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^{-19}C

The mass of an electron (M_{e}) = 9.1 x 10^{-31}
kg

The mass of proton (M_{p}) =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 V_{1}

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 πr^{3} x g

therefore, W = 4/3 πr ^{3}ρg

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 πr^{3}σ g

where 'σ' is the density of air

And,

F= viscous force

i.e. F= 6πηrV_{1}

Using the value of W, U and F in equation (i)

4/3 πr^{3}ρ g = 6πηrV_{1} + 4/3
πr^{3} σ g

4/3 πr^{3}ρ g - 4/3 πr^{3} σ g =
6πηrV_{1}

4/3 πr^{3}(ρ-σ)g = 6πηrV_{1}
<----- (ii)

r^{2} = 9ηV_{1} / 2(ρ - σ)g

therefore, r = √9ηV_{1} / 2((ρ-σ)g <----- (iii)

By knowing the value of η, V_{1}, 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 V_{2}. So net upward
force is equals to net downward force.

i.e. F_{e} + U = F + W <----- (iv)

where F_{e} is the force experienced
by oil drops due to electric field i.e.
F_{e} = 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 πr^{3} σg =
6πηrV_{2} + 4/3πr^{3}ρg

or, QE = 6πηrV_{2} +
4/3πr^{3} (ρ -σ)g <------ (v)

Using equation (ii) in equation (v)

QE = 6πηrV_{2} +
6πηrV_{1}

QE = 6πηr (V_{2} +
V_{1})

i.e. Q = ( 6πηr (V_{2} +
V_{1}) / E ) √9ηV_{1} /
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 (V_{2} +
V_{1}) / E )
√9ηV_{1} / 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 a_{y}
<----- (iv) [a_{x} = 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 a_{y}

therefore, a_{y} = 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)a_{y}t^{2}
<------ (vii)

Substituting the value of
a_{y} and t in equation
(vii) we get,

y = (1/2)eV/md (x/v)^{2}

y = [(1/2) ev/mdv^{2}]
x^{2}

y = kx^{2}
<------ (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/mdv^{2})
x^{2}

y = kx^{2}

y = [(1/2)
eV/mdv^{2}]
D^{2}
<----- (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,

v_{x} = 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,

v_{y} =
a_{y}t
<------ (xii)

v_{y} =
(eV/md)(D/v)
<------ (xiii)

Let,
v_{R}
be the
resultant
velocity.

Then,
v_{R}
=
√(v_{x}^{2}
+
v_{y}^{2})

v_{R}
=
√[v_{x}^{2}
+
(eVD/mdv)^{2}]
<--------- (xiv)

Let, θ
be the
angle of
deflection
at a
point of
emergence
then,

tanθ =
v_{y}
/
v_{x}

=
eVD/mdv/D/v

therefore,
θ=
tan^{-1}
(eVD/mdv^{2})

##### **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 S_{1}
and S_{2} 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 E_{1}, B and v be the velocity of beam of
electron. Then, for no deflection condition,

F_{e} = F_{m}

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) mv^{2}

e/m = v^{2} / 2V <------- (ii)

Equating equation (i) and (ii)

e/m = (E/B)^{2} x 1/2V

therefore, e/m = E^{2} / 2VB^{2}
<------ (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 10^{11} 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.

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