Physics

Electromagnetic Waves and Spectrum

670 Questions

Electromagnetic waves and spectrum questions test a candidate's understanding of radiation frequencies, wavelengths, and the properties of different rays like infrared, ultraviolet, and visible light. Concepts also cover practical applications in astronomy and the fundamental speed of light calculations. This physics topic appears regularly in general science sections of major competitive examinations.

UV rays propertiesElectromagnetic radiation frequencyWavelength identificationSpeed of light calculationsBlack body radiation

Electromagnetic Waves and Spectrum Questions

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The radiation emitted by a star $A$ is $1000$ times that of the sun. If the surface temperatures of the sun and star $A$ are $6000 K$ and $2000 K$, respectively, the ratio of the radii of the star $A$ and the Sun is:

  1. 300:1

  2. 600:1

  3. 900:1

  4. 1200:1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$E\propto A{ T }^{ 4 }$


$\displaystyle \frac { { E } _{ 1 } }{ { E } _{ 2 } } =\frac { 1000 }{ 1 } =\frac { \pi { { r } _{ 1 } }^{ 2 }\times { T }^{ 4 } }{ \pi { { r } _{ 2 } }^{ 2 }\times { T }^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 }\times { T }^{ 4 } }{ { { r } _{ 2 } }^{ 2 }\times { T }^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 }\times { 2000 }^{ 4 } }{ { { r } _{ 2 } }^{ 2 }\times 6000^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 } }{ { { r } _{ 2 } }^{ 2 }\times 81 } $

$\displaystyle \frac { { r } _{ 1 } }{ { r } _{ 2 } } =\sqrt { \frac { 1000 }{ 1 } \times \frac { 81 }{ 1 }  } =284.6:1$

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The amount of thermal radiations emitted from one square centimeter area of a black body in a second when at a temperature of 1000K

  1. 5.67 J

  2. 56.7 J

  3. 567 J

  4. 5670 J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Stefan's law $\Delta Q = \sigma ST^4\Delta t$
$\sigma=5.67E- 8W/m^2K^4;\, S=1E-4m^2;\, T=1000K; \, \Delta t=1s;$
subsutituting value in Stefan's law $\Delta Q=5.67J$
The value can be directly calculated by the Stefan's equation. After substituting the parameters sigma $= 5.67 E-8  W/m^2/K^4, A=10^-4 m^4,$ $T=1000K$ the value comes $5.67$ J
Thus, A is correct answer.
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The rate of radiation of a black body at $0^{\circ}C$ is $E$ J/s. Then the rate of radiation of this black body at $273^{\circ}C$ will be

  1. 16 E

  2. 8 E

  3. 4 E

  4. E
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From the stefan's law: ${E}={\sigma}{A}{T}^{4}$
So, ${E}={\sigma}{A}{273}^{4}$------(1)
Now, for second one ${E} _{1}={\sigma}{A}{546}^{4}$-----(2)
On dividing equation(2) by (1), we get
$\dfrac{{E} _{1}}{E}=\dfrac{{546}^{4}}{{273}^{4}}={16}$

Or, we can say ${E} _{1}={16}{E}$

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The rate of emission of a black body at temperature $27$$^{o}$C is $E _{1}$. If its temperature is increased to $327$$^{o}$C, the rate of emission of radiation is $E _{2}$. The relation between $E _{1} $ and $  E _{2}$ is :

  1. $E _{2}=24E _{1}$
  2. $E _{2}=16E _{1}$
  3. $E _{2}=8E _{1}$
  4. $E _{2}=4E _{1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know the relation for emissive power of body: $\dfrac { { E } _{ 1 } }{ { E } _{ 2 } } =\dfrac { { T } _{ 1 }^{ 4 } }{ { T } _{ 2 }^{ 4 } } $
${T} _{1}=27+273=300\ K$
${T} _{2}=327+273=600\ K$
So, putting all these data in above formulas, we get
${E} _{2}=16{E} _{1}$

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The wave length corresponding to maximum intensity of radiation emitted by a star is $289.8$nm. The intensity of radiation for the star is :

(Stefans constant $=$ 5.6x10$^{-8}Wm^{-2}K^{-4}$, Wien's displacement constant = $2898 \times 10^{-6} mK$ )

  1. 5.67 x 10$^{8}Wm^{-2}$
  2. 5.67 x 10$^{4}Wm^{-2}$
  3. 10.67 x 10$^{7}Wm^{-2}$
  4. 10.67 x 10$^{4}Wm^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, Wavelength corresponding to maximum radiation $=289.8nm$
Stefans constant$=5.6\times 10^{-8}Wm^{-2}K^{-4}$
From Wien's law,
$\lambda _{max} T$= constant (b)
$b=2898\times 10^{-6}$
$\lambda _{max}=289.8$
$\therefore$ $T=\frac{b}{\lambda _{max}}=\frac{2898\times 10^{-6}}{289.8\times 10^{-9}}=10^4 K$

Intensity of radiation E
$E=\sigma T^4=5.6\times 10^{-8}\times 10^{16}= 5.6\times 10^8 Wm^{-2}$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The power radiated by a black body is $P$ and it radiates maximum energy around the wavelength $\lambda  _{o}$ . If the temperature of the black body is now changed so that it radiates maximum energy around a wavelength $3\lambda  _{o}/4$ , the power radiated by it will increase by a factor of :

  1. $4/3$
  2. $16/9$
  3. $64/27$
  4. $256/81$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
From Wein's displacement law, we know that ${\lambda} _{max}{T}={CONSTANT}$

So, let us assume that initial temperature of the body be ${T} _{1}$
and by changing the temperature from ${T} _{1}   \ to \  {T}$, wavelength will change from $\lambda$ to $\dfrac{3{\lambda}}{4}$,so applying the above relation,

${\lambda} _{0}{T} _{1}=\dfrac{3{\lambda} _{0}}{4}{T}$

Hence, ${T}=\dfrac{4}{3}T _1$

and we know that ${P}={e}{{T}^{4}}$

So, ${P'}=\dfrac{256}{81}P$
Multiple choice stefan's law black body radiation heat transfer thermal properties physics

The emissive power of a sphere of radius $5$cm coated with lamp black is $1500$Wm$^{-2}$. The amount of energy radiated per second is.

  1. 15.7 J

  2. 3.14 J

  3. 47.10 J

  4. 4.71 J
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that emissive power is given as: ${E}={\sigma}{T}^{4}$
${T}^{4}=\dfrac{E}{\sigma}$
${T}=\dfrac{1500}{5.67\times{10}^{-8}}=403.29 K$
Now, we know that radiation from the surface is given as $E=\sigma \epsilon A{ T }^{ 4 }$
${E}={5.67\times{10}^{-8}\times4\times{\pi}\times{0.05}^{2}\times{403.29}^{4}}=47.15 J $

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

A black body emits maximum radiation of wavelength $\displaystyle \lambda _{1}=2000A $ at a certain temperature $\displaystyle T _{1} $ On increasing the temperature the total energy of radiation emitted is increased $16$ times at temperature $\displaystyle T _{2} $ If $\displaystyle \lambda _{2} $ is the wavelength corresponding to which maximum radiation emitted at temperature  $\displaystyle T _{2} $ Calculate the value of $\displaystyle \left ( \frac{\lambda _{1}}{\lambda _{2}} \right ) $

  1. $2:1$
  2. $1:2$
  3. $3:4$
  4. $4:3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The rate of radiation, emitted by per unit area of a body at temperature $T$ , is given by Stefan-Boltzmann's law as,

   $E\propto T^{4}$ ,
hence , $E _{1}/E _{2}=(T _{1}/T _{2})^{4}$ ,
given , $E _{2}=16E _{1}$ ,
therefore ,
            $E _{1}/16E _{1}=(T _{1}/T _{2})^{4}$ ,
or         $T _{2}/T _{1}=2$ ,
now by Wein's law ,
            $\lambda _{m}\propto 1/T$ ,
hence  $\lambda _{m1}/\lambda _{m2}=T _{2}/T _{1}$ ,
therefore ,  $\lambda _{m1}/\lambda _{m2}=T _{2}/T _{1}=2$ ,
               

Multiple choice physics electromagnetic waves properties and uses of different radiation electromagnetic spectrum spectrum

If $v _{g}, v _{x}$, and $v _{m}$ are speeds of gamma rays, $X$ rays, and microwaves respectively in vacuum, then:

  1. $V _{g} < V _{x} < V _{m}$
  2. $V _{g}>V _{x}>V _{m}$
  3. $V _{g} < V _{x}> V _{m}$
  4. $V _{g}=V _{x}=V _{m}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The gamma rays, $X-$ rays and the microwaves are the Electromagnetic waves and the speed of all the electromagnetic waves remains the same in vacuum irrespective of other parameters. All will travel at the speed of light.

$V _{g}=V _{x}=V _{m}$ 

 The speed of Electromagnetic waves is always the same in vacuum. So, option $(D)$ is correct.

Multiple choice physics electromagnetic waves properties and uses of different radiation electromagnetic spectrum spectrum

A. wavelength of microwave is greater than that of ultraviolet rays.
B. The wave length of infrared rays is lesser than that of ultraviolet rays.
C. wavelength of microwave is lesser than that of ultraviolet rays
D. Gamma rays has shortest wavelength in the electromagnetic spectrum.
Choose the correct option from the given options.

  1. A and B are true

  2. B and C are true

  3. C and D are true

  4. A and D are true
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The wavelength and the frequency of the Electromagnetic waves are inversely proportional to each other.

$A$. The wavelength of Microwave is greater than UV rays.
The frequency of the microwave is less than that of the Ultraviolet rays, the wavelength of the Microwaves will be longer than the Ultraviolter rays.

$D$. Gamma rays have the shortest wavelength in the EM spectrum.
The frequency of the gamma rays is the highest in the electromagnetic spectrum. So, it will have the shortest wavelength.

The wavelength of the infrared is higher than the Ultraviolet rays and also the wavelength of the microwave is higher than Ultraviolet rays. So, options $B$ and $C$ are not true.
Option $(D)$ is correct.
Multiple choice physics electromagnetic waves properties and uses of different radiation electromagnetic spectrum spectrum

A potential difference of 0.1kV is applied across an X-ray tube. The ratio of the de-Broglie wave lengths of incident electron, striking the target to the shortest wavelength of X-rays produced nearly (e/m of electron= $1.8\times 10^{11}C/kg$) ?

  1. 1:1

  2. 1:10

  3. 1:100

  4. 1:1000
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The de-Broglie wavelength of an electron is lambda = h / sqrt(2meV). The shortest X-ray wavelength is lambda_min = hc / eV. The ratio lambda / lambda_min = (h / sqrt(2meV)) * (eV / hc) = sqrt(eV / 2mc^2). Plugging in values leads to approximately 1:1000.

Multiple choice physics electromagnetic waves properties and uses of different radiation electromagnetic spectrum spectrum

A wave has a wavelength of $0.01A^o$. Name the wave:

  1. X rays

  2. Micro waves

  3. Gamma waves

  4. Radio waves
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electromagnetic radiation is classified into types according to the frequency of the wave. These types include, in order of increasing frequency, radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.
The frequency is the inverse of wavelength. Hence, in the order of increasing wavelength, the waves are Gamma rays ($<1nm$), X-rays ($1-{ 10 }nm$), infra red rays ($700-{ 10 }^{ 5 }nm$), micro waves (${ 10 }^{ 5 }-{ 10 }^{ 8 }nm$), radio waves (${ >10 }^{ 8 }nm$).
Therefore, gamma rays has a wavelength of $0.01 \mathring{A}$.

Multiple choice physics electromagnetic waves properties and uses of different radiation electromagnetic spectrum spectrum

Name the high energetic invisible electromagnetic wave which helps in the study of structure of crystals.

  1. $X$-rays

  2. Cathod rays

  3. Gamma rays

  4. All of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electromagnetic wave which helps in the study of structure of crystals is X-rays.
If a beam of x rays is passed through a crystal, it makes a pattern on a photographic plate. The pattern is called crystallogram (the arrangement of the atoms each crystal has its own crystallogram). It reveals the crystal`s internal structure. 

Multiple choice physics electromagnetic waves properties and uses of different radiation electromagnetic spectrum spectrum

A wave has wavelength $50A^o$. Name the wave.

  1. X-ray

  2. Gamma ray

  3. Radio waves

  4. Micro waves
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electromagnetic radiation is classified into types according to the frequency of the wave. These types include, in order of increasing frequency, radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.
The frequency is the inverse of wavelength. Hence, in the order of increasing wavelength, the waves are Gamma rays ($<1nm$), X-rays ($1-10nm$), infra red rays ($700-{ 10 }^{ 5 }nm$), micro waves (${ 10 }^{ 5 }-{ 10 }^{ 8 }nm$), radio waves (${ >10 }^{ 8 }nm$).
Therefore, X-rays has a wavelength of $50 \mathring{A}$.

Multiple choice physics electromagnetic waves properties and uses of different radiation electromagnetic spectrum spectrum

Name the rays or waves of wavelength nearly $0.1nm$.

  1. IR rays

  2. UV rays

  3. Gamma rays

  4. X-rays
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

X-rays is a form of electromagnetic radiation. Most X-rays have a wavelength ranging from $0.01$ to $10$ nanometers, corresponding to frequencies in the range 30 petahertz to 30 exahertz (${ { 3\times 10 }^{ 16 }\ Hz\ to\ 3\times 10 }^{ 19 }Hz$) and energies in the range 100 eV to 100 keV.
Hence, the rays or waves of $0.1 nm$ is X-rays.