Electromagnetic Spectrum Questions in English

(A) The electromagnetic spectrum in increasing order of frequency is: Radio waves < Microwaves < Infrared < Visible < Ultraviolet < $X-$rays < $\gamma-$rays.
Given frequencies are:
$p$ (frequency of $X-$rays)
$q$ (frequency of $\gamma-$rays)
$r$ (frequency of ultraviolet rays)
Comparing their positions in the spectrum:
$1$. $\gamma-$rays have the highest frequency,so $q > p$.
$2$. $X-$rays have a higher frequency than ultraviolet rays,so $p > r$.
Combining these,we get $q > p > r$.
Looking at the options provided,we evaluate the relations:
For $p < q$ and $q > r$,this is consistent with $q > p$ and $q > r$.
Thus,the correct relation is $p < q$ and $q > r$.

(A) The electromagnetic spectrum in increasing order of frequency (or decreasing order of wavelength) is: Radio waves > Microwaves > Infrared > Visible > Ultraviolet > $X$-rays > Gamma rays.
Given:
$\lambda_{1}$ = Short radio waves
$\lambda_{2}$ = $X$-rays
$\lambda_{3}$ = Ultraviolet waves
Comparing their positions in the spectrum,the wavelength order is: $\lambda_{1} > \lambda_{3} > \lambda_{2}$.
Therefore,the correct decreasing order is $\lambda_{1}, \lambda_{3}, \lambda_{2}$.

(D) Microwave ovens operate by using electromagnetic waves to excite water molecules in food. The standard frequency commonly used for domestic microwave ovens is $2.45 \text{ GHz}$. However,among the provided options,$0.915 \text{ GHz}$ is a frequency also allocated for industrial,scientific,and medical $(ISM)$ applications,including specific types of microwave heating. Given the choices,$0.915 \text{ GHz}$ is the correct answer.

(B) In $LASIK$ (Laser-Assisted In Situ Keratomileusis) eye surgery,an excimer laser is used to reshape the cornea. This laser emits $Ultraviolet$ radiation,specifically at a wavelength of $193 \ nm$. This high-energy radiation is capable of breaking molecular bonds in the corneal tissue,allowing for precise removal of tissue without causing thermal damage to surrounding areas.

(C) speed gun used in cricket matches operates on the principle of the Doppler effect. It emits electromagnetic waves,specifically microwaves,towards the moving ball. When these waves reflect off the ball,their frequency changes due to the motion of the ball. By measuring this frequency shift,the speed gun calculates the velocity of the ball. Therefore,microwaves are the correct type of electromagnetic wave used for this purpose.

(C) $TV$ waves,which are a part of the radio wave spectrum used for television broadcasting,typically operate in the frequency range of $54 MHz$ to $890 MHz$. This range covers both Very High Frequency $(VHF)$ and Ultra High Frequency $(UHF)$ bands used for television signals. Therefore,the correct option is $C$.

(A) To match the electromagnetic waves with their respective wavelength ranges,we refer to the standard electromagnetic spectrum:
$1$. Microwave: The wavelength range is approximately $0.1 \ m$ to $1 \ mm$ (corresponds to $c$).
$2$. Visible light: The wavelength range is approximately $700 \ nm$ to $400 \ nm$ (corresponds to $a$).
$3$. Ultraviolet: The wavelength range is approximately $400 \ nm$ to $1 \ nm$ (corresponds to $d$).
$4$. $X$-rays: The wavelength range is approximately $1 \ nm$ to $10^{-3} \ nm$ (corresponds to $b$).
Thus,the correct matching is $i-c, ii-a, iii-d, iv-b$.

(C) The electromagnetic spectrum is ordered by wavelength. The increasing order of wavelength for the given radiations is:
$\lambda_{\text{X-rays}} < \lambda_{\text{UV-rays}} < \lambda_{\text{IR-rays}} < \lambda_{\text{microwaves}}$
Comparing the given options,microwaves have the longest wavelength among the listed radiations.
Therefore,the correct option is $C$.

(C) Ultraviolet $(UV)$ rays are electromagnetic radiations with wavelengths shorter than visible light but longer than $X$-rays. Due to their high energy and ability to damage the $DNA$ of microorganisms, they are widely used in the food industry to sterilise milk and other liquids by killing bacteria and pathogens.

(D) Microwaves have a short wavelength and travel in straight lines. They do not bend significantly around corners of obstacles in their path. Due to this property,they are highly effective for detection and tracking,which is why they are extensively used in radar systems for remote sensing and navigation.

(B) The electromagnetic spectrum is arranged in order of increasing frequency and decreasing wavelength.
Starting from the longest wavelength to the shortest,the order is:
$1$. Radio waves
$2$. Microwaves
$3$. Infrared rays
$4$. Visible light
$5$. Ultraviolet rays
$6$. $X$-rays
$7$. Gamma rays
Since radio waves appear at the beginning of this sequence,they possess the longest wavelength among the given options.

(B) 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. It is specifically used to detect infrared radiations.

(A) The electromagnetic spectrum in order of increasing frequency is: Radio waves < Microwaves < Infrared < Visible light < Ultraviolet < $x$-rays < Gamma rays.
Comparing the given options:
Option $A$: Infrared < $x$-rays < Gamma rays. This follows the increasing order of frequency.
Option $B$: Gamma rays < Visible light < $x$-rays. This is incorrect as Gamma rays have the highest frequency.
Option $C$: Visible light < Infrared < Ultraviolet. This is incorrect as Infrared has a lower frequency than Visible light.
Option $D$: Visible light < Ultraviolet < Infrared. This is incorrect as Infrared has a lower frequency than Ultraviolet.
Therefore,the correct order is Infrared < $x$-rays < Gamma rays.

(C) The electromagnetic spectrum classifies waves based on their wavelength or frequency.
$X$-rays typically have a wavelength range from approximately $0.01 \,nm$ to $10 \,nm$.
Since $1 \,nm$ falls within this range, light of wavelength $1 \,nm$ belongs to the class of $X$-rays.

(B) The electromagnetic $(EM)$ spectrum is ordered by frequency and wavelength. As shown in the spectrum,frequency increases from left to right,while wavelength increases from right to left.
The order of increasing wavelength (ascending order) is:
Gamma rays < $X$-rays < Ultraviolet < Visible light < Infrared.
Therefore,the correct sequence is: Gamma ray,$X$-ray,ultraviolet,visible light,infrared.

(D) The energy of a photon is given by $E = h\nu = \frac{hc}{\lambda}$.
Given $E = 14.4 \text{ keV} = 14.4 \times 10^3 \times 1.6 \times 10^{-19} \text{ J} = 2.304 \times 10^{-15} \text{ J}$.
Using the relation $\lambda = \frac{hc}{E}$,where $hc \approx 1240 \text{ eV} \cdot \text{nm} = 1.24 \times 10^{-6} \text{ eV} \cdot \text{m}$.
$\lambda = \frac{1240 \text{ eV} \cdot \text{nm}}{14.4 \times 10^3 \text{ eV}} \approx 0.086 \text{ nm} = 0.86 \times 10^{-10} \text{ m}$.
Since the wavelength range for $X$-rays is approximately $10^{-8} \text{ m}$ to $10^{-13} \text{ m}$,this radiation falls into the $X$-ray region.

(C) Microwaves have a short wavelength,which allows them to be transmitted in straight lines with minimal diffraction. Due to this property,they are highly effective for detecting objects and determining their distance,speed,and direction. Therefore,microwaves are extensively used in $Radar$ (Radio Detection and Ranging) systems.

(A) The ozone layer in the Earth's atmosphere plays a crucial role in protecting life by absorbing harmful ultraviolet $(UV)$ radiation from the Sun.
$UV$ radiation is characterized by wavelengths shorter than those of visible light.
Specifically,the ozone layer effectively absorbs $UV$ radiation with wavelengths less than $3 \times 10^{-7} \ m$ (or $300 \ nm$).

(D) The electromagnetic spectrum is ordered by wavelength. $X$-rays have the shortest wavelength among the given options,followed by ultraviolet light,then visible blue light,and finally visible red light.
Comparing the wavelengths: $\lambda_{\text{red}} > \lambda_{\text{blue}} > \lambda_{\text{UV}} > \lambda_{\text{X-ray}}$.
Therefore,visible red light has the longest wavelength.

(B) The electromagnetic spectrum in increasing order of wavelength is: Gamma rays < $X$-rays < Ultraviolet < Visible < Infrared < Microwaves < Radio waves.
$1$. Radioactive source produces Gamma rays $(\lambda \approx 10^{-12} \text{ m})$.
$2$. $X$-ray tube produces $X$-rays $(\lambda \approx 10^{-10} \text{ m})$.
$3$. Sodium lamp produces Visible light $(\lambda \approx 10^{-7} \text{ m})$.
$4$. Magnetron valve produces Microwaves $(\lambda \approx 10^{-3} \text{ m})$.
Therefore, the correct sequence in increasing order of wavelength is: Radioactive source $\rightarrow$ $X$-ray tube $\rightarrow$ Sodium lamp $\rightarrow$ Magnetron valve.

(C) Electromagnetic waves with a frequency of $6 GHz$ fall within the microwave frequency range ($1 GHz$ to $300 GHz$).
Due to their high frequency and ability to penetrate the ionosphere,these waves are primarily used for long-distance,line-of-sight communication.
Specifically,$6 GHz$ is a standard frequency band utilized in satellite communication systems for transponder operations.

(B) The correct matching is as follows:
$A$) $X$-rays are used for the study of crystal structure because their wavelength is comparable to the interatomic spacing in crystals. Thus,$A \rightarrow R$.
$B$) $UV$-rays are used in forensic laboratories for detecting finger prints and analyzing documents. Thus,$B \rightarrow Q$.
$C$) Radio waves are widely used in $TV$ and radio communication systems. Thus,$C \rightarrow S$.
$D$) $IR$-rays (Infrared rays) are used in remote switches for electronic devices like $TV$ remotes. Thus,$D \rightarrow P$.
Therefore,the correct sequence is $A \rightarrow R, B \rightarrow Q, C \rightarrow S, D \rightarrow P$.

(C) The correct matches are as follows:
$A$. The wavelength range $400 \ nm$ to $1 \ nm$ corresponds to $X$-rays,which are produced when inner shell electrons in atoms move from one energy level to a lower level. Thus,$A \rightarrow 4$.
$B$. The wavelength range $> 0.1 \ nm$ (covering radio waves and microwaves) is produced by the rapid acceleration and deceleration of electrons in aerials. Thus,$B \rightarrow 3$.
$C$. The wavelength range $1 \ mm$ to $700 \ nm$ (covering infrared radiation) is produced by the vibration of atoms and molecules in a substance. Thus,$C \rightarrow 2$.
$D$. The wavelength range $< 10^{-3} \ nm$ corresponds to gamma rays,which are produced by the radioactive decay of the nucleus. Thus,$D \rightarrow 1$.
Therefore,the correct match is $A \rightarrow 4, B \rightarrow 3, C \rightarrow 2, D \rightarrow 1$,which corresponds to option $C$.

(C) The energy of a photon is given by $E = h\nu = \frac{hc}{\lambda}$. Since the frequency $\nu$ of electromagnetic waves increases in the order: Visible light < $X$-rays < Gamma rays,the energy of the photons follows the same order.
$1$. For visible light,the energy range is approximately $1.6 \ eV$ to $3.2 \ eV$ $(E_{v} \approx 2.48 \ eV)$.
$2$. For $X$-rays,the energy range is approximately $100 \ eV$ to $100 \ keV$.
$3$. For gamma rays,the energy is typically greater than $100 \ keV$.
Comparing these ranges,we find that $E_{\gamma} > E_{X} > E_{v}$.

(C) Radio waves are produced by the rapid acceleration and deceleration of electrons in aerials.
Microwaves are produced by special vacuum tubes like klystrons, magnetrons, and Gunn diodes.
Infrared waves are produced by hot bodies and molecules due to changes in their vibrational and rotational modes.
$X$-rays are produced when high-energy electrons are suddenly stopped or when inner-shell electrons transition from a higher energy level to a lower energy level.
Therefore, the correct matching is: $A - IV, B - I, C - II, D - III$.

(B) Gamma rays possess high energy and high penetrating power. They are widely used in medical treatments such as radiotherapy to destroy or inhibit the growth of cancer cells by damaging their $DNA$.

(A) The electromagnetic spectrum is arranged in order of increasing frequency or decreasing wavelength.
The general order of wavelengths for these radiations is: $\lambda_{\text{microwave}} > \lambda_{\text{visible}} > \lambda_{\text{X-rays}}$.
Given that $\lambda_m$ represents the wavelength of microwaves,$\lambda_v$ represents the wavelength of visible rays,and $\lambda_x$ represents the wavelength of $X$-rays.
Therefore,the correct relation is $\lambda_m > \lambda_v > \lambda_x$.
Thus,the correct option is $A$.

(D) The power $P$ is given by $P = n E_{photon} = n \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.
Rearranging the formula to solve for wavelength $\lambda$,we get $\lambda = \frac{nhc}{P}$.
Substituting the given values: $n = 2.5 \times 10^{22} \text{ s}^{-1}$,$h = 6.6 \times 10^{-34} \text{ J}\cdot\text{s}$,$c = 3 \times 10^8 \text{ m/s}$,and $P = 15 \text{ kW} = 15 \times 10^3 \text{ W}$.
$\lambda = \frac{(2.5 \times 10^{22}) \times (6.6 \times 10^{-34}) \times (3 \times 10^8)}{15 \times 10^3} = \frac{49.5 \times 10^{-4}}{15 \times 10^3} = 3.3 \times 10^{-7} \text{ m}$.
Converting to nanometers: $3.3 \times 10^{-7} \text{ m} = 330 \text{ nm}$.
The wavelength range for the ultraviolet region is approximately $10 \text{ nm}$ to $400 \text{ nm}$.
Therefore,$330 \text{ nm}$ falls in the Ultraviolet region.

(D) The production mechanisms for electromagnetic waves are as follows:
$1$. Microwaves are produced by electronic devices like Klystron valves or magnetron valves $(A-IV)$.
$2$. Visible light is produced by the transition of electrons in atoms from a higher energy level to a lower energy level $(B-I)$.
$3$. Gamma rays are high-energy electromagnetic radiations produced during the radioactive decay of atomic nuclei $(C-II)$.
$4$. Infra-red rays are produced by the vibration of atoms and molecules $(D-III)$.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.