n \cdot E_{photon} = n \cdot h \nu$,where $n$ is the number of photons per second,$h$ is Planck's constant,and $\nu$ is the frequency.
Given $P = 4 \ kW = 4 \times 10^3 \ W$ and $n = 10^{20} \ photons/s$.
Rearranging for frequency: $\nu = \frac{P}{n \cdot h}$.
Substituting the values: $\nu = \frac{4 \times 10^3}{10^{20} \times 6.63 \times 10^{-34}} \approx 6.03 \times 10^{16} \ Hz$.
The frequency range $6 \times 10^{16} \ Hz$ falls within the $X$-ray region of the electromagnetic spectrum.

(C) $(1)$ Infrared rays are used to treat muscular strain because these are heat rays.
$(2)$ Radio waves are used for broadcasting because these waves have very long wavelengths ranging from a few centimeters to a few hundred kilometers.
$(3)$ $X$-rays are used to detect fractures of bones because they have high penetrating power but cannot penetrate through denser media like bones.
$(4)$ Ultraviolet rays are absorbed by the ozone layer of the atmosphere.
Therefore, the correct matching is $a-i, b-ii, c-iii, d-iv$.

(A) The energy of a photon is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
Since $E \propto \frac{1}{\lambda}$,radiations with shorter wavelengths have higher energy.
The order of wavelengths for the given radiations is: $\lambda_{\text{Radiowave}} > \lambda_{\text{Yellow}} > \lambda_{\text{Blue}} > \lambda_{\text{X-ray}}$.
Therefore,the order of increasing energy is: $E_{\text{Radiowave}} < E_{\text{Yellow}} < E_{\text{Blue}} < E_{\text{X-ray}}$.
This corresponds to the sequence $D, B, A, C$.

(D) The electromagnetic spectrum in increasing order of wavelength is: Gamma rays < $X$-rays < $UV$-rays < Visible light < $IR$ rays < Microwaves < Radio waves.
Evaluating the statements:
$(i)$ Microwaves have a longer wavelength than $UV$-rays. (True)
$(ii)$ $IR$ rays have a longer wavelength than $UV$-rays. (False)
$(iii)$ Microwaves have a longer wavelength than $IR$ rays. (False)
$(iv)$ Gamma rays have the shortest wavelength in the electromagnetic spectrum. (True)
Therefore,statements $(i)$ and $(iv)$ are correct.

(D) An oscillating charged particle produces electromagnetic waves of the same frequency as the oscillation frequency.
Given frequency $f = 10^9 \ Hz$.
Therefore,the frequency of the electromagnetic waves produced is $10^9 \ Hz$.
The wavelength $\lambda$ is given by the formula $\lambda = \frac{c}{f}$,where $c = 3 \times 10^8 \ m/s$ is the speed of light.
$\lambda = \frac{3 \times 10^8 \ m/s}{10^9 \ Hz} = 0.3 \ m$.
Electromagnetic waves with a frequency of $10^9 \ Hz$ (or $1 \ GHz$) fall within the microwave region,which is a part of the radio wave spectrum.

(D) $RADAR$ stands for $Radio$ $Detection$ $And$ $Ranging$.
It operates by transmitting electromagnetic waves,specifically microwaves,towards a target.
These waves reflect off the object and are detected by the radar receiver to determine the distance,speed,and direction of the object.
Therefore,the waves used in radar are microwaves.

(D) The electromagnetic spectrum in increasing order of wavelength is: Gamma rays < $X$-rays < $UV$ rays < Visible light < $IR$ rays < Microwaves < Radio waves.
Analyzing the statements:
$(A)$ Microwaves have a longer wavelength than $UV$ rays. This is true.
$(B)$ $IR$ rays have a longer wavelength than $UV$ rays. This is false.
$(C)$ Microwaves have a longer wavelength than $IR$ rays. This is false.
$(D)$ Gamma rays have the shortest wavelength in the electromagnetic spectrum. This is true.
Therefore,statements $(A)$ and $(D)$ are correct.

(B) The electromagnetic spectrum in descending order of frequency is: $X$-rays > Ultraviolet > Visible light (Red light) > Infrared > Microwaves > Radio waves.
Comparing the given options,the correct descending order of frequencies is: $X$-rays > Red light > Microwaves > Radio waves.
Therefore,the correct option is $B$.

(D) The correct matching is as follows:
$(1)$ $700\, nm$ to $1\, mm$ corresponds to Infrared rays,which are produced by the vibration of atoms and molecules $(i)$.
$(2)$ $1\, nm$ to $400\, nm$ corresponds to Ultraviolet rays,which are produced by inner shell electrons in atoms moving from one energy level to a lower level $(ii)$.
$(3)$ $< 10^{-3}\, nm$ corresponds to Gamma rays,which are produced by the radioactive decay of the nucleus $(iii)$.
$(4)$ $1\, mm$ to $0.1\, m$ corresponds to Microwaves,which are produced by a magnetron valve $(iv)$.
Therefore,the correct sequence is $(1)-(i), (2)-(ii), (3)-(iii), (4)-(iv)$.

(B) $1$. The sodium doublet ($D$-lines) has wavelengths of approximately $589.0 \; nm$ and $589.6 \; nm$,which falls in the visible spectrum. Thus,$I-A$.
$2$. The isotropic radiation filling all space (Cosmic Microwave Background Radiation) corresponds to a temperature of about $2.7 \; K$,which peaks in the microwave region. Thus,$II-B$.
$3$. The $21 \; cm$ line radiation emitted by atomic hydrogen in interstellar space is a radio wave. Thus,$III-C$.
$4$. The radiation arising from the Lamb shift (two close energy levels in hydrogen) corresponds to a wavelength of approximately $30 \; cm$,which is in the microwave region. Thus,$IV-B$.
Therefore,the correct matching is $I-A, II-B, III-C, IV-B$.

(A) The electromagnetic spectrum in increasing order of wavelength is:
Gamma rays $(G)$ < $X$-rays $(X)$ < Ultraviolet $(U)$ < Visible $(V)$ < Infrared $(I)$ < Microwaves $(M)$ < Radio waves $(R)$.
Therefore,the ascending order is $G, X, U, V, I, M, R$.

(A) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
Rearranging for wavelength: $\lambda = \frac{hc}{E}$.
Given $E = 11 \, keV = 11 \times 10^3 \times 1.6 \times 10^{-19} \, J$.
Using $hc = 12400 \, eV \cdot \mathring{A}$,we have $\lambda = \frac{12400 \, eV \cdot \mathring{A}}{11000 \, eV} \approx 1.13 \, \mathring{A}$.
The wavelength range for $X-$rays is approximately $0.1 \, \mathring{A}$ to $100 \, \mathring{A}$.
Since $1.13 \, \mathring{A}$ falls within this range,the radiation belongs to the $X-$ray region.

(A) The electromagnetic spectrum is ordered by frequency. The frequency ranges are approximately as follows:
$1$. $\gamma$-rays $(b)$: $10^{19} \, Hz$ to $10^{24} \, Hz$
$2$. $X$-rays $(a)$: $10^{16} \, Hz$ to $10^{19} \, Hz$
$3$. Ultraviolet rays $(c)$: $10^{15} \, Hz$ to $10^{16} \, Hz$
Comparing these values,we find that the frequency of $\gamma$-rays $(b)$ is the highest,followed by $X$-rays $(a)$,and then ultraviolet rays $(c)$.
Therefore,$b > a > c$.
Looking at the options provided,$a < b$ and $b > c$ is the correct relationship.

(B) The electromagnetic spectrum is arranged in order of increasing wavelength or decreasing frequency. The order of frequency for electromagnetic waves is: $\gamma$-rays > $X$-rays > Ultraviolet $(UV)$ rays > Visible light > Infrared rays > Microwaves > Radio waves.
Given the options:
$(a) = \gamma$-rays
$(b) = X$-rays
$(c) = UV$-rays
Comparing their frequencies,we have: $\text{Frequency of } \gamma$-rays > $\text{Frequency of } X$-rays > $\text{Frequency of } UV$-rays.
Therefore,the correct order is $a > b > c$.

(B) The frequency given is $f = 1\, GHz = 10^9\, Hz$.
According to the electromagnetic spectrum classification:
- Radio waves range from frequencies less than $300\, GHz$ down to $3\, kHz$.
- Microwaves range from $300\, MHz$ to $300\, GHz$.
- Since $1\, GHz$ falls within the range of $300\, MHz$ to $300\, GHz$,it is classified as a microwave,which is a sub-category of radio waves.
- Among the given options,$1\, GHz$ belongs to the region of radio waves.

(C) Electromagnetic radiation is used for communication and the transmission of information. The waves commonly used in communication systems include radio waves,microwaves,infrared radiation,and visible light. Among the given options,microwaves are specifically used in modern telecommunication systems,such as mobile networks and satellite communication.

(C) Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field. $\beta -$ rays consist of a stream of fast-moving electrons (charged particles) emitted during radioactive decay. Since they are composed of matter particles with mass and charge,they are not electromagnetic waves.

(N/A) The energy of a photon is given by the formula:
$E = hv = \frac{hc}{\lambda}$
Where:
$h = \text{Planck's constant} = 6.63 \times 10^{-34} \text{ J s}$
$c = \text{Speed of light} = 3 \times 10^{8} \text{ m/s}$
$\lambda = \text{Wavelength of radiation}$
To convert energy from Joules to $eV$,we divide by $1.6 \times 10^{-19} \text{ J/eV}$.
$E (eV) = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda \times 1.6 \times 10^{-19}} \approx \frac{1.24 \times 10^{-6}}{\lambda} \text{ eV}$.

Part of Spectrum	Wavelength $\lambda$ $(m)$	Energy $E$ (eV)
Radio waves	$10^{3}$	$1.24 \times 10^{-9}$
Microwaves	$10^{-2}$	$1.24 \times 10^{-4}$
Infrared	$10^{-5}$	$0.124$
Visible	$0.5 \times 10^{-6}$	$2.48$
Ultraviolet	$10^{-8}$	$124$
$X$-rays	$10^{-10}$	$1.24 \times 10^{4}$
Gamma rays	$10^{-12}$	$1.24 \times 10^{6}$

The different scales of photon energies are related to the energy level transitions within the source. Higher energy photons (like Gamma rays) correspond to transitions between deep nuclear energy levels,while lower energy photons (like Radio waves) correspond to transitions between atomic or molecular energy levels.

$(a)$ Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum.
$(b)$ Radio waves; it belongs to the short wavelength end.
$(c)$ Temperature, $T = 2.7 \; K$. According to Wien's displacement law, the wavelength $\lambda_{m}$ corresponding to maximum intensity is given by $\lambda_{m} = \frac{b}{T} \approx \frac{0.29 \; cm \cdot K}{2.7 \; K} \approx 0.11 \; cm$. This wavelength corresponds to microwaves.
$(d)$ This is the yellow light of the visible spectrum.
$(e)$ Transition energy is given by $E = h\nu$. For $E = 14.4 \; keV = 14.4 \times 10^3 \times 1.6 \times 10^{-19} \; J$, the frequency $\nu = \frac{E}{h} \approx 3.5 \times 10^{18} \; Hz$. This corresponds to $X$-rays.

(N/A) Nature has endowed the human eye (retina) with the sensitivity to detect electromagnetic waves within a small range of the electromagnetic spectrum.
Electromagnetic radiation belonging to this region of the spectrum (wavelength of about $400 \ nm$ to $750 \ nm$) is called light.
It is mainly through light and the sense of vision that we know and interpret the world around us.

(B) Visible light is the portion of the electromagnetic spectrum that is visible to the human eye.
The range of wavelengths for visible light typically extends from approximately $380 \ nm$ to $750 \ nm$.
Below $380 \ nm$ lies the ultraviolet region,and above $750 \ nm$ lies the infrared region.
Therefore,the correct range is $380 \ nm$ to $750 \ nm$.

(N/A) When Maxwell predicted the existence of electromagnetic waves,only visible light waves were considered electromagnetic. There was no information regarding infrared and ultraviolet rays.
At the end of the $19^{th}$ century,$X$-rays and gamma rays were discovered.
At present,light waves,$X$-rays,gamma rays,radio waves,microwaves,ultraviolet waves,and infrared waves are all part of the electromagnetic spectrum.
When electromagnetic waves are classified based on frequency (or wavelength),it is called the electromagnetic spectrum. In this spectrum,the values of wavelengths are continuously distributed. There is no distinct boundary separating different types of waves.
This classification is based on how these waves are produced or how they can be detected.