2nd Year Chemistry Physics (Ph 211) - Atomic Physics PDF

# 2nd Year Chemistry Physics (Ph 211) - Atomic Physics ## Contents ### Chapter 1 - Page - 1.1 Determination of the electronic charge - 1 - 1.2 Determination of the electron mass - 5 - 1.3 Determination of the electron size - 6 - 1.4 Experimental determination of the speed of an electron - 8 - 1.5...

# 2nd Year Chemistry Physics (Ph 211) - Atomic Physics ## Contents ### Chapter 1 - Page - 1.1 Determination of the electronic charge - 1 - 1.2 Determination of the electron mass - 5 - 1.3 Determination of the electron size - 6 - 1.4 Experimental determination of the speed of an electron - 8 - 1.5 Variations of mass with velocity (Relativistic effects) - 9 - 1.6 The Quantum Theory - 10 - 1.7 The Bohr Theory - 11 - 1.8 Calculations the various parameters of Bohr's model - 13 ### Chapter 2 - 2.1 Electromagnetic waves - 16 - 2.2 Vis-Uv Spectrometer - 17 - 2.3 Х- гау - 19 - 2.4 X-ray production - 19 - 2.5 X-rays spectra types - 21 - 2.5.1 The continuous spectra - 21 - 2.5.2 The characteristic spectra - 22 - 2.6 X-ray absorption - 23 - 2.6.1 Linear absorption coefficient - 24 - 2.6.2 Mass absorption coefficient - 25 - 2.7 X-ray scattering - 25 - 2.7.1 Coherent and incoherent scattering - 26 - 2.8 X-ray Diffraction - 27 - Solved problems - 29 ## 1.1 Determination of the electronic charge Millikan (1910) carried out an experiment to determine the electronic charge as shown in the fig. 1.1. A pair of insulated, parallel, metal plates is kept charged by means of a potential source V. Through a small hall in the upper plate, tiny droplets of oil fall. These come from a spray of oil vapor shot into the space above the plates by an atomizer. The oil droplets may be charged from the spraying. They can also pick up positive or negative ions from the air and become charged. The air is kept ionized for this purpose by occasional bursts of X-rays, or any rays from a radioactive source. Through the microscope, the oil droplet may be observed falling.  If the droplet is charged and the electric field between the plates is now applied by closing the switch, then, the droplet's motion may be accelerated, slowed or even reversed depending on the sign and amount of the charge. By adjusting the field, the droplet can be maintained motionless. Under these conditions, the gravitational force urging the droplet down is just balanced by the electric force attracting it upwards. From these conditions, the charge on the droplet may be calculated. The gravitational force $F_g$ = $mg$ = electric force $F_e$ $mg$ = $EQ$ = $Q \frac{V}{d}$ $Q$ = $mgd/V$ where: - m is the mass of the droplet, - V is the difference of potential, - d is the separation of the plates, - g is acceleration of gravity, and - Q is the charge on the droplet In practice, the determination of the mass of the droplet gave Millikan great difficulty. The droplets used appeared only as points of light in the microscope, and no estimate of their size could be obtained. The difficulty was solved as follows: with electric field turned off, the droplet fell through the air. Since it was very small, the air's viscosity reduced its velocity from an accelerated one (in free fall) to a constant one (in a viscus medium). This latter fact stems from Stocke's law, which states that the retarding force exerted by a viscus medium on sphere is proportional to its velocity, $F_D$ = 6 πηr v where: - $F_D$ is the viscus force exerted on the sphere in Newton - η is the coefficient of viscosity - r is the radius of the sphere in (meter) - v is the velocity in m/sec As the body falls from rest, the constant gravitational force is gradually balanced by the viscus force $F_D$ increases as the velocity of the particle increases. When the two forces are balanced as shown in fig. (a), the velocity of the particle remains constant. The gravitational force is now given by: $F_g$ = $mg$ = $\frac{4}{3}πr^3 (ρ_o-ρ_a)g$ where: - $ρ_o$ is the density of the oil in(kg/m³) - $ρ_a$ is the density of the air in kg/m³ When the two forces are balanced, we obtain from equations 2 & 3: $r$ = $\sqrt{\frac{9 η v}{2g(ρ_o-ρ_a)}}$ Thus by measuring the velocity, Millikan could calculate the radius. From the volume of the droplet, its mass could easily be found. Millikan could determine the mass of his droplet. To balance them precisely with equal gravitational and electric forces was experimentally very difficult, however, so Millikan used a different approach as follows: When the droplet fell freely, its terminal velocity took place under the droplet weight given by equation (3). Because q is negative and E is directed downward, this electric force is directed upward, as shown in fig. (b). If this force is sufficiently great, the drop moves upward and the drag force $F_D$ acts downward. When the upward electric force EQ balances the sum of the gravitational force and the downward drag force $F_D$, the drop reaches a new terminal speed v' in the upward direction. $EQ$ = $\frac{4}{3}$πr² (ρ_o - ρ_a) g + $F_D$ $EQ$ = 6 πηr v’ + 6πηγν’ $EQ$ = 6 πη r (v + v’) Q = $\frac{6πηrd}{(v + v’)}$ where: - v is the downwards velocity and - v’ is the upwards velocity Using this equation, Millikan measured the charge on many thousands of drops. They found the droplet charge Q could be always represented: Q=ne where: - n is an integers, 1, 2, 3 ... etc. - The value of e obtained from such exp. was found to be equal to: e = 1.6 x 10-19 C.  The gravitational force = electric force $f_g$ = $f_e$ $mg$ = $EQ$ = $Q \frac{V}{d}$ $Q$ = $\frac{mgd}{V}$ When electric filled turn of (see figure(a)) $F_v$ = 6πηr$V_d$ $F_v$ = $f_g$ 6πηr$V_d$ = $mg$ = $\frac{4}{3}$πr³ (ρ_o - ρ_a) g $r$ = $\sqrt{\frac{9 ηV_d}{2g(ρ_o-ρ_a)}}$ When electric field turn on (see figure (b)) $f_e$ = $f_g$ + $f_v$ $EQ$ = $\frac{4}{3}$πr² (ρ_o - ρ_a) g + $F_v$ = 6 πrn $V_d$ + 6πηγν' $EQ$ = 6πrn ($V_d$ + $ν_u$) : Q = 6πrn ($V_d$ + $ν_u$) = 6πrn d ($V_d$ + $ν_u$) $E$ $V$ ## 1.2 Determination of the electron mass A stream of electrons is accelerated by an electric field maintained between a pair of slits by a different of potential V as shown in following fig. 1.2. The stream then enters magnetic field of flux density B whose lines of force are normal to the motion of the electrons. In the magnetic field, the electron follows a circular path, whose radius R may be measured by the positioning of the electron detector D.  During the circular part of their motion, a magnetic force on the electrons causes them to be accelerated towards the center, hence the circular trajectory. We note that the force exerted by the magnetic field will be proportional to the field, to the charge & its velocity: F = Bve where: - F is the force exerted on the charged particle in Newton, - B is the magnetic flux density in weber/m², and - v is the velocity in (m/sec) We-are now ready to calculate the charge to mass ratio of electrons, when they pass through the slits they are accelerated by the electronic field to a velocity v such that the K.E. of the electrons is equal to the energy given them by the field. $κ.Ε$ = $\frac{1}{2} m v²$ = $eV$ where: - m is the electrons mass in kg and - V is the different potential in Volt. In the magnetic field, the magnetic force exerted on the electrons produces a centripetal acceleration ($\frac{v^2}{R}$) or: $Bev$ = $\frac{mv^2}{R}$ R is the radius of curvature in meter. From equation (3), we have v = $\frac{BeR}{m}$ Substitution in equation (2), we find: m = $\frac{e}{m}$ = $\frac{R^2B^2}{2V}$ ## 1.3 Determination of the electron size As was the case with the mass, the size of the electron is so small that no direct method of measuring it exists. Suppose the electron is a small sphere of radius ($r_e$). magnative force = centripetal force $BVe$ = $\frac{mv^2}{R}$ $m$ = $BeR$ $\frac{1}{2} m v²$ = $eV$ $v²$ = $\frac{2eV}{m}$ $v$ = $\sqrt{\frac{2eV}{m}}$ $m²$ = $B^2e^2R²$ $\frac{2eV}{m}$ $m$ = $\frac{B^2R²e}{2V}$ $m$ = $B^2R²$ $e$ = $\frac{B^2R²}{2V}$ ## 1.4 Experimental determination of the speed of an electron It was J.J. Thomson who showed that the speed of an electron may be obtained by a comparatively simple experiment. Thomson found that if an electron stream is passed through a region in which magnetic and electric fields are superposed in such a manner as to produce deflections in opposite directions. The field can be so adjusted that the effect of one just cancels the effect of magnitude of the other. The electron for which the fields are thus balanced would go through undeflected (fig. 1.3). For such a balance the magnitude of the force of (Ee) of the electric field E on the electron must be equal to the magnitude of the force of the magnetic field (Bev): $E_e$ = $Bev$ And the fields must be at right angles to each other, then $E$ = $Bv$ v is the velocity (m/sec) is of the electron where E is measured in volts/mand B is measured in web/m². This method is called the method of balanced deflection.  ## 1.5 Variations of mass with velocity (relativistic effects) Although all slow moving electrons are found to have the same mass ($m_e$ = 9.1 x 10-31 kg), when a high voltage applied on the electrons, i.e. the velocity of electrons increase their mass is found to increase (all masses are known to increase with velocity). It is with electrons that this effect is most easily observed, however, & electrons have observed in many experiments designed to test the relativistic mass effect. It is well known that the total energy E of a body of mass is given by the relation: $E$ = $mc^2$ E is the energy in joules and c is the velocity of light in vacuum (3 x 10^8 m/sec). From this expression we can obtain 'm' in terms of the body's total energy that is as a function of its velocity. The mass at any velocity v is given by: $m$ = $\frac{m_o}{\sqrt{1.-v^2/c^2}}$ at v = 0, m becomes $m_o$, the rest mass. ## 1.6 The Quantum Theory The old quantum theory was born in 1900 when Max Plank announced his theoretical derivation of the distribution law for black body radiation. He showed that the results of the experiment on the distribution of energy with frequency of radiation at a given temperature can accounted from by postulating that the vibrating particles of matter do not emit or absorb light continuously, but instead of this it emits or absorbs light in discrete quantities of magnitude hv) which has a magnitude of $h$ = 6.547 x 10^-27 erg.sec In the year 1905 Einstein suggested that the quantity of radiant energy hu can be sent out like a particle. The name like quanta or photon is applied to such portion of energy. Einstein discussed also the photoelectric effect and he found that when light falls on the metal photoelectrons are emitted from it. The maximum speed of these photoelectrons is independent on the intensity of light but only on its frequency. Einstein pointed out that this is to be expected from the quantum theory. The process of photoelectric emission involves the conversion of energy(hu) into the kinetic energy of the photoelectron + the energy required to remove an electron from the metal. ## 1.7 The Bohr Theory In the year 1911, Rutherford discovered the nuclear constitution of the atom. It was clearly evident that the laws of classical machines & electromagnetic theory could not apply to Rutherford hydrogen atom. According to the theory, the electron in hydrogen atom attracted towards the nucleus by the inverse square coulomb's law. The electron can describe an elliptical or circular orbit & the acceleration of the charged particles would lead to the emission of light. With emission of energy, the radius of the orbit would diminish & the mechanical frequency would change, so the emission of radiant energy will be continued until the electron had fallen into the nucleus; but this is not the case because the emitted light has not wide range of frequencies, but instead it has sharp spectral lines of sharply defined frequencies, so Bohr made a modification of Rutherford's theory involving two postulates that were simply acknowledgements of experimental facts. i) The first postulates This postulate is based on the fact of spectral emission, and stated that only certain energy levels could be occupied by the carpeting electrons in the atom. The simplest way of describing these quantized cherry levels is to state the quantized angular moments of the electron corresponding to them, thus the energy of an orbiting electron is quantized such its angular momentum is : L = Iw = nh $\frac{mR^2.v}{R}$ = nh $\frac{h}{2π}$ -> $I$ = $mR^2$ $ω$ = $\frac{v}{R}$ where: - n is a quantum number 1, 2, 3, ....., - ħ is h/2 π, - I is the moment of inertia of electron with respect to the nucleus = mR², - R is the radius of the orbit, and - w is the angular velocity of the electron radius/sec. = v/R ii) The second postulates It acknowledges that the orbiting electron did not emit any energy unless it change orbit (that is, energy level). When an electron dit change from a higher energy orbit E₂ to a lower orbit E1, its loss of energy was emitted as a photon, thus: E₂ - E₁ = hv Those postulates are at odds with the classical theory of mechanics, which as we have seen allows a stable orbit for any radius of curvature and energy. ## 1.8 Calculations the various parameters of Bohr's model The two postulates of Bohr provide a new picture of the atom. We shall now calculate the various parameters of Bohr's model of the atom to determining what photon energies it predicts will emerge spectral emission. i) Radius The second postulate provides for the stability of This is provides by balancing the electrostatic force actii it to the nucleus by the centripetal acceleration of its orbital motion. The expression for this balance was given by: ($Ze$)e $\frac{mv^2}{EoR^2}$ = $R$ where; - Ze is the charge of the nucleus in Coulombs (Z being the atomic number = 1 for H) and - Eo is permittivity. This gives R = $\frac{Ze^2}{Eo mv^2}$ The first postulate states that the energy of the orbits is quantized, or atomic (that is, the energy of would be a function of a number n = 1,2,3,.... etc.) since we may have atom of a quantity, 1 or 2, or 3, but not a fractional number. The precise values of the energy levels may be found from an examination of the spectra lines (as Bohr did). The result is written down more slowly, however, in terms of angular momentum. Bohr thus gave as the mathematical expression of this second postulates. L = Iw = nh $\frac{mR^2.v}{R}$ = nh By substituting eq. (6) in eq. (4), we obtain: R = Con²² $\frac{η^2h^2}{mZe^2}$ In eq. (7), Eo, ħ m and e are known(constant) we may therefore calculate R for various atoms (values of Z) and quantities number n. By way of example, we consider the hydrogen atom, here Z = 1. When n is also put equal to 1 and the values of the other constants substituted in equation (7), we obtain the values 0.53 A This corresponds very well with the radius of about 2.7 Å. Obtained for the diameter of the hydrogen molecule obtained by mean-path measurements. ii) Tangential velocity By substituting equation (7) in equation (6), we obtain: V= $\frac{Ze^2}{Ερnh}$ For the first Bohr orbit of the hydrogen atom, we substitute Z=1 and n = lalong with the other constants to obtain the value 2.18 x 10^8 m/sec, in the second the velocity is 1.09 x 10^6 m/sec, and so on, the velocities in succeeding orbits (decreasing in the ratios 1, $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$ , this is also indicated by appropriate vectors. We note that although the highest velocity of an electron in the first orbit, its value of 2.18 x 10^8 m/sec. is only about 1/150 the velocity of light, and so the relativistic correction for the mass of the electron should be taken into account. ii) Angular velocity By substituting equations (6) & (7) for R and v in the relation w = v/R, we obtain: $ω$ = $\frac{mz^2e^4}{εo^2h^2}$ W = $\frac{V}{R}$ For the first Bohr orbit in hydrogen substitutions of the various constants gives a value for the angular velocity of 4.14 x 10^16 rad/sec. In the second orbit, the angular velocity is 5.17 x 10^15 rad/sec, and so on, the rate of rotation decreasing quickly in the ratio 1, 1/8, 1/27, 1/64, ...... The ultimate value of n is for this last orbit of the electron, in the hydrogen stem is infinite, and the tangential and the angular velocities zero. The electron is no longer associated with nucleus. ## 2.1 Electromagnetic waves If we could freeze an electromagnetic wave at a given instant of time propagation would be as shown in following figure. Let us summarize the properties of electromagnetic waves: 1- Electromagnetic waves travel through empty space at the speed of light c 2-The components of the electric E and magnetic fields B of plane electromagnetic waves are perpendicular to each other and perpendicular to the direction of wave propagation. 3- Electromagnetic waves obey the principle of superposition. 4- The magnitudes of E and B in empty space are related by the expression E/B = c.  Electromagnetic waves transport energy through empty space, stored in the propagating electric and magnetic fields. The various types of electromagnetic waves are listed as: | Frequency, Hz | Wavelength | |---|---| | 10^22 | 1 pm| | 10^21 | | | 10^20 | | | 10^19 | Gamma rays | | 10^18 | | | 10^17 | X-rays | | 10^16 | 1 nm | | 10^15 | Ultraviolet| | 10^14 | 1 μm | | 10^13 | Visible light| | 10^12 | Infrared | | 10^11 | 1 mm | | 10^10 | Microwaves | | 10^9 | 1 cm | | 10^8 | 1 m | | 10^7 | TV, FM | | 10^6 | Radio waves | | 10^5 | AM | | 10^4 | 1 km | | 10^3 | Long wave | ## 2.2 Vis-Uv Spectrometer The fundamental properties of electromagnetic waves are the same, whether the frequency approaches zero or approaches infinity. To detect these waves, however we must use various methods that depend on the frequency, such as antennas for low frequencies, the human eye for intermediate frequencies, scintillating crystals for high frequencies. We show in fig. (2.1) a highly schematic diagram of a spectrometer suitable for use in the visible and ultraviolet region of the spectrum. The spectrometer is consisted of an arrangement of the following: a- A white source of light is focused by lens (1) on to a narrow slit arranged perpendicularly to the plan of the paper. The beam of light then made into a parallel by lens (2). b- Sample container: it is filled with a substance under investigation to obtain the spectrum of it. After the beam is passing through the sample, it is separated into its constituent frequencies by a prism and is then focused on to a photographic plate by lens (3).  If the sample container is empty) the photographic plate should ideally show an even blackening over the whole range of frequencies covered from point A to point B. If the sample space is filled with a substance having only two possible energy levels E₁ and E2 then the photographic plate will show a blackening at all points expectat frequency v= (E2-E₁)/h . Since energy at this frequency will have been absorbed by the sample in raising each molecule from state one to state two. If there are many possible energy levels E1, E2, E3,.........,Ej,Ek, .....it's available to the sample, a series of absorption lines will appear on the photographic plate at frequencies given by the general equation: v =(Ε; - Εκ )/h ## 2.3 X-ray X-rays have wavelengths in the range from approximately 10-8 m to 10-12 m The most common source of X-rays is the stopping of high-energy electrons upon bombarding a metal target. X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. Because X-rays damage or destroy living tissues and organisms, care must be taken to avoid unnecessary exposure or overexposure. X-rays are also used in the study of crystal structure because X-ray wavelengths are comparable to the atomic separation distances in solids (about 0.1 nm). ## 2.4 X-ray production X-ray production occurs whenever electrons of high energy strike a heavy metal target, like tungsten or copper. When electrons hit this material, some of the electrons will approach the nucleus of the metal atoms where they are deflected because of their opposite charges (electrons are negative and the nucleus is positive). This deflection causes the energy of the electron to decrease, and this decrease in energy then results in forming an X-ray. X-rays are also produced when electrons make transitions between lower atomic energy levels in heavy elements. X-ray tube consists of :- 1. an evacuated chamber with a tungsten filament at one end of the tube, called the cathode (fig. 2.2) 2. a metal target at the other end, called an (anode • Electrical current is run through the tungsten filament, causing it to glow and emit electrons. • A large voltage difference (measured in kilovolts) is placed between the cathode and the anode, causing the electrons to move at high velocity from the filament to the anode target.  ## 2.5 X-rays spectra types X-rays spectra which produced by the X-ray tube are two kinds :- ### 2.5.1 The continuous spectra It consists of a continuous range of frequencies up to the maximum possible for a given voltage between cathode and anode. According to modern theory, the continuous spectrum is produced by deceleration of incident electrons through inelastic collisions. with atoms of the target material. Energy lost in each such collision (a-single-electron-may-undergo-a-series-of--collisions before finally losing all its energy-or-coming-at-rest) is emitted as an X-ray photon corresponding energy. The X-ray spectrum produced in this way will thus consist of continuous range of range of energies between the; limits of zero and the maximum energy electron. ### 2.5.2 The characteristic spectra The characteristic spectrum is consisting of one or more narrow bands of frequency characteristic of target material, resulting from disturbances within the atom of the target produced by the bombarding electrons as in fig. 2.3. They would then appear as peaks of considerable intensity superposed on the continuous background of general radiation. Characteristic or line spectrum is produced whenever the electrons have energies in excess of a critical excitation energy which varies from element in the target. Hence, incident electrons have sufficient energy to excited electrons from the inner sub shells of the target atoms to higher energy orbital levels. Such excitation can take place only to a vacant energy level. The excited atom is unstable and regains its initial ground state by transfer of high energy, each such transmition releases energy which is then emitted as an X-ray photon.  ## 2.6 X-ray absorption When an X-ray passes through any matter it is progressively attenuated, or reduced in intensity, as a consequence of a complex series of interactions between the X-ray photons and the atoms of the attenuating medium. X-ray suffer intensity losses a combined result of several different processes as follows: a) They may be photoelectrically absorbed by interaction with absorbing atoms, leading to either: i) The emission of photoelectron, or ii) The emission of X-ray characteristic spectrum of the absorbing atoms. b) They may be scattered either: i) Coherently, with no loss in energy, or ii) Incoherently, with a small energy loss and hence a slight increase in wavelength. ### 2.6.1 Linear absorption coefficient Consider an infinitely thin layer of thickness of the absorbing material, and suppose that the intensity I of a monochromatic incident beam of X-ray is reduced by dl on passing through dL. The intensity loss dI will be proportional to the absorbing thickness dL, i.e dI a- IdL dI = -µ I dL where u is a constant called the linear absorption coefficient. It incorporates the combined effects of all photoelectric and scattering process i.e. μ = τ+ σ $\frac{dI}{I}$ = -µdL $I_L$ = $I_₀e^{-µL}$ ### 2.6.2 Mass absorption coefficient The extent of absorption depends on the density p of the absorbing medium and the path length L in a given system. The absorption equation can be written: $I$ =$I_₁$ = $I_₀e^{-(µ/p) PL}$ where (µ/p) is a constant is known as the mass absorption coefficient. ## 2.7 X-ray scattering The scattering coefficient include two terms, corresponding to coherent scattering (Rayleigh scattering) and incoherent scattering (Compton effect). In fact: σ = Zf² + (1-f²) In which the first term describes the extent of the coherent scatter and the second the incoherent scatter. f is the atomic scattering factor which known as the ratio of the amplitude of the wave scattered by an atom to that scattered by a free electron. Scattering by an aggregate of atoms, as in a crystal structure, is expressed by the structure, is expressed by the structure factor F, which is the integral of the atomic scattering factors (taking phase differences into account). ### 2.7.1 Coherent and incoherent scattering Coherent scattering arises when an X-ray photon collides with an atom and is deviated without loss in energy. An electron in an alternating electromagnetic field, such as is generated by an X-ray photon, will oscillate at the same frequency as the field. The oscillating electron then re-emits electromagnetic radiation with the frequency of its oscillation, i.e. an X-ray photon with the same energy as the incident photon, and for scattering by a single atom there are no constraints on the direction of re-emission. Alternatively, the photon may lose part of its energy to the scattering electron, particularly if some of the latter are loosely bound fig. 2.4. Since total momentum must be maintained, the wavelength of the scattered photon must change, increasing (λ_o ≤ 1') in accordance with the expression: λ'-λ_o= (1-cos θ) $\frac {h}{m_ec}$ => λ'-λ_o=0.0243 (1-cos θ) where (h/m_ec)=0.0243 Å is called the Compton wavelength of the electron and θ is the angle of scatter. This is the physical basis of incoherent scattering.  ## 2.8 X-ray Diffraction In 1912 W.L. (Brage proposed a simple way of understanding X-ray diffraction by crystalline materials. He pointed out that, for any crystal, one can draw a set of equidistant parallel planes that pass through all of the atoms in the crystal, and that there are many different sets of such "Bragg" planes. As shown in fig. 2.5. When a beam of X-rays strikes one of the ions in the crystal, the X-rays are diffracted in all directions. In general these diffracted waves from different ions will be, on average, out of phase and cancel out. However, Bragg showed that, for an X-ray of wavelength λ and angle of incidence θ with respect to a Bragg plane (not the normal to the plane), scattered waves from the various ions lying in a single Bragg plane will be coherent if the angle of reflection equals the angle of incidence, If the path difference is either zero or some integer multiple of the wavelength, then the two waves are in phase and constructive interference results. pathdifference = ηλ ...2d sinθ = ηλ where: - n is integer and - is called the order number, This is the Bragg law of diffraction.  ## Solved problems 1. In a Millikan exp. an oil drop of radius 10^-6 m & density 8 x 10^2 kg/m^3 carries a charge 3.2 x 10^-19 C, find the intensity of the electric field to keep the drop in equilibrium state? At equilibrium: eE=mg E = mg/e = (4/3 π R³p g)/e = (4 x 3.14 x 10^-18 x 8 x 10^2 x 9.8)/(3 x 3.2 x 10^-19) = 102 KV/m 2. In a Millikan exp., the P.D between the electrodes is 1500 volts & the distance between them is 0.03m. Calculate the radius of the drop carries on elementary unit charge & the oil density is 900 kg/m³? e E = mg = 4/3 πr²pg r³= (3 e E)/(4 πp g) = (3 eV)/(4 πp g d) = (3 x 1.6 x 10^-19 x 1 500) / (4 x 3.14 x 900 x 9.8 x 0.03) r = 10^7 √227 = 6 x 10^-7 m 3. An electron accelerated by a P.D. of 5000V enters a uniform magnetic field of 2 x 10^-2 Wb/m² perpendicular to its direction motion. Determine the path of the electron? Κ.Ε of the electron = 1/2 m v² = 5000 x 1.6 x 10^-19 joules (1/2 x 9.1 x 10^-31) v² = 5000 x 1.6 x 10^-19 v = 4.4 x 10^7 m/sec. Bev = $\frac{mv^2}{R}$ $\frac{1}{2}mv^2$ = $eV$ R= (mv)/(Be) = (9.1 x 10^-31x 4.4 x 10^7)/(2 x 10^-2x 1.6 x 10^-19) = 1.25 x 10^2 m. 4. Taking 10 KV electrons, calculate: a- The speed of electrons. b- The strength of the magnetic field in which they travel in circular path-of-radius 10 cm, (e/m = 1.76 x 10^11 C/kg)? a- v = √(2eV)/m = 6 x 10^5 √V = 6 x 10^5 √10000 = 6 x 10^7 m/sec b- B = (mv) / (eR) (from Bev = mv²/R) = (6 x 10^7)/(1.76 x 10^11 x 0.1) = (6/1.76) x10^-3 wb/m² 5. Calculate the radius of curvature of electron of velocity 1/20 of velocity of light if it enters a magnetic field of magnetic intensity 4x1 0^-3 wb/m² & perpendicular to the path of the electrons? Bev = mv²/R , v =3 x1 0^8 x (1/20) e/m =1.76 x 10^11 C/kg R = mv/Be = 3 xl0^8/(20 x 4 x10^-3 x 1.76 x 10^11) = 0.021 m. 6. If electrons enter a magnetic field perpendicular to its path with a velocity of 3 x10^7 m/s it makes a circular path of radius 10 cm, calculate the intensity of the magnetic field? B = (mV)/(eR) (Bev = mv²/R) = 3 x 10^7/(1.76 x 10^11 x 0.1)=1.7 x 10^3 wb/m² = 17 gauss (wb = 10^+ gauss) 7. Find the percentage increase in mass of an electron accelerated through a P.D. of 5000