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 physics oscillations and waves huygen's wave theory refraction of water waves reflection and refraction at plane surfaces theories on light

What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen spectrum. (take hc = 1240 eV nm)

  1. 102 nm

  2. 150 nm

  3. 82 nm

  4. 122 nm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Find out $\lambda$ of light

$hc=1240eVmm$
Lymen series of the hydrogen spectrum
Solution
For any series the transistion that produces the least energetic photon is the transition between the home base level that defines the series and the level immediately above it. For the hymen series, the home-base level is at $n=1$
So the transition that produces the least energetic photon is the transition from the $n=2$ level to then $n=1$ level.
$\therefore$ The wavelength for the least energetic photon is
$\lambda=\cfrac{hc}{E _2-E _1}$
Here $hc=1240eVnm$
$E _1=-\cfrac{13.6}{1^2}eV\ \quad=-13.6eV$
(as $E _n=-\cfrac{13.6}{n^2}eV)$
$E _2=-\cfrac{13.6}{2^2}eV\ \quad=-3.4eV\ \therefore\lambda=\cfrac{1240eVnm}{-3.4eV-(-13.6eV)}\ \quad=\cfrac{1240eVnm}{102eV}\ \quad=122nm$

Multiple choice physics energy production

Ozone layer blocks the radiations of wave length

  1. less than $3 \times 10^{-7}$ m
  2. equal to $3 \times 10^{-7}$ m
  3. more than $3 \times 10^{-7}$ m

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

The ozone layer absorbs harmful ultraviolet radiation, specifically UV-C and some UV-B, which have wavelengths shorter than approximately 300 nm (3 * 10^-7 m).

Multiple choice

How do scientists use the electromagnetic spectrum?

  1. by sending radio waves and reading their response

  2. they use to communicate with aliens

  3. by examining visible light rays

  4. they do not... they use light years
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Scientists use the electromagnetic spectrum to study objects in space by detecting the various wavelengths of radiation they emit, such as radio waves, which can penetrate dust clouds.

Multiple choice

Fluence refers to the . . .

  1. Power density

  2. irradiance multiplied by exposure time. (j/cm2)

  3. Microwave amplification by stimulated emission of radiation

  4. Level of laser radiation to which a person may be exposed
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Fluence is defined as the total energy delivered per unit area, calculated as irradiance (power density) multiplied by the exposure time.

Multiple choice evs - i communication and mass media understanding communication and impact of mass media satellites in communication importance of transport system artificial satellite

A ground receiver station is receiving a signal at (a) 5 MHz and (b) 100MHz, transmitted from a ground transmitter at a height of 400m located at a distance of 125km. Identify whether it is coming via space wave or sky wave propagation or satellite transponder. Radius of earth = $6.4\times { 10 }^{ 6 }m$; maximum number density of electrons in ionosphere = ${ 10 }^{ 12 }{ m }^{ -3 }$ 


  1. $9MHz$
  2. $5MHz$
  3. $4MHz$

  4. All of these.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Height of the ground transmitter $=30\ cm$
Range of transmission $=100\ km$
maximum distance covered by space wave propagation is
$d=\sqrt{2\ Rh}$
$=\sqrt{2\times 64\times 10^{6}\times 300}$
$=62\ km$
critical frequency for ionospheric propagation is, 
$F _{C}=9(N _{max})^{112}$
$=9(10^{12})^{112}$
$=9\times 10^{6}=9\ MHZ$
Multiple choice botany photosynthesis electron transport system modern concept of photosynthesis photosynthesis in higher plants
Which range of wavelength (in nm) is called photosynthetically active radiation (PAR)? 

  1. 100-390

  2. 390-430

  3. 400-700

  4. 760-10,000
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Sunlight or solar radiations reaching the earth have wavelength between 300 nm (in the ultra volt rang) to 2600 nm (in the infa-red range). Part of the spectrum used in photosynthesis has a wavelength between 400-700 nm, it is called as photo synthetically active radiation (PAR). 

Multiple choice modelling gases - the kinetic model ideal gases kinetic theory of gases physics

Solar radiation reaches the earths atmosphere at a rate of $1353 Wm^{-2}$. If 36% of this
radiation is reflected back into space and 18% is absorbed by the earths atmosphere. The
radiant emittance is given by $\sigma T^{4}$
 where $\sigma$ is the Stefan-Boltzmanns constant and T is the
absolute temperature. What maximum temperature would an isolated black body on the
earths surface be expected to attain?
$(\sigma  = 5.67 x10^{-8} Wm^{-2}K^{-4})$.

  1. $120^{o}C$
  2. $63.9^{o}C$
  3. $50.7^{o}C$
  4. $31.4^{o}C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Of the solar radiation reaching the earths atmosphere
36% is reflected back into space
18% is absorbed by the earths atmosphere
This implies that only 46% reaches the earths surface
If absorbed by a black body, expect that the maximum temperature of this body is given by
where $I _{sc} = 1353 Wm^{-2}$

Multiple choice botany photosynthesis action spectrum and absorption spectrum spectrum of electromagnetic radiation chloroplast and pigments of photosynthesis site of photosynthesis

Which of the is the range for UV in spectroscopy?

  1. 400-700nm

  2. 500-800nm

  3. 200-300nm

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

UV spectroscopy (400-700nm) is a type of absorption spectroscopy in which light of ultra-violet region is absorbed by the molecule resulting in the excitation of the electrons from the ground state to higher energy state. It obeys the Beer-Lambert law that states that when a beam of monochromatic light passes through a solution of an absorbing substance, the rate of decrease of intensity of radiation with the thickness of the absorbing solution is directly proportional to the incident radiation as well as the concentration of the solution.

So, the correct answer is '400-700nm'

Multiple choice botany photosynthesis action spectrum and absorption spectrum spectrum of electromagnetic radiation chloroplast and pigments of photosynthesis

Electromagnetic radiation with wavelengths between 0.4 and 0.7 micrometers is called:

  1. Ultraviolet light

  2. Visible light

  3. Infrared light

  4. Microwaves
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The visible spectrum of electromagnetic radiation ranges from approximately 0.4 micrometers (violet) to 0.7 micrometers (red).

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

Wavelength of first line in Lyman series is $\lambda $. The wavelength of first line in Balmer series is:

  1. $\dfrac { 5 }{ 27 } \lambda $
  2. $\dfrac { 32}{ 27 } \lambda $
  3. $\dfrac { 27 }{ 5 } \lambda $
  4. $\dfrac { 27 }{ 32 } \lambda $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Bohr, the wavelength emitted when an electron jumps from ${ n } _{ 1 }^{ th }$ to ${ n } _{ 2 }^{ th }$ orbit is
$E=\dfrac { hc }{ \lambda  } ={ E } _{ 2 }-{ E } _{ 1 }$
$\Rightarrow \dfrac { 1 }{ \lambda  } =R\left( \dfrac { 1 }{ { n } _{ 1 }^{ 2 } } -\dfrac { 1 }{ { n } _{ 2 }^{ 2 } }  \right) $
For first line in Lyman series
$\dfrac { 1 }{ { \lambda  } _{ L } } =R\left( \dfrac { 1 }{ { 1 }^{ 2 } } -\dfrac { 1 }{ { 2 }^{ 2 } }  \right) =\dfrac { 3R }{ 4 } $                        ......(i)
For first line in Balmer series,
$\dfrac { 1 }{ { \lambda  } _{ B } } =R\left( \dfrac { 1 }{ { 2 }^{ 2 } } -\dfrac { 1 }{ { 3 }^{ 2 } }  \right) =\dfrac { 5R }{ 36 } $                      ......(ii)
From equations (i) and (ii)
$\therefore \dfrac { { \lambda  } _{ B } }{ { \lambda  } _{ L } } =\dfrac { 3R }{ 4 } \times \dfrac { 36 }{ 5R } =\dfrac { 27 }{ 5 } $
$\therefore { \lambda  } _{ B }=\dfrac { 27 }{ 5 } \lambda $               $\left( \because { \lambda  } _{ L }=\lambda  \right) $

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

If photon energy $E$ and an electron have same energy $E$ (kinetic energy) and De-Broglie wavelength of an electron is $\lambda _{e}$ and De-Broglie wavelength of photon is $\lambda _{p}$. The correct relation between $\lambda _{e}$ and $\lambda _{p}$ is 

  1. $\lambda _{p} \propto \lambda _{e}$
  2. $\lambda _{p} \propto \sqrt{\lambda _{e}}$
  3. $\lambda _{p} \propto \dfrac{1}{\sqrt{\lambda _{e}}}$
  4. $\lambda _{p} \propto \lambda _{e}^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For photon

$E = \dfrac{{hc}}{{{\lambda _p}}}$---------------$(1)$
$ \Rightarrow {\lambda _p} = \dfrac{{hc}}{E}$

$E = \dfrac{1}{2}m{v^2}$

$ \Rightarrow v = \sqrt {\dfrac{{2E}}{m}} $

${\lambda _e} = \dfrac{h}{{mv}} = \dfrac{h}{{n\sqrt {\dfrac{{2E}}{m}} }} = \frac{h}{{\sqrt {2Em} }}$

$ \Rightarrow {\lambda _e}^2 = \dfrac{{{h^2}}}{{2Em}}$

$ \Rightarrow E = \dfrac{{{h^2}}}{{2m{\lambda _e}^2}}$
$\therefore \dfrac{{{h _e}}}{{{\lambda _p}}} = \dfrac{{{h^2}}}{{2m{\lambda _e}^2}}$

$ \Rightarrow {\lambda _p}\alpha \,{\lambda _e}^2$
Hence,
option $(D)$ is correct answer.

Multiple choice physics light and shadow formation of image by a pinhole camera pinhole camera shadow

Which one statement is correct -

  1. Speed of light in free space $= \frac{1}{\sqrt{\mu _0 \varepsilon _0}}$
  2. Speed of light in any medium $=\frac{1}{\sqrt{\mu \varepsilon}}$
  3. $\frac{E _0}{B _0}=2C$
  4. $\frac{B _0}{E _0}=4C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The speed of light in free space is given by c = 1 / sqrt(mu_0 * epsilon_0), which is a fundamental result of Maxwell's equations.