Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

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

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

  1. travel through vacuum

  2. cast shadow

  3. get deflected by electric and magnetic field

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

The fact that cathode rays are deflected by electric and magnetic fields proves they consist of charged particles (electrons), as neutral waves would not be affected by these fields.

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

The particle nature of cathode rays is proved by

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

This is essentially the same question as 431450. Cathode rays (electrons) have charge and mass, which is proven by their deflection in electric and magnetic fields. Option A is correct. Colour of glow (B) depends on gas, not cathode rays. Straight-line propagation (C) and all options (D) don't specifically prove particle nature.

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

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

  1. $A^{0}$

  2. Js

  3. Ns

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

de Broglie wavelength,

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

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  1. a, b, c

  2. a, b, d

  3. b, c

  4. a,b
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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).