The electron nd Xrays

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 due to the fluorescence of glass produced by the invisible rays coming from the cathode are called cathode rays. Since cathode ray is a stream of electrons and electromagnetic waves are beams of photons. Different forms of matter/energy, with totally different properties.The cathode ray is made by ionizing a cathode, releasing its electrons in a stream. Photons are exchanged between atoms constantly, at a rate of many billions of times per second.

It was found that cathode when an electric field was applied perpendicular to the length of the discharge tube the cathode rays were found to be moving in a direction opposite of electric field. This showed that the cathode rays consist of negatively charged particles i.e.; the electron.

Properties of cathode ray:

1. It is emitted from the surface of cathode electrode

2. It can penetrate thickness of matter

3. It ionize gas through which it passes

4. It is deflected by both electric and magnetic field because it carries negative charge.

5. It produce heat when it strike on the metal surface

6. It travel in straight line and cast shadow when object is place in the path

7. It is not 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.

Xrays

X – Rays are electromagnetic waves of short wavelength in the range of 10A° to 0.5

The discovery of X – rays goes to 1895 when W. Roentgen discovered while working with a discharge tube. He found that when the pressure in the discharge tube was reduced to about 10-3mm of Hg and the electric discharge was passed between the cathode and anode, the glass wall of the discharge tube behind the cathode began to glow with greenish yellow color. He also observed that a surface coated with barium platinum cyanide placed outside the discharge tube emitted light even when it was shielded from direct visible and ultraviolet light emitted by the discharge tube. After performing a series of experiment, Roentgen concluded that highly penetrating radiation of unknown nature are produced when a beam of fast moving electrons strikes a solid target such as tungsten. He named these radiations X – rays which led to the discovery of X – rays. Diagram of X – ray tube is:

Fig:

Different between X-ray and Ordinary light:

Ordinary light	x-ray
These are visible	These are invisible
They have heating effect	They have no heating effect
The wavelength of ordinary light range from 4 x 10-7m to 7.6 x 10-7m.	The wavelength of x-ray range from 10-9m to 10-12m.
It does not have high penetrating power. Ordinary light penetrates only transparent substance not opaque.	They have high penetrating power and power of transparency for opaque substance

The important properties of X – rays are:

(i) The X – rays are the electromagnetic wave of wavelength 10A° to 0.5°

(ii) The X – ray travels in vacuum with the speed of light ie. 3 * 108m/s.

(iii) They affect high penetrating power.

(iv) They have photographic effect.

(v) They are not deflected by electric and magnetic field.

X-rays are produced by the two methods

a. When fast moving electron having sufficient energy strikes the metal surface of high atomic number,it knock out the some electron from the inner orbit of the target metal due to vacancies are created so, to fill the vacant space electron from the higher energy level jump into these spaces emitting the radiation whose energy is equal to the difference in the energy of two orbits .thus obtained radiations by the heavy metals are X –rays.

b. When the fast moving electron strike the target they are heavily retarded by the coulomb repulsive force due to the charges of the electron of the atom .As a result retarding energy emits the radiation called X-rays.

Bragg’s law:

 Statement: when a monochromatic X-rays impinge upon the atom in the crystal lattice, each atom acts as source of scattering radiation of the same wavelength. The crystal acts as a series of parallel reflecting plane. Then the intensity of the reflected beam at certain angle will be maximum when the path difference between two reflected waves from two different planes is an integral multiple of λ

Explanation of law:

Let us consider a set of parallel plane of atom point at a spacing d between the successive plates. Let a narrow monochromatic X-ray beam of wavelength λ be incident on the first plane at a glance angle ɵ consider the ray PQ incident on the first plane. The corresponding reflected ray QR must be also be incident at the same angle ɵ to the plane. Since X-ray are must more penetrating then that of ordinary light there is only partial reflected at each plane The complete absorption take place only after penetrating several layers. Consider two parallel rays PQR and P’Q’R’ in the beam which reflected by two atom Q and Q’. is vertically below Q. the ray P’Q’R’ has longer path than the ray PQR, To compute the path difference between the two rays from Q draw normal QT and QS on P’Q’ and Q’R’ respectively. Then the path difference =TQ’ +Q’S=d sinɵ +d sinɵ= 2dsinɵ

Hence the two ray will reinforce each other and produce maximum intensity, if

2dsinɵ=nλ where n=1, 2, 3, 4,……………

The integer n gives the order of the scattered beam; λ is the length of X-ray used. This equation is called bragg’s law

It is originated when the fast moving electron strike the target, they are heavily retarded. As the electrons go on being retorted continuously, the frequency of radiation emitted also goes on changing continuously. Hence, continuous X-rays are obtained.

The characteristics spectrum of X – rays origin is explained using Bohr’s postulate i.e. when an electron jumps from higher shell to lower, it emits energy in the form of radiation. So, when the fast moving electron strikes a metal target having high atomic number the electrons can knock out electrons from the inner shells of the atom, thus creating vacant spaces for its origination.

 Some of the features of continuous spectrum of X – rays are:

(i) It consists of all possible wave length within a range.

(ii) The value of maximum intensity increases as potential increases.

Some of the features of characteristics X – ray are:

(i) The line of characteristics X – ray spectra occurs in various groups.

(ii) For each target material, there is minimum potential below which there is no radiation.

Production of X-rays:

X-rays are produce when fast moving electron are suddenly stopped by a solid target. A Coolidge tube is shown in figure.

Figure:

The tube is exhausted to the beast possible vacuum of the order of 10-5mm of Hg. The cathode consist of a tungsten filament (F) heated by a low tension battery. Thermionic electrons emitted by the filament are accelerated toward the target (T) by high P.D. maintained between F and T. the filament is placed inside a metal cup G to focus the electrons on the target. The target must be cooled to remove the heat generated in it by continuous electron- bombardment. The usual method is to mount the target material on a hollow copper tube through which cold water is continuously circulated. The target is made of material like tungsten or molybdenum having a high m.p. and high atomic no. metal with atomic no. give more energetic and intense X-rays when used as target. In the Coolidge tube the intensity and frequency of x-ray can be easily controlled.

The intensity of X-ray is depending upon the no. of electrons striking the target per second. The no. of the electrons give out by filament is directly propatational to its temperature, which can adjust by varying the current in the filament circuit. Therefore the intensity of X-ray varies with the filament current.

The frequency of X-ray emitted depends on the voltage between cathode and anode .let V be the accelerating potential across the tube. If e be the charge on the the electron the the work done on the electron in moving from cathode to anticathode is eV. Then the electron thus gain K.E. which is converted into x-ray when the electron strikes the target.

If υmaxbe the maximum frequencies of the x-ray produce

Then h υmax=eV

or, hC/λmin=eV

or, λmin=hC/eV  where C= velocity of light and h= plank’s constant .

Crystal diffraction

Let us consider a set of parallel plane of atom point at a spacing d between the successive plates. Let a narrow monochromatic X-ray beam of wavelength λ be incident on the first plane at a glance angle ɵ .consider the ray PQ incident on the first plane. The corresponding reflected ray QR must be also be incident at the same angle ɵ to the plane. Since X-ray are must more penetrating then that of ordinary light there is only partial reflected at each plane, The complete absorption take place only after penetrating several layers. Consider two parallel rays PQR and P’Q’R’ in the beam which reflected by two atom Q and Q’. is vertically below Q. the ray P’Q’R’ has longer path than the ray PQR, To compute the path difference between the two rays from Q draw normal QT and QS on P’Q’ and Q’R’ respectively. Then the path difference =TQ’ +Q’S=d sinɵ +d sinɵ = 2dsinɵ

Hence the two ray will reinforce each other and produce maximum intensity, if

2dsinɵ=nλ where n=1, 2, 3, 4,

The integer n gives the order of the scattered beam; λ is the length of X-ray used. This equation is called Bragg’s law.