Physics

Electromagnetic Waves and Spectrum

659 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 detection and recording of x-ray images option c: imaging physics

A photon collides with an electron and gets scattered through an angle of $90^{0}$. The electron recoils and moves in another direction. The compton wavelength is $(h=6.62 \times 10^{-34}Js.)$

  1. $0.121\times 10^{-11}m$
  2. $0.486\times 10^{-11}m$
  3. $2.4\times 10^{-11}m$
  4. $0.243\times 10^{-11}m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Compton wavalength $=\dfrac{h}{m _{e}c}(1-cos\theta )$


                             $=\dfrac{6.62\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}(1-cos\ 90^{0})$

                             $=0.243\times 10^{-11}(1-0)\ ( \because cos\ 90^{0}=0)$
                            $=0.243\times 10^{-11}m$
So, the answer is option (D).

Multiple choice detection and recording of x-ray images option c: imaging physics

The wavelength of scattered radiation when it undergoes compton scattering at an angle of $60^o$ by graphite is $2.54 \times 10^{-11}$m, then the wavelength of incident photon is

  1. $4.2\times 10^{-11}m$
  2. $1.12\times 10^{-11}m$
  3. $1.21\times 10^{-11}m$
  4. $2.42\times 10^{-11}m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Compton formula
$\Delta \lambda = \lambda _f-\lambda _i= \dfrac{h}{m _ec}  (1- cos  \theta )$


$2.54\times 10^{-11}-\lambda _1= \dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^8}  \big(1- cos  60^{\circ }\big)$

$\lambda _i= 2.54\times 10^{-11}
- \dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^8}  \big(1- 1/2 \big) \ \ \ \  \big(\because cos  60^{\circ} =  1/2\big)$

$= 2.54\times 10^{-11}-0.12\times 10^{-11}$
$=2.42\times 10^{-11}m$
So, the answer is option (D).

Multiple choice detection and recording of x-ray images option c: imaging physics

In a Compton effect experiment, the wavelength of incident photons is 3$A^{0}$.If the incident radiation is scattered through $60^{0}$ , the wavelength of scattered radiation is nearly (given$h=6.62\times 10^{-34}Js$, $m _{o} = 9.1 \times 0^{-31}$ kg, $c = 3 \times 10^{8}$ m/s)

  1. 3.024 $A^{0}$
  2. 3.012 $A^{0}$
  3. 3.048$A^{0}$
  4. 2.988 $A^{0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${\lambda }'-\lambda =\dfrac{h}{m _{e}c}(1-cos\theta)$


${\lambda }'=\lambda +\dfrac{6.62\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}(1-cos60^{\bullet})$

$=\lambda +0.0121\ A^{0}$
$=3A^{\bullet}+0.0121$
$=3.012\ A^{\bullet }$
So, the answer is option (B).

Multiple choice detection and recording of x-ray images option c: imaging physics

The maximum increase in X-ray wavelength that can occur during Compton scattering is

  1. $5.84\times 10^{-12}m$
  2. $6.84\times 10^{-3}m$
  3. $7.84\times 10^{-10}m$
  4. $4.84\times 10^{-12}m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know compton formula is
$\Delta\lambda =\ \lambda _{f}-\lambda _{i}   =  \dfrac{h}{m _{e}C}  (1-cos\theta )$

For maximum increase $ cos\theta =   -1$

so $\Delta \lambda =  \dfrac{h}{m _{e}C} (1-(-1))$

$=  \dfrac{2h}{m _{e}C}$

$=  \dfrac{2\times 6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}$

$=  0.484 \times 10^{-11}m$
$=  4.84 \times 10^{-12}m$

So, the answer is option (D).

Multiple choice detection and recording of x-ray images option c: imaging physics

X-rays of 1.0$A^{0}$ are scattered from a carbon block. The wavelength of the scattered beam in a direction making $90^{0}$ with the incident beam is

  1. 1.024$A^{0}$
  2. 2.024$A^{0}$
  3. 3.024$A^{0}$
  4. 4.024$A^{0}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Compton effect formula

$\Delta \lambda = \lambda _{f}-\lambda _{i} = \dfrac{h}{m _{e}C}   (1-cos \theta)$

$\lambda _{f}-1.0 A^{\circ} = \dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}\ (1-cos  90^{\circ})$

$\lambda _{f} = 1  A^{\circ}+ \dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}  (1-0)$             $(\because  cos  90^{\circ}=0)$

$= 1  A^{\circ}+  0.24 \times 10^{-11}m$
$= 1  A^{\circ}+  0.024  A^{\circ}   (\because  10^{-10}m= 1  A^{\circ})$
$= 1.024  A^{\circ}$

So, the answer is option (A).

Multiple choice detection and recording of x-ray images option c: imaging physics

A photon recoils back after striking a free electron. Then the value of compton shift is

  1. 0.0242 $A^{0}$
  2. 0.0484 $A^{0}$
  3. 0.0121 $A^{0}$
  4. 0.242 $A^{0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\lambda -\lambda \ '=\dfrac{h}{m _{e}c}(1-cos\theta )$


$\Delta \lambda =\dfrac{6.62\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}(1-cos180)$

$=0.0484\ A^{0}$
So, the answer is option (B).

Multiple choice detection and recording of x-ray images option c: imaging physics

The $X-$ray beam emerging from an $X-$ray tube

  1. is monochromatic

  2. contains all wavelength smaller than a certain maximum wavelength

  3. contains all wave length larger than a certain minimum wavelength

  4. contains all wave length lying between a minimum and a maximum wavelength
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

X-ray tubes produce a continuous spectrum of wavelengths starting from a minimum wavelength (lambda_min) determined by the accelerating voltage, extending to all larger wavelengths.

Multiple choice physics propagation of sound waves longitudinal vs transverse wave sound and light comparison of speed of sound with speed of light

A laser signal sent towards the moon returns after T seconds.If c is the speed of light, then  the distance of the moon from the observer is given by

  1. cT
  2. $c{T^{ - 2}}$
  3. $\frac{{ct}}{2}$

  4. 2 cT
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The total distance traveled by the signal is 2 * distance (to and back). Since distance = speed * time, 2 * d = c * T, therefore d = c * T / 2.

Multiple choice physics propagation of sound waves longitudinal vs transverse wave sound and light comparison of speed of sound with speed of light

Speed of light in air is about ....... times greater than the speed of sound in air.

  1. Million

  2. Trillion

  3. Billion

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The speed of light in air is $\displaystyle 3\times { 10 }^{ 8 }{ m }/{ s }$ which is about a million times greater than the speed of sound in air (i.e. 330 m/s at $\displaystyle { 0 }^{ \circ  }C$

Multiple choice physics superposition of waves coherence young's double slit experiment interference

A light of wavelength $400\overset{o}{A}$ after travelling a distance of $2\mu m$ produces a phase change of:

  1. Zero
  2. $3\pi$
  3. $\displaystyle\frac{\pi}{2}$
  4. $\displaystyle\frac{\pi}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The number of wavelengths the wave travels is $=\dfrac{l}{\lambda}$ $=\dfrac{2\times 10^{-6}}{400\times 10^{-10}}$ $=50$     which is an integer.

Hence, the point is at a distance which is integral multiple of wavelength.
Thus the phase difference between the points is zero.

Multiple choice nanochemistry nanotechnology applied chemistry materials chemistry

Which of the following is / are true?

  1. When a quantum dot is irradiated with UV light, it emits visible light, the wavelength of which depends on the size of nanoparticles.

  2. The rainbow colours are emitted by the nanoparticles of different size of single substance.

  3. The wavelength of light emitted decreases from red to violet in the rainbow spectrum as the size of the nanoparticles decreases.

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

All of the given statements are correct.
When a quantum dot is irradiated with UV light, it emits visible light, the wavelength of which depends on the size of nanoparticles. For example, a 3nm cadmium selenide particle emits green light of wavelength 520 nm whereas 5.5 nm particle of the same substance emits red light at the wavelength of 620 nm.
The rainbow colours are emitted by the nanoparticles of different size of single substance.
The wavelength of light emitted decreases from red to violet in the rainbow spectrum as the size of the nanoparticles decreases.

Multiple choice evs light, shadows and images what makes things visible nature and sources of light light travels in straight line

What is the wavelength range of visible light?

  1. $ 4 \times 10^{-7} m $ to $ 8 \times 10^{-7} m$
  2. $ 4 \times 10^{-9} m $ to $ 8 \times 10^{-9} m$
  3. $ 4 \times 10^{-5} m $ to $ 8 \times 10^{-5} m$
  4. $ 4 \times 10^{-6} m $ to $ 8 \times 10^{-6} m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The wavelength of visible light lies between $400\,{\rm{nm}}$ to $800\,{\rm{nm}}$.

Or, $4 \times {10^{ - 7}}$ to $8 \times {10^{ - 7}}\,{\rm{m}}$

 

Multiple choice geography environmental changes elements of weather and climate : temperature process of heat transfer in the atmosphere environmental regions (zones)

Electromagnetic spectrum consists of

  1. Alpha radiation

  2. Gamma radiation

  3. Beta radiation

  4. All of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electromagnetic spectrum is the range of all possible frequencies of electromagnetic radiation. There are 7 regions in the electromagnetic spectrum and they are gamma rays, x-rays, ultraviolet, visible light, infrared, microwaves and radio waves. All types of electromagnetic radiation are transverse waves and they all travel at the same speed in a vacuum.  The regions of the electromagnetic spectrum are explained below in order of increasing wavelength and decreasing frequency.

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

A lossless coaxial cable has a capacitance of $7\times { 10 }^{ -11 }$ F and an inductance of $0.39\mu H$. Calculate characteristic impedance of the cable.

  1. 65

  2. 75

  3. 66

  4. 77
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here,$C=7\times { 10 }^{ -11 }F$,
        $L=0.39\times { 10 }^{ -6 }H$

         ${ Z } _{ o }$ As the cable is lossless,
        $\therefore { Z } _{ o }\sqrt { \dfrac { L }{ C }  } =\sqrt { \dfrac { 0.39\times { 10 }^{ -6 } }{ 7\times { 10 }^{ -11 } }  } =75ohm$