CLASS 12 PHYSIC NOTES ON ELECTRON

 ELECTRON CLASS 12 NOTES 

Electrons and photons both respond to the electromagnetic force, but they differ in:

	Electron	Photon
Mass (m)	9×10-31 kg	Zero
Charge (q)	−1.6*10-19 Coulombs	Zero
Specific charge (q/m)	1.7 6 * 1011C/kg	Zero
Speed	May be zero, or less than speed of light.	In vacuum speed is equal to light but slight less in denser medium.

 

Besides electrons, other particles of matter like protons, positrons and mesons also

 carry an electric charge, and have a position and velocity in space. They differ in

 many ways from each other, but their electric charge also makes them respond 

to the electromagnetic force.

Millikan’s experiment

The principle of the working of Millikan oil drop experiment is to measure the 

terminal velocity of the charged oil droplets under the effect of

i. gravity alone and

ii. combined effect of gravity and electric field.

Knowing the terminal velocity we can easily determine the charge on the droplets

 and it is found that the charge on the charge droplet is integral multiple of the basics charge.

 

Experimental arrangement

the apparatus consist of two optically plane parallel plate

 P and Q separated by insulating rods of glass. The

 lower plate Q is earthen while the upper plate P can

 be charged to a positive potential of order of 1000V by

 high tension battery. This observation chamber is situated

 in a bigger chamber R which is completely surrounded by

 a constant temperature bath of oil E. Tiny drop of the 

heavy non- volatile oil are spray into the chamber

 by an atomizer A. some of the oil drops are enter the 

space between the plate that the pin hole H in the top of 

the plate P. These are charged, due to the fractional effect 

at the nozzle of the atomizer. The air between the plates 

can ionize by allowing X-ray to pass through it. Then 

the drops may pick up additional charges. The drop

 are illuminated by from a arc lamp L the oil drop 

observed by the means of short-focus telescope T

 provided with the millimeter scale in eyepiece.

 

Theory 1

Motion of oil drop under gravity only

If no electric field applied then oil drop fall due to gravity and the velocity of oil 

drop goes on increase and after certain time the viscous force on the oil drop is

 equal to resultant weight of oil drop at that condition the oil drop move with 

constant velocity v is called terminal velocity.

Let r be the radius of the oil drop, m be the mass of oil drop, ρ be the density of

 oil drop and σ be the density of air.

Then the volume of the oil drop (V)=43∏r3$\frac{4}{3}{\prod }^{{\mathrm{r}}^3}$

Now the weight of the oil drop (W)=mg=43∏r3ρ$\frac{4}{3}{\prod }^{{\mathrm{r}}^3}\rho$g (ρV=m)

Let U be the up thrust force which is equal to the wt. of the air displace by oil

 drop and given by U==43∏r3σ$\frac{4}{3}{\prod }^{{\mathrm{r}}^3}\sigma$g

 And the force which is pulling up the oil drop with velocity ‘v’ in upward 

direction is viscous force and given by F=6∏ηrv1

After certain time, the viscous force on the oil drop is equal to resultant weight 

of oil drop at that condition the oil drop move with constant velocity v1.

Then, F + U =W

Or, 6∏ηrv1+43∏r3σ$\frac{4}{3}{\prod }^{{\mathrm{r}}^3}\sigma$g = 43∏r3ρ$\frac{4}{3}{\prod }^{{\mathrm{r}}^3}\rho$g

Or, 6∏ηrv1= 43∏r3ρ$\frac{4}{3}{\prod }^{{\mathrm{r}}^3}\rho$g- 43∏r3σ$\frac{4}{3}{\prod }^{{\mathrm{r}}^3}\sigma$g (F=W-U)………..1

Or, 6∏ηv1= 43∏r2$\frac{4}{3}{\prod }^{{\mathrm{r}}^2}$g(ρ$\rho$ - σ$\sigma$)

Or, r =6πηv143πg(ρ−σ)−−−−−−−−√$\sqrt{\frac{6\pi \eta {\mathrm{v}}_1}{\frac{4}{3}\pi \mathrm{g}\left(\rho -\sigma \right)}}$=9ηv12g(ρ−σ)−−−−−−−√$\sqrt{\frac{9\eta {\mathrm{v}}_1}{2\mathrm{g}\left(\rho -\sigma \right)}}$ on knowing 

the value of oil drop m, density of oil drop ρ and density 

of air σ we can find the radius of oil drop.

Theory 2

Motion of the oil drop under the influence of the 

electric field:

When strong electric field is applied between the two plate P and Q then the oil drop moves P to Q i.e. in downward or earthen with velocity v2 and experience electrostatic force of attraction to the positive plate P.

At equilibrium condition

The sum of electrostatic force of attraction and up

 thrust force is equal sum of viscous force and weight 

of oil drop i.e. Fe+U=W+F’

Or, Fe=(W-U)+F’……………….2 

From 1 and 2 we have

Eq=F + F’

Or, Eq=6∏ηrv1+6∏ηrv2

Or, q=6∏ηr(v1+v2)/E

Or, ne=6πηr(v1+v2)E$\frac{6\pi \eta \mathrm{r}\left({\mathrm{v}}_1+{\mathrm{v}}_2\right)}{\mathrm{E}}$

On putting the value from above theory i.e. theory 1

We get

 

Or, e=6πη(v1+v2)Ex9ηv12g(ρ−σ)−−−−−−−√$\frac{6\pi \eta \left({\mathrm{v}}_1+{\mathrm{v}}_2\right)}{\mathrm{E}}\mathrm{x}\sqrt{\frac{9\eta {\mathrm{v}}_1}{2\mathrm{g}\left(\rho -\sigma \right)}}$. …………..3

Hence knowing all the right side value of equation 3 we can determine the value of the charge by millikan’s oil drop method.

Cathode rays: The rays which glow the glass tube because of the fluorescence of glass produced with the

aid of using the invisible rays coming from the cathode are known as cathode rays. Since

cathode ray is a move of electrons and electromagnetic waves are beams of photons. Different

types of remember/energy, with completely specific properties.The cathode ray is made with

the aid of using ionizing a cathode, freeing its electrons in a move. Photons are exchanged among atoms constantly, at a price of many billions of instances in line with second. It turned into located that cathode whilst an electric powered area turned into carried

out perpendicular to the period of the release tube the cathode rays had been located to be transferring in a route contrary of electrical area. This confirmed that the cathode rays encompass negatively charged debris i.e.; the electron. Properties of cathode ray: 1. It is emitted from the floor of cathode electrode 2. It can penetrate thickness of remember three. It ionize fueloline thru which it passes 4. It is deflected with the aid of using each electric powered and magnetic area as it incorporates bad fee. 5. It produce warmth whilst it strike at the metallic floor 6. It tour in instantly line and solid shadow whilst item is vicinity withinside the direction 7. It isn't always a electromagnetic wave.

Discharge of electricity through gas:

Discharge tube  is a strong glass tube having diameter 0.04m and length 0.5mand

 able to provide low pressure and conduct electric through gas.

 

a.

At a pressure of about 10mm of Hg: when the pressure of discharge tube is reduced 

to 10mm of Hg. Discharge of gas take place inside the tube in the form of luminous steaks between the two electrode i.e. cathode and anode called blue stremes with cracking sound.            

 b.

At a pressure of 5mm of Hg: when the pressure is reduce to 5mm of Hg then the 

blue steaks broaden out into a luminous column which is bright and steady. The

 luminous colur which is bright and steady called Geisster’s discharge. Color of

 discharge depend upon the nature of gas which is fill in the discharge tube.

 

c.

At pressure 2mm of Hg: when the pressure of discharge is reduced to 2mm of Hg 

then a long luminous column appear between the two electrode i.e. anode and

 cathode called positive column. Color of discharge depend upon the nature of gas 

inside the discharge tube.

 

d.

At 1mm of Hg: when the preasure is reduce to 1mm of Hg then positive column

 seprated from cathode and move towards the anode.cathode beings to glow blue luminous called negative glow then then a space appear between positive column 

and negative glow is called faraday’s dark’s space.

e.

At the pressure of 0.5mm of Hg: when the pressure is reduce to0.5mm of Hg then

 the positive column get shorten and the negative glow move away from cathode and another glowing space again appear on the cathode called cahtod glowing and sepratin 

of the distance between glowing and negative glow called crook’s dark space.

              

f.

At the pressure of 0.05mm of Hg: when the pressure is reduce to 0.05mm of Hg 

then positive column get more and more shorten and breaks into alternative bright

 and dark like disc shape called striations.

 

g.

At the pressure 0.01mm of Hg: when the pressure is reduced to 0.01 then the striation , positivecloumn, negative column and cathode glow all are disappear and the tube is

 filled with crook’s dark space and this stage luminous ray come out of cathode called cathode ray and travel from cathode to anode.

Thomson’s experiment:

Experimental arrangement of Thomson’s experiment for determination of specific 

charge (e/m) of electron is given below. It consist an evacuate discharge tube having cathode ( C) and anode (A) on the same side. When high p.d is set up between the cathode and anode then the emission of electron take place from cathode and passing through the hole of anode and finally passes between two parallel plate P and Q. when

 the uniform magnetic field is applied horizontally to the plate then electron deflected

 of electron beam take place in upward direction but if magnetic field is applied perpendicular then electron beam get deflected in downward direction

Theory:

since the beam of electron passes between two plate P and Q on which uniform 

magnetic field is applied i.e. field is normal to the beam then the electron move in the circular path with velocity v in the magnetic field B and electron experience force 

called magnetic force Fm=Bev

Since field if perpendicular then the electron move in the circular path having radius 

in that condition magnetic force is equal to centripetal force

Magnetic force = centripetal force

Bev=mv2/r

Or, e/m=v/rB

On knowing the value of v, r, and B we find the value of e/m.

Measurement of velocity of electron:

When both the magnetic field and electric field is applied at the same time then the 

beam don’t get deflected and strike at point S. at that condition

Electrostatic force =magnetic force

Bqv =Eq

Or, v=E/B

Measurement of r

When the electron come out from the uniform magnetic field it get deflected and 

strike at the point S’ instead of S. let ɵ be the deflection angle between the original 

path and deflected path as shown in figure

Figure

If we draw tangent from P and Q point of circle at O then from ∆PSS’

We have, tanɵ=SS’/PS

Since ɵ is very small then tanɵ≈ɵ=y/d …………..1 (from figure)

We also have tanɵ≈ɵ=arc PQ/length of OQ

Or, ɵ=L/r…………2 (from figure)

From eq. 1 and 2 we have

 L/r = y/d

Or, r = (L x d)/y…………..3

On knowing the value of L, d, and y. we find the value of r. J.J. Thomson’s find the value of e/m =1.77 x 1011C/kg.

 

Cross – field:

The electric filed and magnetic fields are set up perpendicular to each other so that

 the force exerted by electric and magnetic field on the moving charge (electrons)

 inside it cancel each other is called cross – field.