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Observing the force and electron beam tubes - class-XII

Description: observing the force and electron beam tubes
Number of Questions: 19
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Tags: physics electromagnetic forces charged particles
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Metallic sphere of radius is removed from other bodies,  and is irradiated with the radiation of wavelength Lambda, so that it is capable of causing Photoelectric effect.

  1. The maximum potential gained by the sphere will be independent of its radius

  2. Net positive charge appearing on the sphere after a long time will depend on the radius of the sphere

  3. The kinetic energy of the most energetic electron emanating from the Sphere will keep on declining with time

  4. The kinetic energy of the most energetic electrons emitted from this sphere initially will be independent of the radius of the sphere


Correct Option: A

Monochromatic light of wavelength 440 nm is produced. The power emitted by light is 18 mW. The number of photons emitted per second by light beam is $(h=6.6\times { 10 }^{ -34 }Js)$

  1. $3\times { 10 }^{ 16 }$

  2. $4\times { 10 }^{ 16 }$

  3. $3\times { 10 }^{ 18 }$

  4. $4\times { 10 }^{ 18 }$


Correct Option: B
Explanation:

Given,

$\lambda=440nm$
$h=6.6\times 10^{-34} Js$
$I=18m W$
The energy of monochromatic light,
$E=\dfrac{hc}{\lambda}$
$E=\dfrac{6.6\times 10^{-34}\times 3\times 10^8}{440\times 10^{-9}}$
$E=0.045\times 10^{-17}J$
The number of photon emitted per second by the light beam,
$n=\dfrac{I}{E}$ 
$n=\dfrac{18\times 10^{-3}}{0.045\times 10^{-17}}$
$n=4\times 10^{16}$
The correct option is B

In a hydrogen atom, an electron moves in an orbit of radius $5.0\times 10^{-11}\ m$ with a speed of $2.2\times 10^{6}\ ms^{-1}$. The equivalent current is :

  1. $11.2\times 10^{-3}\ A$

  2. $1.9\times 10^{-3}\ A$

  3. $1.12\times 10^{-3}\ A$

  4. $11.2\times 10^{4}\ A$


Correct Option: C
Explanation:
Current $=$ Charge passing per unit time $q$
$I=\dfrac { q }{ t } $
$t=\dfrac { 2\pi r }{ v } $
$I=\dfrac { q }{ 2\pi r } .v=\dfrac { 1.6\times { 10 }^{ 19 }\times 2.2\times { 10 }^{ 6 } }{ 5\times { 10 }^{ -11 }\times 2n } $
$I=\dfrac { 1.6\times 2.2 }{ 5\times 2\pi  } \times { 10 }^{ -2 }=1.12\times { 10 }^{ -3 }A$
$\boxed { I=1.12mA } $

Statement -1 :When ultraviolet light is incident on a photo cell, its potential is $ V _0 $ and the maximum kinetic energy of the photoelectrons is $ K _{max} $. When the ultravoilet light is replaced by X-rays, both $ V _0 $ and $ K _{max} $ increase. 

Statement-2: Photoelctrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

  1. Statement -1 is true,statement-2 is true ; statement -2 is correct explanation of statement-1.

  2. Statement -1 is true,statement-2 is true ; statement -2 is not correct explanation of statement-1.

  3. Statement -1 is false,statement-2 is true

  4. Statement -1 is true,statement-2 is false


Correct Option: D

If an electron covers 1/4th of revolution in a circle having radius R and in the presence of perpendicular magnetic field B, the amount of energy acquired by electron will be :

  1. $\dfrac{1}{4}mv^2$

  2. $0$

  3. $\dfrac{1}{8}mv^2$

  4. $(\pi\times R/2) \times (Bev)$


Correct Option: B
Explanation:

Magnetic Force is perpendicular to electron's velocity, so it can only change direction of electron's velocity but cannot change its acceleration

Velocity ratio of the two cathode rays is 1:2. They are applied to the same electric field. What is the deflection ratio of the two cathode rays

  1. 1 : 2

  2. 1 : 4

  3. 4 : 1

  4. 8: 1


Correct Option: A

In a photocell bichromatic light of wavelengths  $2475 \mathring { A } $  and  $6000 \mathring { A } $  are incident on cathode whose work function is  $4.8 e V.$  If a uniform magnetic field of  $3 \times 10 ^ { - 5 }$  Tesla exists parallel to the plate, the radius of the path describe by the photoelectron will be (mass of electron  $=9\times 10^{ { -31 } }{ kg }$ )

  1. $1 cm$

  2. $5 cm$

  3. $10 cm$

  4. $25 cm$


Correct Option: B

The minimum amount of energy released in annihilation of electron-positron is.

  1. $1.02\ MeV$

  2. $0.58\ MeV$

  3. $185\ MeV$

  4. $200\ MeV$


Correct Option: A
Explanation:
Energy released during the annihilation of ${ e }^{ - },{ e }^{ + }$ pair is,
$E=2{ m } _{ c }{ C }^{ 2 }$
$E=2\times 9.1\times { 10 }^{ -31 }\times { \left( 3\times { 10 }^{ 8 } \right)  }^{ 2 }J$
$E=\dfrac { 2\times 9.1\times 9\times { 10 }^{ -15 } }{ 1.6\times { 10 }^{ -13 } } MeV$
$E=1.02MeV$

A spherical drop of radius $10$ has absorbed $40$ electrons. The energy required to given an additional electron to it is.

  1. $9.2 \times 10^{-21} J$

  2. $5.7 \times 10^{-21} J$

  3. $9.21 \times 10^{-23} J$

  4. $4Ke^2$


Correct Option: D
Explanation:
We know that,
Energy $=$ Potential $\times $ Charge
$E=\left( \dfrac { k\quad q }{ r }  \right) \times e$
$E=\dfrac { k\left( 40e \right) \left( e \right)  }{ 10 } $
$E=4k{ \left( e \right)  }^{ 2 }$
$\boxed { Energy=4{ ke }^{ 2 } } $

Cathode rays enter an electric field normal to the field. Then their path in the electric field is :

  1. A parabola

  2. A circle

  3. A straight line

  4. An ellipse


Correct Option: A
Explanation:

Electric field applies a constant force  $eE$ on the electrons (i.e. cathode rays) perpendicular to the beam of electrons due to which the electrons will trace a parabolic path.

When the energy of the incident radiation is increased by 20%, The kinetic energy of the photelectrons emitted from a metal surfce in creased from 0.5V to 0.8eV The work function of the metal is 

  1. 0.65 eV

  2. 1.0 eV

  3. 1.2 eV

  4. 1.5 eV


Correct Option: C

Out of the following transitions, the frequency of emitted photon will be maximum for 

  1. $n = 5$ to $n = 3$

  2. $n = 6$ to $n = 2$

  3. $n = 2$ to $n = 1$

  4. $n = 1$ to $n = 2$


Correct Option: C

When electron is incident on molybdenum they by changing energy of electron:-

  1. $\lambda _{\text{min}}$ changes

  2. $\lambda _{\text{min}}$ remains constant

  3. $\lambda _{K _{\alpha}}, \lambda _{K _{\beta}}$ changes

  4. $\lambda _{\text{min}}, \lambda _{K _{\alpha}}$ and $\lambda _{K _{\beta}}$ all changes


Correct Option: A

An electron beam moving with a speed of $2.5\times 10^{7}ms^{-1}$  enters into the magnetic field directed perpendicular to its direction of motion. The magnetic induction of the field is $4\times 10^{-3}  wb/m^2$. The intensity of the electric field applied so that the electron remains undeflected due to the magnetic field is.

  1. $10^{4}N/C$

  2. $10^{5}N/C$

  3. $10^{7}N/C$

  4. $10^{3}N/C$


Correct Option: B
Explanation:

For no deflection
VB = E
$\rightarrow E = 2.5*10^7*4*10^{-3}$
$= 10*10^4$
$= 10^5 {N}/{C}$

J.J. Thomson's cathode-ray tube experiment demonstrated that:

  1. cathode rays are streams of negatively charged ions

  2. all the mass of an atom is essentially in the nucleus

  3. the e/m of electrons is much greater than the e/m of protons

  4. the e/m ratio of the cathode ray particles changes when a different gas is placed in the discharge tube


Correct Option: A
Explanation:

(A) is the correct option. Since, in Thomson's first experiment, he discovered the cathode rays and the change they deposited were intrinsically linked together. In the second experiments, he discovered that the cathode ray were negative. He deducted that the cathode rays were made up of negatively charged particles.

A particle of mass $1\times 10^{-26} \ kg $ and charge $1.6\times 10^{-19} \ C$ travelling with a velocity $1.28\times 10^6 \ m/s$ along the positive $X$-axis  enters a region in which a uniform electric field $\vec E$ and a uniform magnetic field of induction $\vec B$ are present. if $\vec E=-102.4 \times 10^3 \ \hat k \ NC^{-1}$ and $\vec B=8 \times 10^{-2} \ \hat j \ Wbm^{-2}$, the direction of motion of the particles is 

  1. along the positive X-axis

  2. along the negative X-axis

  3. at $45^{O}$ to the positive X-axis

  4. at $135^{O}$to the postive X-axis


Correct Option: A
Explanation:

Write Force
F = q(vXB) + qE.
F =0N.
Hence the particle will continue moving towards positive X-axis.

A proton and an $\alpha$ particle enter a magnetic field in a direction perpendicular to it. If the force acting on the proton is twice that acting on the $\alpha$- particle, the ratio of their velocities is

  1. $4 : 1$

  2. $1 : 4$

  3. $1 : 2$

  4. $2 : 1$


Correct Option: A
Explanation:

We know that Force $(F _m)$ on a particle with charge "q" moving with a velocity of v at an angle $\theta$ to the magnetic field $(B)$ can be expressed as

$F _m=qvB \sin{\theta}$
Given, $\cfrac{F _m (p)}{F _m(\alpha)}$=2
$q _{\alpha}=2 \times q _{p}$
$\Rightarrow \ \cfrac{q _{p}v _pB}{q _{\alpha}v _{\alpha}B}=2$
We get 
$\cfrac{v _{p}}{v _{\alpha}}=4$

When cathode rays moving horizontally pass through an electric field directed vertically downward, the rays would be deflected ________ .

  1. downward

  2. upward

  3. backward

  4. none of the above


Correct Option: B
Explanation:

The electric field is directed vertically downward i.e. the upper plate is positively charged and the lower plate is negatively charged . Because $ \vec{E} $ points in a direction from positive to negative.

When a cathode ray particle  i.e. negatively charged particle enters the electric field, it will be following a parabolic trajectory and would be directed towards the  positive plate. Since the positive plate is upward directed. So cathode ray will be directed upwards.

When a charged particle of charge $x\ C$ moves through a potential difference of $y\ V$, the gain in kinetic energy is equal to $xy \ J$.
An electron and an alpha particle have their masses in the ratio of 1 : 7200 and charges in the ratio of 1 : 2. If they start moving from rest through the same electrical potential difference then the ratio of their velocities will be ______. 

  1. 1 : 60

  2. 60 : 1

  3. 1 : 20

  4. 20 : 1


Correct Option: B
Explanation:

Given : If particle has charge $x$  coulombs and potential  difference $ y $ volts . Then gain in K.E. = $ xy $ joules.

Using similar analysis,

$ \dfrac{\text{(mass)electron}}{\text{(mass) alpha particle}} = \dfrac{1}{7200} $

$ \dfrac{\text{(charge)electron}}{\text{(charge) alpha particle}} = \dfrac{1}{2} $

We need to find $ \dfrac{\text{(velocity)electron}}{\text{(velocity) \alpha particle}} $

Since initially there at rest, initial K.E. = 0;

They move through same potential difference. Let potential difference be $ y $ volts.

So, gain in K.E = Final K.E – Initial K.E

= Final K.E – 0

So, gain in K.E = $ charge \times potential  \, difference $

Ratio of gain in K.E of electron and $\alpha$ particle = Ratio of final K.E of electron and $\alpha$ particle

 

$ \dfrac{\dfrac{1}{2} m v _e^2 \times y}{\dfrac{1}{2} m v _\alpha^2 \times y} $

 

$ = \dfrac{1}{2} \times 7200 = 3600 $

$ \dfrac{v _e}{v _\alpha} = \sqrt{3600} = 60 $

 

 

 

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