Chapter :

Electron

Chapter Outline

➢Introduction

➢Fundamental Properties of an Electron

➢Motion of Electron Beam in Electric and Magnetic Fields

➢Determination of Charge of an Electron by Millikan’s Oil

Drop Experiment

➢Determination of Specific Charge (e/m) of an Electron by

J.J. Thomson Experiment

ElectronIntroduction:

The electron is a subatomic particle that carries a negative electric charge. It is not known about its internal structure up to now so it is a fundamental particle. The electron was identified as a subatomic particle in 1897 by J.J. Thomson.

Fundamental Properties of an Electron

Some fundamental properties of an electron are as follows:

• Its rest mass is 9.1 × 10−31kg.

• It is a negatively charged particle carrying charge e = −1.6 × 10−19C .

• It has both particle and wave nature.

• It is deflected in electric and magnetic fields.

• Its mass is about 1836 times less than that of proton.

Lorentz force(F)

The sum of forces on moving charge q due to electric and magnetic field is called Lorentz force and is given as,

= 𝐅𝐞 + 𝐅𝐦

Where 𝐅𝐞 = 𝐞𝐥𝐞𝐜𝐭𝐫𝐢𝐜 𝐥𝐨𝐫𝐞𝐧𝐭𝐳 𝐟𝐨𝐫𝐜𝐞 = 𝐪 𝐄

and

𝐅𝐦 = 𝐦𝐚𝐠𝐧𝐞𝐭𝐢𝐜 𝐥𝐨𝐫𝐞𝐧𝐭𝐳 𝐟𝐨𝐫𝐜𝐞 = 𝐪 𝐯 × 𝐁

Magnitude of magnetic force is qvBsinθ

Where v be the velocity of the charge particle and B be the magnetic field intensity or

magnetic flux density .

**Motion of Electron in Uniform Magnetic field **

A) When a beam of electron enters in a magnetic field perpendicularly i.e. 𝛉 = 𝟗𝟎°

Let us consider an electron beam is

moving with speed ‘v’ horizontally which

enters a uniform magnetic field of

magnitude ‘B’ acting perpendicularly inward

to the direction of motion as shown in figure.

When electron enters into the

magnetic field it experiences a magnetic

force which is given as

F = Bev ……….(i) ( since θ = 90°)

The direction of this force is always

perpendicular to both B and v, which is given

by Fleming’s Left Hand Rule .Therefore, the

speed and energy of the electron remains

unaltered but its path is deflected from its

original path.

If force continuously acts perpendicular to the velocity of a particle then it changes the direction of the velocity but not its magnitude which is the condition of uniform circular motion. Thus on this basis we can conclude that and electron moving normally to the magnetic field follows circular path of radius r and arc PQ .The magnetic force Bev provides necessary centripetal force to move the electron in a circular path of radius ‘r’.

** Magnetic force = Centripetal force**

Bev = mv2/r

where m = mass of an electron .

Or r =mv/Be =momentum/Be

Which gives the radius of the circular path made by the electron in a magnetic field.

Again , v/r=Be/m

ω =Be/m

since v = ωr

Where ω is angular velocity of electron.

2π/T=Be/m

T=2πm/Be

Note : T = circumference of the circle/speed along circl

Here , T is the time period of revolution of electron in the uniform magnetic field . The frequency of revolution of electron is ,

(cyclotron frequency) f=1/T=Be/2πm

Here we saw that time period T and frequency f are independent with the velocity of electron .

**B) When electron enters the field parallelly or antiparallelly, i.e. 𝛉 = 𝟎° 𝐨𝐫 𝟏𝟖𝟎°**

The magnitude of magnetic Lorentz force is

F = Bev sinθ

For θ = 0° or 180°, sin θ = 0

∴ F = 0

The electron does not experience any magnetic force, if it enters

into the field parallelly or antiparallelly .

Hence the electron moves in a straight path.

C) When electron enters at any oblique angle 𝛉 (𝛉 ≠ 𝟎°, 𝟏𝟖𝟎°𝐨𝐫 𝟗𝟎°)

In this condition, the component of velocity which is parallel to field tends to move the electron in linear path, whereas the perpendicular component of velocity tends to move the electron in circular path. Due to the combined effect, electron obeys a helical path

(i.e. spiral path ) as shown in figure below .

r

Magnetic

field(B)

Velocity of electron (v)

𝜃

𝑣𝑥 = 𝑣 𝑐𝑜𝑠𝜃

𝑣𝑦 = 𝑣 𝑠𝑖𝑛𝜃

x

To move in circular path necessary centripetal force is provided by the magnetic force .

i.e. **Centripetal force = Magnetic force**

mvy2/r= evyB

or Radius of Helical path r = mvy

eB

or r =

mvsinθ

eB … … . . i

Let T be the time period then

T = circumference of the circle/speed along circle = 2πr/vsinθ

By using equation (i) we get

Time period T = 2πm /eB

Hence, it is clear that time of revolution and frequency (f=1/T) of electron is independent

of speed of the particle as well as angle of projection but depends on m, B and e.

The linear distance travelled by the particle during this time period is called pitch of helix.

Also, the pitch is defined as linear distance between two consecutive turns of a helical

path. So,

Pitch(x) = horizontal velocity × time period

∴ x = 2πmvcosθ / eB

**Conclusions:**

- The possible natures of path in uniform magnetic field are straight line, circular and spiral.
- In magnetic field, no force acts on a stationary charged particle but only on moving charged particle.
- Deflection is possible only when the motion of the particle is other than parallel or anti-parallel to the field.

**LQ Show that , the path of an electron moving through a transverse uniform electric field is parabola.**

Motion of Electron in Uniform Electric Field

Let us consider a horizontal beam of electrons which are perpendicular with the

direction of the electric field E are moving with velocity ‘u ‘ passes between two

parallel plates which are separated by a distance ‘d’ as shown in figure above .

The upper plate is connected with positive terminal and lower plate is connected

with negative terminal of a battery with a potential difference V .

Consider an electron is at a point P(x , y) at any instant of time ‘t’ .

For Horizontal Motion

Sx = uxt +1/2axt

or x = ut … … … … …( i)

Electrostatics force acts only in vertical direction therefore horizontal

acceleration of the electron is zero i.e. ax = 0. If there is no force along

horizontal direction then the horizontal component of velocity of the electron

remains same

i.e. ux = vx = u

For vertical motion

Sy = uyt +1/2ayt2

or y = 0 × t +1/2ayt2

[Since initial vertical component of velocity is zero ]

or y = 1/2*ayt2 … … . . (ii)

From Newton’s second law,

Vertical acceleration of electron( ay) = Force in vertical direction/mass of an electron

ay = Fe/m

ay = eE/m=e𝐕/md [Since Fe = eE and E = 𝐕/d]

∴ Equation ii becomes

y =1/2e𝐕t2/md

By using equation i we get

y = 1/2e𝐕/mdx/u = 1/2 e𝐕/mdu2x2 = kx2 … (iii)

where k = e𝐕/mdu2 = constant

Equation (iii) represents the equation of the parabola . Hence the path of an electron in the

electric field is parabolic in nature. √

When the electron just passes the plates, or just grazes the end of the plate, x = D . Then

equation (iii) can be written as

y =1/2e𝐕/mdu2D2

After crossing the electric field, the electron continues in a straight path tangent to the parabola.

The time t for which the electron is between the plates is given by t =D/u

Thus , the component of the velocity vy , gained in the direction of the field during this time is

given by

vy = uy + vertical acceleration × time =e𝐕D/mdu

Total velocity when the electron just emerges from the field is given as

v = vx2 + vy2 = u2 + (e𝐕D/mdu)2

The angle θ at which the beam emerges out the field is given by

tanθ = vy/vx = e𝐕D/mdu2

or θ = tan−1 (e𝐕D/mdu2)

where m = mass of electron = 9.1 × 10−31kg

Gain in Kinetic energy when it emerges from the electric field or change in K.E

# = final K.E – initial K.E

1/2mv2 −1/2mu2 =1/2m(e𝐕D/mdu)2

;E =𝐕/d

**Note**

1).Gain in kinetic energy is

inversely proportional to mass if

other quantities are constant. i.e.

Gain in kinetic energy ∝𝟏𝐦.

2).In the above situations, force due to

gravity and resistance by air are not taken into

consideration.

3).The possible path in uniform electric field is

a) straight line and

b) parabolic

SQ** Beams of electrons and protons having the same kinetic energy enter normally intoan electric field, which beam will be more curved, Justify.**

**Ans.**The transverse deflection of an electron in an electric field is ,

y =1/2e𝐕/mdu2*x2

or y = 1/4d*e𝐕(1/2*mu2)*x2

For constant e, V, d and t (identical condition) then

𝐲 ∝ 𝟏/𝐊. 𝐄

If kinetic energy is also equal for both electron and proton , trajectory of particles is also equally

curved.

**SQ An electron and a proton enter a transverse electric field with the same****velocity . Name the particle whose trajectory is more curved ?**

Hint y =1/2d*e𝐕/(mu2)x*2

y ∝1/m

Since other quantities are taken as constant .

Mass of proton is greater than the mass of an electron so path of electron is

more curved than the path of proton .

Mass of electron = 9.1 × 10−31kg

Mass of proton = 1.67 × 10−27kg

**Cross Fields**

A region of uniform electric fields and magnetic field applied simultaneously perpendicular to each

other such that a charge particle entering normally into this region passes without any deviation is called cross field.

This is possible when, magnetic force is equal to electric force.

i.e. Magnetic force = Electrostatic force

qvB = qE

Or v= E/B

So, on such fields , the velocity of particle is equal to the ratio of electric field to magnetic field.

**Specific Charge (q/m)**

The ratio of charge and mass of a charged particle is

called its specific charge.**Note**

The ratio of charge of electron to mass of electron is called the specific

charge of an electron . Similarly, the ratio of charge of proton to mass of proton

is called the specific charge of the proton. According to JJ Thomson’s

experiment , the specific charge of an electron is

𝑒/𝑚= 1.76 × 1011C/kg

The ratio of charge and mass of a charged particle is

called its specific charge

Determination of specific charge (e\m) of an electron by J.J Thomson’s Experiment

This experiment is based on principle of cross fields. The simplified form of the apparatus for

determining the specific charge of an electron is as shown in the figure above.

The apparatus consists of a cathode C and anode A which are enclosed in an evacuated discharge

tube. When a high potential difference V is applied between cathode and anode , a beam of cathode rays emerges out from the cathode and passes through the small hole in the anode and passes between the two parallel plates P1and P2

A uniform electric field E is applied between the plates P1 and P2

separated by a distance d then the electrons will be deflected towards positive plate and finally strikes the fluorescent screen at point S2

The magnitude of electric force experienced by the electron in upward direction is given as,

Fe = eE =e𝐕′/d… … … . . i

The electric field is switched off and a uniform magnetic field B is applied in the perpendicularly

inward direction. As soon as the electrons beam enters the region of uniform magnetic field, a magnetic

force(Fm) acts in the downward direction as decided by Fleming’s left hand rule. The beam is then deflected

downwards and strikes the fluorescent screen at point S1.

If v be the velocity of electron with which it enters the field region then the magnetic force

experienced by it is given as,

Fm = evB … … … . . (ii)

Finally, electric and magnetic fields are applied simultaneously perpendicular to each other and the fields are so adjusted

that, the electron beam is neither deflected upward nor downward. Rather it passes straight along axis of tube and strikes

the fluorescent screen at point S, where it had stroked in the absence of both fields. In this condition,

Magnetic force = Electrostatic force

Fm = Fe

evB = eE

v = E/B

v = 𝐕/dB … . . iii [since E =𝐕/d]

When the electron is accelerated between the potential difference V of electrodes C and A, it gains kinetic energy given by ,

K. E = e𝐕

1/2*mv*2 = e𝐕*e/m=v2/**2𝐕*

From equation (iii) we get ,

e/m = 𝐕′2/2B2d*2𝐕 … … . iv

Thus, by knowing the values of V’, B, V and d we can calculate the value of e/m . The value of e/m for electron calculated by

J.J Thompson was 1.76 × 1011C/Kg

Millikan’s Oil Drop Experiment

The experimental arrangement used by Millikan to determine the charge of an electron is shown in figure below . It is based on Stoke’s law in viscosity. It consists of two metal circular plates A and B about 20cm in diameter and 1.5cm apart with a small hole H in the center of the upper plate A.

Fig Experimental arrangement of Millikan’s Oil

Drop Experiment

The upper plate A is connected to a high tension battery ( order of 10,000V) while

the lower plate B is earthed. The plates are arranged inside a double walled chamber. The

chamber is maintained at constant temperature by circulating water between the walls

which avoids convection current of air and may protect the oil drop to move in zig-gag way

and hence the error may not arise.

Through the hole in the upper plate H , clock oil ( a non volatile and viscous liquid

which preserved its geometry during the motion ) is sprayed by means of atomizer. These

drops get charged due to friction and carry a few electronic charges. The window W1 is

used to illuminate the oil drops by providing enough light. The window W2 is used to let X-ray pass in to the space between the plates in order to ionize the oil drops in case the oil

drops are not ionized by friction. The microscope is provided with a crosswire and a

micrometer scale so that the motion of the drop can be observed and measured. The

motion of the oil drop was studied under following two cases.

i) Motion under the effect of gravity alone

ii) Motion under the combined effect of gravity and the electric field**i) Motion of Oil drop under gravity alone**

Suppose the electric field is not applied. As the oil drop falls under gravity

its velocity goes on increasing. A stage comes when the viscous force on the

oil drop becomes equal to its resultant weight. The oil drop now moves with

a constant velocity v1 called terminal velocity.

Let

r= radius of oil drop

m= mass of oil drop

σ = density of air

And ρ= density of the oil

Then ,

Volume of the oil drop = 4/3*πr3

Weight of the oil drop (W) = 4/3*πr*3ρg

Upthrust due to the air (U)= weight of the air displaced by the drop = 4/3*πr*3σg

The viscous force on the oil drop in upward direction (F) = 6πηrv1

When the oil drop is moving with terminal velocity v1, then

Total upward force = Total downward force

Or F+U = W

Or F = W-U

Or 6πηrv1= 4/3*πr*3ρg − 4/3*πr*3σg =4/3πr*3 ρ − σ g……..(*)*

* ∴ r = 9ηv1 2 ρ − σ g … … … … … … … … … … … … . i*

* Knowing the value of ρ, η, σ, v1and g the radius of the oil drop can be calculated.*

*ii) Motion of the oil drop under electric field*

* A strong electric field is applied between the plates in such a direction that force on the negatively charged oil drop due to the electric field acts in the vertically upward direction. Now the oil drop starts moving upward and soon attains a terminal velocity v2 in upward direction.*

* Let E be the electric field intensity . As the drop carries a charge q then, electrostatic force on oil drop in upward direction (Fe) = qE*

* Viscous force in downward direction (F) = 6πηrv2 *

*When the oil drop is moving with terminal velocity v2 then *

*Total downward force = Total upward force *

*Or W+F = Fe+ U *

*Or W-U +F = Fe *

*Or Fe= W-U+F *

*Or qE= 6πηrv1+ 6πηrv2 [ By using equation(*)]

Or qE = 6πηr(v1+v2)

Or q=6πηr(v1+v2)/E

Or q=6πη(v1+v2)/E × 9ηv1/2 ρ−σ g……………(ii)

Knowing all the quantities on the right hand side, the value of charge q on the oil drop can

be determined .The experiment is repeated with many other drops. It was found that the

charge of the drops is always found to be approximately equal to integral multiple of 1.6 ×

10−19C. The decrease in the terminal velocity of the drop when positive potential is given

to the upper plate shows that the drop has negative charge. Hence, it can be concluded

that the basic charge ( the charge of electron ) is −1.6 × 10−19C.

Note

If the oil drop falls down with a small terminal velocity v2 even though the electric field is applied, then the viscous force acts upward and the charge on the oil drop will be

q=6πηr(v1− v2)/E

q=6πη(v1− v2)×9ηv1/2 ρ−σ g

**Estimation of Mass of an electron**

Millikan’s Oil drop experiment gave the value of e and J.J

Thomson’s Experiment gave the value of 𝑒/𝑚

Dividing charge of an electron by its specific charge we get mass of the

electron . i.e.

Mass of an electron (m) = 𝑒/𝑒/𝑚=1.6×10−19/1.75×1011= 9.11 × 10−31𝐾𝑔

**Importance of Millikan’s Oil Drop Experimen**t

1. Millikan’s experiment showed that electronic charge was

the smallest possible charge on a charged particle or ion.

2. Millikan’s experiment has proved the quantization of

charge i.e. A body can carry an integral multiple of

minimum charge e. i.e. q=±𝑛𝑒 , where n=1,2,3…

3. There is no direct method to find the mass of an electron .

Knowing the charge of an electron and specific charge, the

mass of the electron can be determined .

Important Short questions

1)** What is clock oil?**

• It has a low vapor pressure (vapor pressure is an indication of a liquid’s evaporation rate)

[Non-volatile and viscous liquid, Does not evaporate easily]

• Has quite small droplets which acquire terminal velocity quickly

2) **Water drop cannot use in Millikan’s experiment in place of clock oil why?**

• It can evaporate during experiment as it has high vapor pressure.

• It cannot give quite small droplets so that it is hard to attain terminal velocity.

3) **The Millikan’s apparatus is surrounded by a constant temperature enclosure. Why ?****Ans.** The Millikan’s apparatus is surrounded by a constant temperature enclosure because

• It avoids convection current of air which may move the drop and cause error

• It shields the apparatus from a current of air.

4)** Millikan’s experiment can not be performed with bigger drops. Why?****Ans.** If bigger drops are used in experiment the Electric field required would be of very high

magnitude and it may be very difficult to produce it . Furthermore, the terminal velocity of bigger

drops can not be obtained soon after the drops are sprayed. Then the Stoke’s formula can not be

applied. Hence bigger drops can not be used .

SQ ) **Cathode rays cannot be regarded as electromagnetic waves. Why?**

**Ans. **Cathode rays are streams of negatively charged particles which are

deflected by electric and magnetic fields but electromagnetic waves are

charge less which are not deflected by electric and magnetic fields. Hence

cathode rays are not electromagnetic waves.

SQ )** What property of the cathode rays indicates that they consist of electrons?****Ans**: Cathode rays are deflected in electric and magnetic fields. When the specific

charge (e/m) of cathode rays are measured, it is found exactly equal to the

specific charge of an electron which confirms that the cathode rays are the

streams of elections.

SQ )** The value of 𝒆/𝒎 is constant for cathode rays, but not for positive rays.Explain**.

**Ans.**Cathode ray is beam of electrons. The mass and charge of electron is

always constant due to which specific charge of cathode (e/m) is

always constant. But positive rays are stream of positive ions. Masses and

charges of positive ions varies for different substances. Hence the value of (e/m)

is not constant for positive rays.

SQ1.** What will happen on the nature of path followed bycathode rays, when they move normally with electric field?**

**Ans**. Cathode rays are the beam of electrons moving from

cathode to anode in discharge tube. Electrons are negative

charge particles. When electrons enter normally in an electric

field then they are attracted towards positive plate due to

electrostatic force. The path followed by the electrons when

enters normally in the electric field is parabolic in nature.

SQ2.** An electron and a proton are projected in an electric fieldat right angles to field with same velocity. Which particle hastrajectory more curved?**

A

**ns**. The vertical displacement produced on path of charged

particle inside the electric field E and potential difference V is

given by

𝑦 =1/2𝑑*𝑒𝑽/(𝑚𝑢2)*𝑥2

𝑦 ∝1/𝑚

Since d, e, V, u and x are taken as constant.

Here charge of electron and proton is same. Mass of proton is

greater than the mass of an electron so path of electron is more

curved than the path of proton. (If y is greater then the curve is

more curved and vice versa. For electron y is more since mass is

less. Hence path of electron is more curved)**Note**

Mass of electron = 9.1 × 10−31𝑘𝑔

Mass of proton = 1.67 × 10−27𝑘𝑔

(The mass of proton is 1837 times that of an electron.)

SQ3. **The mass of proton is 1837 times that of an electron. Anelectron and a proton are projected in a uniform electric fieldwith same initial kinetic energy. Whose path will have morecurvature?**

**Ans.**The transverse deflection of an electron in an electric field

is,

𝑦 =1/2*𝑒𝑽/𝑚𝑑𝑢2*𝑥2

𝑜𝑟 𝑦 =1/4𝑑*𝑒𝑽/(1/2*𝑚𝑢2)*𝑥2

Here charge(e) of an electron and a proton is same.

For constant e, V, d and t (identical condition) then

𝒚 ∝𝟏/𝑲.𝑬

If kinetic energy is also equal for both electron and proton,

trajectory of particles is also equally curved.

SQ4.** An electron and proton are injected into a uniform electricfield at right angles to the direction of the field with equalmomentum. Then, which will trace greater curved together?**

**Ans.**The vertical displacement produced on path of charged

particle inside the electric field E and potential difference V is

given by

𝑦 =1/2𝑑*𝑒𝑽/(𝑚𝑢2)*𝑥2

𝑦 =1/2𝑑*𝑒𝑽𝑚/(𝑚𝑢)2*𝑥2

Here for both electron and proton d, e, V, x and

momentum(mu) are same then

𝑦 ∝ 𝑚

That is the particle of greater mass has larger deflection or

more curved trajectory as both moving with same momentum.

That is proton trajectory will be more curved.

SQ5. **Light is not deflected by electric field, why?****Ans.** Light (electromagnetic radiation) consists of photons

which are chargeless particles. Therefore, light is not deflected

by electric as well as magnetic field.

**Very important short questions**

SQ1. **An electron and a proton move with the same speed in a uniform magneticfield of equal magnitude. Compare the radii of their circular paths.**

**Ans**. When a charge particle enters in a magnetic field normally then its path

becomes circular. For circular path

Magnetic Force = Centripetal force

𝑒𝑣𝐵 =𝑚𝑣2/𝑟

Or 𝒓 =𝒎𝒗/𝑩𝒆

Here speed v, charge e and magnetic field B are same for both electron and

proton. Therefore, radius of the circular path only depends upon their mass. i.e.

𝑟 ∝ 𝑚

In this case, radius of the circular path is directly proportional to the mass of the

particle.

Mass of the proton is greater than the mass of the electron. Therefore, radius of

the circular path made by the proton is greater than that of the electron. The

ratio of radius of proton to the radius of electron is given as,

𝑟𝑝/𝑟𝑒=𝑚𝑝/𝑚𝑒

SQ2**. A charged particle is fired into a cubical region of space, where there isuniform magnetic field. Outside this region, there is no magnetic field. Is itpossible that the particle will remain inside the cubical region? Explain.**

**Ans.**Yes, it is possible that the particle will remain inside the cubical region if it

enters perpendicularly to the magnetic field and the diameter of circular path is

less than the side of the cube.

SQ3.** A charged particle moves through a region of space with constant velocity.If external magnetic field is zero in this region, can we conclude that theexternal electric field in the region is also zero? Explain**.

**Ans.**Yes, the velocity is constant only in the region of zero electric field because

the velocity of charged particle will be constant in a field free space and in cross

field space. Here, magnetic field is zero and hence for constant velocity electric

field also be zero.

**An electron and proton are injected into a uniform magnetic field withsame kinetic energy at right angles to the field direction. Then, which will tracegreater curved trajectory?**

**Ans:** When a charged particle enters to the region of uniform magnetic field B

with speed v, the path traced by it is circular. The radius of circular path is given

by 𝑞𝑣𝐵 =𝑚𝑣2/𝑟

Or 𝑟 =𝑚𝑣/𝑞𝐵 = √2𝑚𝐾.𝐸/𝑞𝐵 [ mv= momentum(p) and 𝐾. 𝐸 = *𝑃2/2𝑚]

For same kinetic energy (K.E), K. E= constant

Here charge q (=e), K.E, v and B are constant.

Radius r only depends upon mass of the particle moving in the field.

Here radius of the circular path is directly proportional to the square root of

the mass of the particle. i.e.

𝑟 ∝ √𝑚

Particle having higher mass has greater radius. Mass of proton is greater than the

mass of the electron. Therefore, proton traced greater circle than the electron as

both moving with same kinetic energy.

** Which will have the more radius in the circular path, when 𝜶 particle and aproton entering a region of uniform magnetic field normally having same kineticenergy?**

**Ans**. When a charged particle enters to the region of uniform magnetic field B

with speed v, the path traced by it is circular. The radius of circular path is given

by 𝑞𝑣𝐵 = 𝑚𝑣*2/𝑟

Or 𝑟 = 𝑚𝑣/𝑞𝐵 = √2𝑚𝐾.𝐸/𝑞𝐵 [ mv= momentum(p) and 𝐾. 𝐸 = 𝑃*2/2𝑚]

For same kinetic energy (K.E), K. E= constant.

Here K.E and B is constant. Therefore, radius of the circular path depends upon

both mass and charge.

Charge of alpha particle = 2 times charge of proton

Mass of alpha particle = 4 times mass of proton

The ration of radius of alpha particle to proton is given as,

𝑟𝛼/𝑟𝑝 = √𝑚𝛼/𝑞𝛼

//√𝑚𝑝/𝑞𝑝 = √𝑚𝛼/𝑞𝛼 × 𝑞𝑝/√𝑚𝑝 = √4𝑚𝑝/2𝑞𝑝×𝑞𝑝/√𝑚𝑝 = 1

Or 𝑟𝛼 = 𝑟𝑝

Therefore, both alpha particle and proton will have same radius in the circular

path.

** An electron and proton are injected into a uniform magnetic field at rightangle to its direction with the same momentum. Then which one’s path is morecurved?**

**Ans.**When a charged particle enters to the region of uniform magnetic field B

with speed v the path traced by it is circular path and is given as

𝑟 =𝑚𝑣/𝑞𝐵

According to the question momentum(p)=mv= constant

For proton and electron q (=e) is also same. Here q, p and B are same for both

electron and proton. Therefore, radius also constant for both charge particle

proton and electron. Hence in this case electron and proton are equally curved in

magnetic field.** Write down expression for acceleration of a moving charge Q in paralleland perpendicular magnetic field?**

**Ans.** When charge Q moves parallel to the magnetic field B the force experience

by the charge particle is given by Lorentz force which is given as

F = QvBsin𝜃

For parallel motion of charge with B, 𝜃 = 0°

∴ 𝐹 = 0

From Newtons second law F=ma

Therefore ma =0

Or a=o

Hence the charge particle moves in straight path with uniform speed.

When charge Q moves perpendicular to the magnetic field B the force

experienced by the charge particle is given by Lorentz force and is given as

F= QvBsin𝜃

Or F=QvB [ since 𝜃 = 90°]

Or ma = QvB [since F=ma]

Or a=

𝑸𝒗𝑩/𝒎

Hence for constant Q, v, B and m. Acceleration(a) is also constant.

SQ8.** What is the difference between the deflection of the electron due toelectric and magnetic fields?**

**Ans.**The difference between the deflection of the electron due to electric and

magnetic fields are as follows.

**Why do** **we use cross field in Thomson’s experiment?**

**Ans:** in Thomson’s experiment the beam of electron must be un deviated under the action of electric and magnetic fields. when the electric field is applied perpendicular to the magnetic field, the deviation produced on electron on electron beam by one field is cancelled by the deviation produced on it by other field. It means, the force on electrons due to electric field is equal and opposite to the force on electron due to magnetic field and hence the electron beam is un deviated .Hence, cross field is used in Thomson’s experiment.

**A charged particle is fired into a cubical region of space where there is uniform magnetic field outside the region, there is no magnetic field. Is it possible that the particle will remain inside the cubical region ?Explain.**

**Ans:** Yes, it is possible that the particle will remain inside the cubical region if it enters perpendicularly to the magnetic field and the diameter of circular path is less then the side of the cube.

**The value of specific charge (e/m) is constant for cathode (negative) rays but not for positive (canal) rays why?**

**Ans:** cathode ray are the stream of fast moving electrons. The value of mass and charge of am electron are constant. so the specific charge (e/m) of cathode ray is constant. However the positive ray consist of positive ions which have different masses for different gases. Therefore , the value of e/m of positive rays is not constant.

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full but unedited

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