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## Quantization of Charge, Light, and Energy

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**1. Quantization of Electric Charge;**• 2. Blackbody Radiation; • 3. The Photoelectric Effect • 4. X-rays and the Compton Effect.**Three great quantization discoveries:**1. Quantization of Electrical Charge 2. Quantization of Light Energy 3. Quantization of energy of Oscillating Mechanical System**Quantization of Electric Charge**Early measurements of e and e/m. The first estimates of the magnitude of electric charges found in atoms were obtained from Faraday law. Faraday passed a direct current through weakly conducting solutions and observed the subsequent liberation of the components of the solution on electrodes. Faraday discovered that the same quantity of electricity, F, called latter one faraday, and equal to about 96,500 C, always decomposed 1 gram-ionic weight of monovalent ions.**Quantization of Electric Charge**1F (one faraday) = 96,500C 1Fdecomposed1gram-ionic weightof monovalent ions. Example: If96,500 Cpass through a solution ofNaCl,23gof Naappears at the cathode and35.5gof Clappearsat the anode. 1F =NAe- Faradays Law of Electrolysis whereNA– Avogadro’s number e– minimum amount of charge, that was called an electron**Discovery of electron: Thomson’s Experiment.**Many studies of electrical discharges in gases were done in the late 19th century. It was found that the ions responsible for gaseous conduction carried the same charge as did those in electrolysis. J.J. Thomson in 1897 used crossed electric and magnetic fields in his famous experiment to deflect the cathode-rays. In this way he verified that cathode-rays must consist of charged particles. By measuring the deflection of these particles Thomson showed that all the particles have the same charge-to-mass ratioq/m. He also showed that particles with this charge-to-mass ratio can be obtained using any material for a source, which means that this particles, now called electrons, are a fundamental consistent of all matter.**Thomson’s tube for measuring e/m.**Electrons from the cathodeCpass through the slits atAandBand strike a phosphorescent screen. The beam can be deflected by an electric field between platesDand Eor by magnetic field. From measurements of the deflections measured on a scale on the tube’s screen,e/mcan be determined.**FE=qE**_ Crossed electric and magnetic fields. When a negative particle moves to the right the particle experience a downward magnetic forceFB=qvBand an upward electric forceFE=qE. If these forces are balanced, the speed of the particle is related to the field strengths by v=E/B - - q FB=qvB FE=FB qE = qvB**Thomson’s Experiment**In his experiment Thomson adjustedE┴ Bso that the particles were undeflected. This allowed him determine the speed of the particleu =E/B. He then turnedoff theBfield and measured the deflection of the particles on the screen.**Deflection of the Electron Beam.**Deflection of the beam is shown with the top plate positive. Thomson used up to200 Vbetween the plates. A magnetic field was applied perpendicular to the plane of the diagram directed into the page to bend the beam back down to its undeflected position.**With the magnetic field turned off, the beam are deflected**by an amounty=y1+y2. y1occurs while the electrons are between the plates,y2after electrons leave the region between the plates. Letsx1be the horizontal distance across the deflection plates. If the electron moves horizontally with speedv0when it enters the region between the plates, the time spent between the plates ist1=x1/v0, and the vertical component of velocity when it leaves the plates is whereEyis the upward component of electric field. The deflection y1is:**The electron then travels an additional horizontal**distancex2in the field-free region from the deflection plates to the screen. Since the velocity of the electron is constant in this region, the time to reach the screen ist2=x2/v0 , and the additional vertical deflection is:**The total deflection at screen is therefore**This equation cab be used to determine the charge to-mass ratioq/mfrom measured deflectiony.**Example**Electrons pass undeflected through the plates of Thomson’s apparatus when the electric field is3000 V/mand there is a crossed magnetic field of1.40 G. If the plates are4-cmlong and the ends of the plates are30 cmfrom the screen, find the deflection on the screen when the magnetic field is turned off. me=9.11 x 10-31kg; q=e=-1.6 x 10-19C**The Mass Spectrometer**The mass spectrometer, first designed by Francis William Aston in 1919, was developed as a means of measuring the masses of isotopes. Such measurements are important in determining both the presence of isotopes and their abundance in nature. For example, natural magnesium has been found to consist of 78.7 %24Mg; 10.1%25Mg; and 11.2%26Mg. These isotopes have masses in the approximate ratio24:25:26.**Schematic drawing of a mass spectrometer.**Positive ions from an ion source are accelerated through a potential differenceΔVand enter a uniform magnetic field. The magnetic field is out of the plane of the page as indicated by the dots. The ions are bent into a circular arc and emerge atP2(a plane of photographic plate or another ion detector). The radius of the circle is proportional to the mass of the ion**The Mass Spectrometer**In the ion source positive ions are formed by bombarding neutral atoms withX-raysor a beam of electrons. (Electrons are knock out of the atoms by theX-raysor bombarding electrons). These ions are accelerating by an electric field and enter a uniform magnetic field. If the positive ions start from rest and move through a potential differenceΔV, the ions kinetic energy when they enter the magnetic field equals their loss in potential energy,q|ΔV|:**The ions move in a semicircle of radiusR. The velocity of**the particle is perpendicular to the magnetic field. The magnetic force provides the centripetal forcev2/Rin circular motion. We will use Newton’s second low to relate the radius of semicircle to the magnetic field and the speed of the particle. If the velocity of the particle isv, the magnitude of the net force isqvB, sincevandB are perpendicular.**The Mass Spectrometer**The speedvcan be eliminate from equations and Substituting this forv2: Simplifying and solving for (m/q):**Separating Isotopes of Nickel**• A58Niion charge+eand mass9.62 x 10-26kgis accelerated through a potential drop of3 kVand deflected in a magnetic field of0.12T. • (a) Find the radius of curvature of the orbit of the ion. • (b) Find the difference in the radii of curvature of58Niions and60Niions. (assume that the mass ratio is58:60).**Blackbody Radiation**• One unsolved puzzle in physics in late nineteen century was the spectral distribution of so calledcavity radiation, also referred to asblackbody radiation. • It was shown by Kirchhoff that the most efficient radiator of electromagnetic waves was also a most efficient absorber. A “perfect” absorber would be one that absorbed all incident radiation. • Since no light would be reflected it is called ablackbody.**A small hole in the wall of the cavity approximating an**ideal blackbody. Electromagnetic radiation (for example, light) entering the hole has little chance of leaving before it is completely adsorbed within the cavity.**Blackbody Radiation**• As the walls of the cavity absorb this incoming radiation , their temperature rises and begin to irradiate. • In 1879, the Austrian physicist J.Stefan first measured the total amount of radiation emitted by blackbody at all wavelengths and found it varied with absolute temperature. • It was latter explained through a theoretical derivation by Boltzman, so the result became known as the Stefan-Boltzman radiation Law: R = σT4 whereRis the power radiated per unit time and per unit area of blackbody;Tis the Kelvin temperature; and σis the Stefan-Boltzman constant,σ=5.672 x 10-8 W/m2K4.**Blackbody Radiation**R = σT4 • Note that the power per unit area radiated by blackbody depends only on the temperature, and not of other characteristic of the object, such as its color or the material, of which it is composed. • Rtells as the rate at which energy is emitted by the object. For example, doubling the absolute temperature of an object increases the energy flows out of the object by factor of24=16. • An object at room temperature(300 K)will double the rate at which it radiates energy as a result of temperature increase of only570.**Radiation emitted by the object at temperature T that passed**through the slit is dispersed according to its wavelength. The prism shown would be an appropriate device for that part of the emitted radiation in the visible region. In other spectral regions other types of devices or wavelength-sensitive detectors would be used.**Spectral distribution functionR(λ)measured at different**temperatures. TheR(λ)axis is in arbitrary units for comparison only. Notice the range inλof the visible spectrum. The Sun emits radiation very close to that of a blackbody at5800 K. λmis indicated for the5000-K and6000-Kcurves.**Blackbody Radiation**The German physicist W.Wien derived a relationship for maximum wavelength and absolute temperature, known as Wien’s displacement law: λmT = constant=(2.898 x 10-3m)K Example: The wavelength at the peak of the spectral distribution for a blackbody at4300 Kis 674 nm(red). At what temperature would the peak be420 nm(violet)? Solution: From the Wien’s law, we have λ1T1 = λ2T2 (674 x 10-9m)(4300 K) = (420 x 10-9m)(T2) T2=6900 K**This law is used to determine the surface temperatures of**stars by analyzing their radiation. It can also be used to map out the variation in temperature over different regions of the surface of an object. Such a map is called thermograph. • For example thermograph can be used to detect cancer because cancerous tissue results in increased circulation which produce a slight increase in skin temperature.**Radiation From the Sun. The radiation emitted by the surface**of the sun emits maximum power at wavelength of about500 nm. Assuming the sun to be a blackbody emitter, (a) what is it surface temperature? (b) Calculateλmaxfor a blackbody at room temperature,T=300 K.**The calculation of the distribution functionR(λ)involves**the calculation of the energy density of electromagnetic waves in the cavity. The power radiated out of the hole is proportional to the total energy densityU(energy per unit volume) of the radiation in the cavity. The proportionality constant can be shown to bec/4,wherecis the speed of the light: R = (1/4)cU Similarly, the spectral distribution of the power proportional to the spectral distribution of the energy density in the cavity.**Ifu(λ)dλis the fraction of the energy per unit volume in**the cavity in the rangedλ, thenu(λ)and R(λ)are related by R(λ)=(1/4)cu(λ) The energy density distribution functionu(λ)can be calculated from classical physics. We can find the number of modes of oscillation of the electromagnetic field in the cavity with wavelength λ in the intervaldλand multiply it by average energy per mode. The result is that the number of modes of oscillation per unit volume,n(λ),is independent of the shape of cavity and is given by: n(λ) = 8πcλ-4**Rayleigh-Jeans Equation**The number of modes of oscillation per unit volume: n(λ) = 8πcλ-4 According to the classical kinetic theory, the average energy per mode of oscillation is kT, the same as for a one-dimensional harmonic oscillator, where k is the Boltzman constant. Classical theory thus predicts for the energy density spectral distribution function u(λ) = kTn(λ) = 8πckTλ-4**Rayleigh-Jeans Equation**u(λ) = kTn(λ) = 8πckTλ-4 This prediction, initially derived by Rayleigh, is called the Rayleigh-Jeans Law. At very long wavelength the Rayleigh-Jeans law agrees with experimentally determined spectral distribution, but at short wavelength this law predicts thatu(λ)becomes large, approaching infinity asλ→0, whereas experiment shows that the distribution actually approaches zero asλ→0. This enormous disagreement between experimental measurements and classical theory for short wavelength was called theultraviolet catastrophe.**Comparison of the Rayleigh-Jeans Law with experimental data**at T=1600 K. The u(λ)axis is linear.**In 1900 the German physicist Max Plank by making some**unusual assumptions derived a functionu(λ)that agreed with experimental data. Classically, the electromagnetic waves in the cavity are produced by accelerated electric charges in the walls vibrating like simple harmonic oscillators. The average energy for simple harmonic oscillator is calculated classically from Maxwell-Boltzman distribution function:f(E) = Ae-E/kT whereAis a constant andf(E)is the fraction of oscillators with energyE.**Maxwell-Boltzman distribution function**f(E) = Ae-E/kT The average energy is then found, as is any weighted average: Plank found that he could derive his empirical function assuming the energy of oscillators, and hence the radiation that they emitted, was a discrete variablethat could take only the valueso, ε, 2ε, 3ε,…..,nεwherenis an integer, and that ε is proportional to the frequency of the oscillators, and, thus, the radiation.**Plank therefore wrote the energy as**En=nε=nhf n=0,1,2,….. wherehis a constant now calledPlank constant. The Maxwell-Boltzmandistribution than becomes: fn= Ae-E/kT = Ae-nε/kT whereAis determined by normalization condition that the sum of all fractionsfnmust be equal 1.**The normalization condition**The average energy of oscillation is then given by discrete-sum equivalent To solve this equation let put wherey=e-x.**This sum is a series expansion of(1-y)-1, so**(1-y)-1=1+y+y2+y3+…, thenΣfn=A(1-y)-1=1givesA=1-yand Note that so we have since**Plank’s Law**Multiplying this sum byhfand usingA=(1-y),the average energy is: Multiplying the numerator and the denominator byexand substituting forx, we obtain:**Plank’s Law**Multiplying this result by the number of oscillators per unit volume in the intervaldλgiven byn(λ)=8πcλ-4(the number of modes of oscillation per unit volume) we obtain the energy distribution function for the radiation in cavity: This function is calledPlank’s Law.**The value of Plank’s constanthcan be determined by**fitting the function to the experimental data, although the direct measurement is better, but more difficult. The presently accepted value of Plank’ constant is: h = 6.626 x 10-34 J·s= 4.136 x 10-15 eV·s Plank’s Law**Comparison of Plank’s Law and the Rayleigh-Jeans Law with**experimental data at T=1600 K. The u(λ)axis is linear.**A dramatic example of an application of Planck’s law is**the test of the predictions of the so-called Big Bang theory of the formation and expansion of the universe. Current cosmological theory suggests that the universe originated in an extremely high-temperature explosion, one consequence of which is to fill the infant universe with radiation that can be approximate with black body spectral distribution. In 1965, Arno Penzias and Robert Wilson discovered radiation of wavelength7.35 cmreaching the Earth with same intensity from all directions in space. It was recognized soon as a remnant of the Big Bang (relict radiation). Plank’s Law**The energy density spectral distribution of the cosmic**microwave background radiation. The solid line is Plank’s Law with T=2.735 K. The measurements were made by the Cosmic Back Ground Exploder (COBE) satellite.**Problem 1. Thermal Radiation from the Human Body. The**temperature of the skin is approximately 35°C. What is the wavelength at which the peak occurs in the radiation emitted from the skin?**Problem 2.The Quantized Oscillator. A 2-kg mass is attached**to a massless spring of force constant k=25N/m. The spring is stretched 0.4m from its equilibrium position and released. (a) Find the total energy and frequency of oscillation according to classical calculations. (b) Assume that the energy is quantized and find the quantum number, n, for the system. (c) How much energy would be carried away in one-quantum change?**Problem 3.The Energy of a “Yellow” Photon. What is the**energy carried by a quantum of light whose frequency equals 6 x 1014 Hz yellow light? What is the wavelength of this light?**The Photoelectric Effect**It is one of the ironies in the history of the science that in the same famous experiment of Heinrich Hertz in1887 in which he produced and detected electromagnetic waves, thus confirmed Maxwell’s wave theory of light, he also discovered the photoelectric effect led directly to particle description of light. It was found that negative charged particles were emitted from a clean surface when exposed to light. P.Lenard in 1900 detected them in a magnetic field and found that they had a charge-to-mass ratio of the same magnitude as that measured by Thompson for cathode rays: the particles being emitted were electrons.**Schematic diagram of the apparatus used by P.Lenard to**demonstrate the photoelectric effect and to show that the particles emitted in the process were electrons. Light from the source Lstrikes the cathode C. Photoelectrons going through the hole in anode Aare recorded by the electrometer connected to α.A magnetic field, indicated by the circular pole piece, could deflect the particles to an electrometer connected to β, enabling the establishment of the sign of their charge and their q/mratio.