Tag: option c: imaging

Questions Related to option c: imaging

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

Show answer
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).

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 $(h=6.62 \times 10^{-34}Js.)$

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

  1. $0.121\times 10^{-11}m$

  2. $0.486\times 10^{-11}m$

  3. $2.4\times 10^{-11}m$

  4. $0.243\times 10^{-11}m$

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Correct Option: D
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).

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

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

  1. $4.2\times 10^{-11}m$

  2. $1.12\times 10^{-11}m$

  3. $1.21\times 10^{-11}m$

  4. $2.42\times 10^{-11}m$

Show answer
Correct Option: D
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).

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)

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

  1. 3.024 $A^{0}$

  2. 3.012 $A^{0}$

  3. 3.048$A^{0}$

  4. 2.988 $A^{0}$

Show answer
Correct Option: B
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).

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

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

  1. $5.84\times 10^{-12}m$

  2. $6.84\times 10^{-3}m$

  3. $7.84\times 10^{-10}m$

  4. $4.84\times 10^{-12}m$

Show answer
Correct Option: D
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).

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

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

  1. 1.024$A^{0}$

  2. 2.024$A^{0}$

  3. 3.024$A^{0}$

  4. 4.024$A^{0}$

Show answer
Correct Option: A
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).