Home | Courses | Back to Chapter | Any Question? |

Physics-12 Complete Syllabus Course

Chapter 17 - Electron

INTRODUCTION

## Electron

Electron is the sub-atomic particles having negative charge and orbiting around the nucleus. It is denoted by 'e'.

The value of charge of an electron (e) = -1.6 x 10-19C
The mass of an electron (Me) = 9.1 x 10-31 kg
The mass of proton (Mp) =1.67 x 10-27 kg.

### Milikan's Oil drop experiment

The Milikan's determine the value of charge of an electron by oil drop method.
It consist of two circular metallic plate having diameter 20 cm and separated by 1.5 cm apart. The two plates A and B are kept in the double walled chamber. The upper plate A is connected with H.T.B whereas lower plate is being earthed. Window W1 is used to provide the sufficient light inside the chamber and Window W2 is used to provide X-rays. The clockoil is sprayed by the means of atomizer through the hole of upper plate. The oil drops get charged due to friction, so it have some electronic charge. If the oil drop doesnot charge due to friction, so X-rays is used to ionized the oil drop to create electronic charge.

Theory 1: The motion of oil drop under gravity.

When the electric field is not applied, the oil drops fall down due to gravity and its velocity increases. At the stage comes when the weight of oil drop becomes equal to viscous force then the oil drop falls down with constant velocity known as terminal velocity V1
When the oil drop attain terminal velocity then net downward force becomes equal to net upward force
i.e. W = F + U <-------- (i)
Since, W= weight of oil drop
W = m x g
W= ρ x v x g
W = ρ x$\frac{4}{3}$ πr3 x g
therefore, W = 4/3 πr 3ρg
where, ρ is the density of oil drop,
Since U is upthrust due to air
U = m x g
U = v x σ x g
U = 4/3 πr3σ g
where 'σ' is the density of air
And,
F= viscous force
i.e. F= 6πηrV1
Using the value of W, U and F in equation (i)
4/3 πr3ρ g = 6πηrV1 + 4/3 πr3 σ g
4/3 πr3ρ g - 4/3 πr3 σ g = 6πηrV1
4/3 πr3(ρ-σ)g = 6πηrV1 <----- (ii)
r2 = 9ηV1 / 2(ρ - σ)g
therefore, r = √9ηV1 / 2((ρ-σ)g <----- (iii)

By knowing the value of η, V1, ρ, σ, on RHS, the value of radius of oil drop can be determined.

Theory 2: Motion of oil drop under the effect of electric field

When the electric field is applied between the end of the plates in such a way that the force acting on the negative charge oil drops due to electric field must act upward and attain terminal velocity V2. So net upward force is equals to net downward force.
i.e. Fe + U = F + W <----- (iv)
where Fe is the force experienced by oil drops due to electric field i.e. Fe = Q.E
where 'Q' is charge of oil drop and 'E' is electric field strength.

Using these values in equation (iv) we get,
QE + 4/3 πr3 σg = 6πηrV2 + 4/3πr3ρg
or, QE = 6πηrV2 + 4/3πr3 (ρ -σ)g <------ (v)
Using equation (ii) in equation (v)
QE = 6πηrV2 + 6πηrV1
QE = 6πηr (V2 + V1)
i.e. Q = ( 6πηr (V2 + V1) / E ) √9ηV1 / 2((ρ-σ)g <----- (vii)

By knowing all the values of RHS of equation (vii) we can determine the value of charge of oil drop. And experimentally, its value was found to be 1.6 x 10-19 C.

NOTE: Q = ( 6πηr (V2 + V1) / E ) √9ηV1 / 2((ρ-σ)g , this is the charge of oil drops when the electric field is applied between the plates, the oil drops doesnot more upward and moves in downward.

#### Motion of an electron in an electric field

Let us consider a beam of an electron is projected horizontally with uniform velocity 'v' which is perpendicular to the electric field. Let 'V' be the potential difference and 'd' be the separation between the plates, then electric field strength is given by,
E = V/d <----- (i)
The force experienced by beam of an electron due to electric field is given by:
F = eE <----- (ii) , where e=electronic charge
Using equation (i) and equation (ii) we get,
F/e = V/d
F = Ve/d <----- (iii)
From Newton's second law of motion we have,
F = m x ay <----- (iv) [ax = 0]
Since, the electric field is in vertical. So, no any force experienced on horizontal direction. Therefore, acceleration in horizontal direction becomes zero.
From equation (iii) and (iv) we get,
eV/d = m x ay
therefore, ay = eV/md <----- (v)
Let, the beam of an electron is at point P(x,y) at time 't'.
Then horizontal distance and vertical distance is given by,
x = vt + 0
t = x/v <------ (vi)
Horizontal distance (x)= vt and
Vertical distance (y)= (1/2)ayt2 <------ (vii)
Substituting the value of ay and t in equation (vii) we get,
y = (1/2)eV/md (x/v)2
y = [(1/2) ev/mdv2] x2
y = kx2 <------ (viii), where k is constant
Equation (viii) is the equation of parabola. Hence the path followed by beam of an electron in an electric field is parabolic in nature.

Special Cases:
1. When the beam of an electron just passes the plates. Then x = D, so equation,
y = 1/2 (eV/mdv2) x2
y = kx2
y = [(1/2) eV/mdv2] D2 <----- (ix)
2. At time 't' the beam of an electron is between the plates.
then, t = D/v
3. Thus, the horizontal velocity at the point of emergence is given by,
vx = v <------ (xi)
Thus the beam of electron moves in a straight line as shown in figure and vertical velocity at the point of emergence is given by,
vy = ayt <------ (xii)
vy = (eV/md)(D/v) <------ (xiii)
Let, vR be the resultant velocity.
Then, vR = √(vx2 + vy2)
vR = √[vx2 + (eVD/mdv)2] <--------- (xiv)
Let, θ be the angle of deflection at a point of emergence then,
tanθ = vy / vx
= eVD/mdv/D/v
therefore, θ= tan-1 (eVD/mdv2)

##### J.J Thomsom method for determination of value of specific charge of an electron. (e/m)

Principle: If the beam of an electron is subjected in electric and magnetic field it experience the force by adjusting magnitude and direction of both field, the net force can be made zero.
It consist of evacuated glass tube with two electrodes A and C called anode and cathode. The high potential difference is applied between the electrodes. A narrow beam of electron emmitted from cathode and passes through the small hole anode and moves to the plate P and Q. Electric field is applied vertically downward along the plane of paper and magnetic field is applied perpendicular to plane of paper.

Working: When both field is not applied the beam of electron move in straight line and strike at a point S on the fluorscent screen. If beam of electron is applied separately in electric and magnetic field then beam of electron strike at a point S1 and S2 respectively.
When electric and magnetic field is applied simultaneously in such a way that the net force becomes zero. In this condition, the force experienced by electron due to electric field is equal to magnetic force experienced by beam of electron but in opposite direction. Therefore, beam of electron moves in a straight line without deflection.
Let E1, B and v be the velocity of beam of electron. Then, for no deflection condition,
Fe = Fm
eE = Bev
v = E/B <------- (i)
when strong electric field is applied between the electrodes then, potential energy P.E. at cathode is converted into gain in kinetic energy K.E. Therefore, P.E. at cathode = Gain in K.E. by electron.
P.E. = eV
K.E. = (1/2) mv2
e/m = v2 / 2V <------- (ii)
Equating equation (i) and (ii)
e/m = (E/B)2 x 1/2V
therefore, e/m = E2 / 2VB2 <------ (iii)
By knowing all value of RHS, the specific charge of an electron can be determined. Experimentally its value is observed as 1.81 x 1011 C/kg.

Some important derivation:
1. Explain Milikan's oik drop experiment.
2. Explain J.J thomsom method for determination of value of specific charge of an electron.
3. Explain motion of an electron in an electric field.

In case of any problem ask me in qustions section!!!!