Tag: deflection of electron beam by magnetic field

Questions Related to deflection of electron beam by magnetic field

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 $