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 energy in wave motion oscillation and waves waves physics

The maximum potential energy / length increases with:

  1. Amplitude

  2. Wavelength

  3. Frequency

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

If $y= A \sin (\omega t- kx)$ is the equation of a wave through a string, then the slope of the wave is $\dfrac{dy}{dx}=- Ak \cos (\omega t- kx)$. 

The maximum potential energy will be $T \times \Delta x \times (\dfrac{dy}{dx})^2A^2k^2=A^2k^2 \cos ^2(\omega t -  kx)T \Delta x$

The maximum potential energy will be obtained if cos (\omega t - kx)=1. Thus, maximum potential energy = $4 \pi^2A^2T f \times T \times \Delta x $; T is the tension in the string

We also know that $T=\mu v^2$. Substituting, we get, 

Maximum potential energy = $4 \pi^2 f^2A^2 \mu$
Thus maximum potential energy depends on frequency and as frequency increases, potential energy also increases

The correct option is (c)

Multiple choice physics kirchhoff's law circuit problems kirchoff's law and problems on it equivalent resistance in series and parallel connection

"A good absorber of a given wavelength of radiation is also a good emitter of that wavelength." This is a statement of:

  1. Stefan-Boltzmann's law

  2. Wien's Law

  3. Kirchoff's Law

  4. The First Law of Thermodynamics
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Kirchhoff's Law of radiation states that for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.

Multiple choice physics dual nature of matter and radiation davisson and germer experiment and its conclusion matter waves wave nature of matter

The wavelength of $L _\alpha$ line in $X-ray$ spectrum of $Pt^{78}$ is $1.32\mathring { A } $ then wavelength of $L _\alpha $ line $X-ray$ spectrum of another unknown element is $4.17\mathring { A } .$ If screening constant for $L _\alpha $ line is $7.4,$ then atomic number of the unknown element is -

  1. $78$
  2. $47$
  3. $40$
  4. $35$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Moseley's Law for X-rays: 1/lambda = R * (Z-sigma)^2 * (1/n1^2 - 1/n2^2). Since the transition is the same (L_alpha), (Z1-sigma)^2 * lambda1 = (Z2-sigma)^2 * lambda2. Plugging in Z1=78, sigma=7.4, lambda1=1.32, lambda2=4.17, we solve for Z2.

Multiple choice relativistic mechanics option a: relativity physics

Radiation with energy that is easily detected as quanta _______________.

  1. $1$ eV
  2. $1$ KeV
  3. $1$ MeV
  4. $10^{-10}$ eV
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Radiation with energy in the MeV range (gamma rays) is typically detected as discrete quanta using devices like scintillation counters. Lower energies are often treated as waves or are harder to detect as individual quanta in standard laboratory settings.

Multiple choice perceive colours resolution of optical instruments lenses option c: imaging physics

The dispersion of a medium for wavelength $\lambda$ is D. Then the dispersion for the wavelength $2\lambda$ will be:

  1. $(D/8)$
  2. $(D/4)$
  3. $(D/2)$

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

As we know, Cauchy's Dispersion formula is :
$ \mu $= $A + \dfrac{B}{ \lambda^{^{2}}} $
And dispersion
D= - $ \dfrac {d\mu}{d\lambda}$
Therefore, from the above 2 equations:
D = $ -(-2\lambda ^{3})B $= $ \dfrac {2B}{\lambda^{3}}$
This implies that
D $ \alpha \dfrac{1}{\lambda^{3}}$
Hence,
$ \dfrac {{D}'}{D}$ = $( \dfrac{\lambda}{{\lambda}' })^{3}$
As $ {\lambda}' = 2\lambda$
Therefore,
$ {D}' =D/8$

Multiple choice physics bharat and science scope and progress of physics recent advancement in medical technology contribution of physics in technology and society

The Giant Metrewave Radio Telescope (GMRT) is related to _________________.

  1. Superconductivity

  2. Generation, propagation and detection of electromagnetic waves

  3. Detection of cosmic radio waves

  4. Trapping and cooling of atoms by laser beams and magnetic fields
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Giant Metrewave Radio Telescope (GMRT) is related to Detection of cosmic radio waves.

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

A beam of ultraviolet radiation having wavelength between $100nm$ and $200nm$ is incident on a sample of atomic hydrogen gas. Assuming that the atoms are in ground state, which wavelengths will have low intensity in the transmitted beam? 

  1. $104nm$
  2. $103nm$
  3. $105nm$
  4. $100nm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy corresponding to wavelength 100nm, $\dfrac{1242eV}{100}=12.42 eV$

Energy corresponding to wavelength 200nm, $\dfrac{1242eV}{200}=6.21 eV$
Energy required from ground state to first excited stage:
$E _2-E _1=13.6-3.4=10.2 eV$
Energy required from ground state to second excited stage:
$E _3-E _1=13.6-13.6-1.5=12.1  eV$

Energy required from ground state to third excited stage:
$E _3-E _1=13.6-0.85=12.75  eV$

At 10.2 eV,
$Wavelength 1=\dfrac{1242}{10.2}=122nm$

At 12.1 eV,
$Wavelength 2=\dfrac{1242}{12.1}=103nm$



Multiple choice gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

A gamma ray photon creates an electron-positron pair. If the rest mass energy of an electron is $0.5MeV$ and the total kinetic energy of the electron-positron pair is $0.78 MeV$, then the energy of the gamma ray photon must be

  1. $0.78MeV$
  2. $1.78MeV$
  3. $1.28MeV$
  4. $0.28MeV$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy of $\gamma $-rays photon $=$ Rest mass energy $+$ $K.E$ 

                                         $=2\left( 0.5 \right) +0.78\ = 1.78\ MeV$

Multiple choice gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

The wavelength of emitted $\gamma $ rays are in the other

  1. ${ \lambda } _{ \gamma 2 }\quad >\quad { \lambda } _{ \gamma 3 }\quad >{ \quad \lambda } _{ \gamma 1 }$
  2. ${ \lambda } _{ \gamma 3 }\quad >\quad { \lambda } _{ \gamma 2 }\quad >{ \quad \lambda } _{ \gamma 1 }$
  3. ${ \lambda } _{ \gamma 1 }\quad >\quad { \lambda } _{ \gamma 2 }\quad >{ \quad \lambda } _{ \gamma 3 }$
  4. ${ \lambda } _{ \gamma 3 }\quad >\quad { \lambda } _{ \gamma 1 }\quad >{ \quad \lambda } _{ \gamma 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As the frequencies of $\gamma-rays$ are in the order:

$\nu _{\gamma 1} > \nu _{\gamma 3} > \nu _{\gamma 2}$
Thus as wavelength is inversely proportional to frequency.
$\lambda _{\gamma 2} > \lambda _{\gamma 3} > \lambda _{\gamma 1}$
Hence option A is correct

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

Kirchoffs law states that

  1. a body absorbs radiation of shorter wavelengths and emits radiation of higher wavelength

  2. a body absorbs radiation of any wavelength but emits radiation of specific wavelengths

  3. a body absorbs and emits radiation of same wavelengths

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In thermodynamics, Kirchhoff's law of thermal radiation refers to wavelength-specific radiative emission and absorption by a material body in thermodynamic equilibrium, including radiative exchange equilibrium.

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

Certain substance emit only the wavelength $\lambda  _{1},\lambda  _{2},\lambda  _{3}  \ and  \ \lambda  _{4}$ when it is at a high temperature. When this substance is at a colder temperature, it will absorb only the following wavelength :

  1. $\lambda _{1}$
  2. $\lambda _{2}$
  3. $\lambda _{1}$ and $\lambda _{2} $
  4. $\lambda _{1},\lambda _{2},\lambda _{3}$ and $\lambda _{4} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

With the help of kirchhoff's law  we can say that $\varepsilon =\alpha $
${\varepsilon}= emissivity $
$\alpha=absorptivity $
so kirchhoff's law state that total emissivity of body is equal to total absorptivity.
so ${\lambda} _{1},{\lambda} _{2}, {\lambda} _{3}, {\lambda} _{4}$ will be absorbed.  

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

In a dark room with ambient temperature $T _o$, a black body is kept at a temperature $T$. Keeping the temperature of the black body constant (at $T$), sunrays are allowed to fall on the black body through a hole in the roof of the dark room. Assuming that there is no change in the ambient temperature of the room, which of the following statement(s) is/are correct?

  1. The quantity of radiation absorbed by the black body in unit time will increase

  2. Since emissivity = absorptivity, hence the quantity of radiation emitted by black body in unit time will

    increase

  3. Black body radiates more energy in unit time in the visible spectrum.

  4. The reflected energy in unit time by the black body remains same.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Since the radiation is continuously falling on the black body, the quantity radiation absorbed per second will increase.

The reason given in question is self explanatory.
With an increase in temperature, the entire Plank's curve shifts upwards and hence radiation in any spectrum will increase.
Reflected energy per unit time will be zero since black body has 0 reflectivity and hence it will remain constant.