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

The minimum value of Compton wavelength shift is:

  1. $h/2 m _{0}c$
  2. $h/m _{0}c$
  3. $2h/m _{0}c$
  4. $zero$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\lambda -\lambda ^{1}  =  \dfrac{h}{m _e c}\ (1+cos  \theta )$

$\Delta \lambda ^{1}  =  \dfrac{h}{m _e c}\ (1+cos  \theta )$

$cf   \theta = 0^{0}$

then  $\Delta \lambda = 0$

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

The intensity of X-rays of wavelength $0. \mathring{A}$ reduces to one fourth on passing through $3.5 \ mm$ thickness of a metal foil. The coefficient of absorption of metal will be:-

  1. $0.2 \ mm^{-1}$
  2. $0.4 \ mm^{-1}$
  3. $0.6 \ mm^{-1}$
  4. $0.8 \ mm^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice detection and recording of x-ray images option c: imaging physics

In Compton effect, the quantity $\dfrac{h}{m _{e}c}$ is called

  1. Compton recovery wavelength

  2. Scattered wavelength of photon

  3. Compton wavelength of electron

  4. Compton wavelength of photon
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\lambda -\lambda ^{1}=\dfrac{h}{m _e c}(1-cos  \theta )$

where $\dfrac{h}{m _e c}$ is called compton wavelength.

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

The compton wavelength shift depends on

  1. Wavelength of the incident photon

  2. Material of the scatterer

  3. Energy of the incident photon

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

$\lambda -\lambda ^{1}=\dfrac{h}{m _e c}\ (1+cos  \theta )$
where $\theta$ is scattering angle.

So, the answer is option (D).

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

Given $h = 6.62 \times 10^{-34}$ Js, $m _e$ $= 9.1 \times 10^{-31}$ kg, $c = 3 \times 10^{8}$ m/s, the value of Compton wavelength is:

  1. 0.0121 $A^{0}$
  2. 0.0484 $A^{0}$
  3. 0.0242 $A^{0}$
  4. 0.0363 $A^{0}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Compton wavelength $= \ \dfrac{h}{m _{e}c}$

$= \ \dfrac{6.62\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^{8}}$

$= \ 0.242\times 10^{-11}m$

$= \ 0.0242\times 10^{-10}m$

$= \ 0.0242\ A^{\circ}$

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

In Compton scattering process, the incident X-radiation is scattered at an angle $60^o$. The wavelength of the scattered radiation is $0.22 A^o$. The wavelength of the incident X-radiation in $A^o$

  1. 0.508

  2. 0.408

  3. 0.232

  4. 0.208
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

$0.22\ A^{\circ}-\lambda \ _{i}= 0.024A^{\circ}(1-\dfrac{1}{2}) \ \ \ \ (\because  cos60^{\circ}= 1/2)$

$\lambda   _{i}= 0.22-0.012$
$= 0.208  A^{\circ}$

So, the answer is option (D).

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

If the scattering angle of the photon in Compton effect is $180^{0}$, the Compton shift is

  1. Equal to the Compton wavelength of the electron

  2. four times the Compton wavelength of the electron

  3. two times the Compton wavelength of the electron

  4. half the compton wavelength of the electron
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\lambda -\lambda ^{1}  =  \dfrac{h}{m _e c}(1-cos  \theta )$

where
$\lambda $ is initial wavelength
$\lambda^{1} $ is the wavelength after & scattering
$h$ is the plank constant
$m _e$ is the electron rest mass.
$c$ is the speed of light
$\theta $ is the scattering angle.

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

The value of Compton wavelength of electron is

  1. $0.0243$ $A^{0}$
  2. $0.243$$A^{0}$
  3. $2.43 $$A^{0}$
  4. $24.3 $$A^{0}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The quantity $(\dfrac {h}{m _0c})$ is known as the  Compton wavelength of the electron; it is equal to $2.43\times 10^{-12}$ $m$ or $0.0243 A$. 


Here, $m _0$ is the ,electron rest mass.

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

How would you relate the new frequency to original one when an X-ray photon collides with an electron and bounces off ?

  1. Is lower than its original frequency

  2. Is same as its original frequency

  3. Is higher than its original frequency

  4. Depends upon the electrons frequency
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 From Scattering Formula
$\lambda'-\lambda=\dfrac{G}{M _{e}c}(1-cos \theta )$
We can see, $\lambda '> \lambda $
So $v'< v$

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

The apparent wavelength of the light from a star moving away from the earth is 0.2% more than its actual wavelength. Then the velocity of the star is 

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

The Doppler shift formula for light is delta_lambda / lambda = v / c. Given delta_lambda / lambda = 0.2% = 0.002, then v = 0.002 * c. Using c = 3 * 10^8 m/s, v = 0.002 * 3 * 10^8 = 6 * 10^5 m/s.

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

X-rays of wavelength of $22\ pm$ are scattered from a carbon target at an angle of $85^0$ to the incident beam. The compton shift for X-rays is $(cos\ 85^0=0.088)$

  1. $2.2\ pm$
  2. $1.1\ pm$
  3. $0.55\ pm$
  4. $4.4\ pm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Compton shift is given as $\Delta \lambda =\dfrac{h}{mc}(1-cos \phi )=\dfrac{6.62\times 10^{-34}Js}{9.1\times 10^{-31} Kg\times 3\times 10^8m/s}(1-cos 85^0)$ 

where $m$ is the mass of electron and $h$ is Planck's constant with $c$ as speed of light in air.
On calculation we get the shift as $\Delta \lambda =2.2\times 10^{-12}meter=2.2pico meter$
Option A is correct.

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

A photon of frequency f under goes compton scattering from an electron at rest and scatters through an angle $\theta$. The frequency of scattered photon is ${ f }^{ ' }$ then

  1. ${ f }^{ ' } > f$
  2. ${ f }^{ ' } = f$
  3. ${ f }^{ ' } < f$

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

After scattering the wavelength of scattered photon increases due to the loss of energy and hence, the frequency decreases.
So, $f'<f$
So, the answer is option (C).

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

In an experiment on Compton scattering, wavelength of incident $X-ray$ is $1.872$ A.U. Then, the wavelength of the $X-ray$ scattered at an angle of $90^{0}$ is 

  1. $1.872$ A.U
  2. $1.896$ A.U
  3. $1.848$ A.U
  4. $0.024$ A.U
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to the Compton's Equation


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


where $ \lambda $= inital wavelenth,
$ \lambda'  $ = final wavelength,
h=Planck's Constatnt,
$M _e$=Mass of electron,
${\theta}$=angle of scattering,
Since ${\theta}$=90, Cos${\theta}$=1,
 hence RHS =0
hence $\lambda'=\lambda=1.872 A.U$ 

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

The minimum wavelength X-ray produced in an X-ray tube operating at 18 kV is compton scattered at $45^{\circ}$ (by a target). Find the wavelength of scattered X-ray.

  1. 68.8 pm

  2. 68.08 pm

  3. 69.52 pm

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

If electrons are accelerated to a velocity v by a potential difference V and then allowed to collide with a metal target, the minimum wavelength is given by:


$\lambda _{ min }=\displaystyle\dfrac { 1240*{ 10 }^{ -9 } }{ 18*{ 10 }^{ 3 } } =68.8*{ 10 }^{ -12 }m$

The change in wavelength in compton scattering is given by:
$\triangle \lambda =2.4*{ 10 }^{ -12 }(1-\cos { \phi  } )$
$=2.4*10^{-12}(1-.7)$
$=.72*10^{-12}m$
So, the wavelength of scattered X-ray is given by:
$\lambda^{'}min = (68.8+.72)*10^{-12}m = 69.52 * 10^{-12}m$.
So, the answer is option (C).

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

X-rays of energy 50 KeV are scattered from a carbon target. The scattered rays are at $90^o$ from the incident beam. The percentage of change in wavelength is
(given $m _{e}= 9 \times 10^{-31}Kg, C= 3 \times 10^{8}$m/s)

  1. 10%

  2. 20%

  3. 5%

  4. 1%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\theta = \ 90^{\circ}$
so, $cos \theta =  0$
$\Delta \lambda  =  \dfrac{h}{m _{e}C}(1-cos \theta )  =  \dfrac{h}{m _{e}C}(1-0)  =  \dfrac{h}{m _{e}C}$


percentage of change in wavelength 

$ \dfrac{\Delta \lambda }{\lambda _{i}}\times 100$ % $ \ \ \ \ (\Delta \lambda = \dfrac{h}{m _{e}C})$

$= \dfrac{h/{m _{e}c}}{hc/{energy}}\times 100 \ \ \ \  (energy = \dfrac{hc}{\lambda})$

$= \dfrac{energy}{m _{e}C^{2}}\times 100$

$= \dfrac{50\times 10^{3}\times 1.6\times 10^{-19}\times 100}{9\times 10^{-31}\times 3\times 10^{8}\times 3\times 10^{8}}$

$=  1\times 10$
$= 10$%
So, the answer is option (A).