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 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).

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.