Tag: physics

Questions Related to physics

A strong argument for the particle nature of cathode rays is that they

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

  1. travel through vacuum

  2. cast shadow

  3. get deflected by electric and magnetic field

  4. produce fluroscence

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Correct Option: C

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

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

  1. ${ f }^{ ' } > f$

  2. ${ f }^{ ' } = f$

  3. ${ f }^{ ' } < f$

  4. None of these

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Correct Option: C
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).

The particle nature of cathode rays is proved by

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

  1. Their deflection under magnetic/ electric field

  2. Colour of glow in gas discharge tube.

  3. Their propagation along a straight line.

  4. All of these.

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Correct Option: A

In the case of Compton effect, which of the following is applicable ?

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

  1. Energy conservation

  2. Momentum conservation

  3. Charge conservation

  4. All of the above

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Correct Option: B

If h is planks constant, $m _o$ is rest mass of electron and c is the speed of light in vacuum, the S.I unit of $\dfrac{h}{m _{0}C}$ is

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

  1. $A^{0}$

  2. Js

  3. Ns

  4. m

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Correct Option: D
Explanation:

de Broglie wavelength,

$\lambda =\dfrac { h }{ p } =\dfrac { h }{ mv } $
and unit of $\lambda$ is $m$.
So, the answer is option (D).

In Compton effect, if the incident x-rays have low energy and the scattering atom has high atomic number then the electrons appear as

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

  1. bound with no measurable Compton shift

  2. free with measurable Compton shift

  3. bound with measurable Compton shift

  4. free with no measurable Compton shift

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Correct Option: A
Explanation:

If atomic number of an atom is high- suggests that more energy is required to eject the electron or electron is more tightly bound . If energy of x rays is low- suggests that x-ray photon possesses insufficient energy to cause any measurable effect on electron. Thus, considering both these factors, A is the correct option

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 

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

  1. $1.872$ A.U

  2. $1.896$ A.U

  3. $1.848$ A.U

  4. $0.024$ A.U

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Correct Option: A
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 $\lambda'=\lambda=1.872 A.U$ 

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.

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

  1. 68.8 pm

  2. 68.08 pm

  3. 69.52 pm

  4. None of these

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Correct Option: C
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).

In Compton scattering
a) The modified line occurs because of scattering with a single electron
b) The unmodified line occurs because of scattering with the entire atom
c)The electron can recoil at an angle greater that $90^o$ .
d) The scattering photon and recoil electron can be projected on the same side of the incident direction

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

  1. a, b, c

  2. a, b, d

  3. b, c

  4. a,b

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Correct Option: A
Explanation:

The modified line occurs due to collision of photon with single electron.
Compton scattering usually refers to the interactive involving only the electrons of atoms,if  photon does not collide with any of electron of an atom, then it shows unmodified lines.

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)

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

  1. 10%

  2. 20%

  3. 5%

  4. 1%

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Correct Option: A
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).