Origin of Quantum Mechanics PDF

# Origin of Quantum Mechanics ## Many optical phenomena can be explained on the basis of the wave nature of electromagnetic radiations. ## Many other phenomena suggest that electromagnetic radiations should be regarded not as a wave but as a stream of particles. In this chapter we shall discuss tho...

# Origin of Quantum Mechanics ## Many optical phenomena can be explained on the basis of the wave nature of electromagnetic radiations. ## Many other phenomena suggest that electromagnetic radiations should be regarded not as a wave but as a stream of particles. In this chapter we shall discuss those experiments which yield information about the particle nature of electromagnetic radiations. ## These experiments involve the absorption or scattering of radiation in matter and show that transfer of radiation takes place in discrete quanta of energy. ## While discussing these experiments we shall introduce the concept of photon as a particle of electromagnetic radiation. ### 1.1 Discovery of Photoelectric Effect The photoelectric effect was originally discovered by Hertz in the year 1887. He was experimenting with electric discharge between two electrodes A and B connected to a high voltage source as shown in Fig. 1.1. He subjected the negative electrode A to a beam of ultraviolet light. It was observed that the spark could jump greater distances than it could in the absence of ultraviolet light. Soon after Hallwachs showed that the ultraviolet light falling on the electrode A causes emission of electrons from it. These emitted electrons help the spark in jumping greater distances. The ejection of electrons from a material by the action of light is called the photoelectric effect. The emitted electrons are called photoelectrons because of the method of their production. Most metallic surfaces show photoelectric effect with ultraviolet light. Alkali metals like lithium, sodium, potassium, rubidium and cesium show photoelectric effect with visible lights. ### 1.2 Experimental Arrangement For Observing The Photoelectric Effect The modern arrangement for studying the photoelectric effect is shown in Fig 1.2. It consists of: * an evacuated glass tube G having two electrodes C and A. * The electrode C made of photosensitive material is connected to negative terminal of potential divider arrangement. It is known as photocathode. * The electrode A connected to positive terminal acts as an anode. * A quartz window W to allow monochromatic ultraviolet light to fall on photocathode C. * A microammeter M to measure the photoelectric current. There are three basic experimental variables in the photoelectric experiment. These are: * the intensity of light * the frequency of light * the voltage between cathode and anode. ### 1.2 Modern Physics #### Experiment 1. Study of variation of photoelectric current with intensity. Apply some positive potential to the anode and keep it fixed. Light of suitable fixed frequency is allowed to fall on the photocathode. Electrons are emitted and the photoelectric current is measured by the microammeter. Now vary the intensity of light continuously and note the corresponding current. Plot the graph between photoelectric current and intensity. It has the shape as shown in Fig. 1.3. #### Conclusions. 1. The graph between the photoelectric current and intensity is a straight line. This shows that the photoelectric current increases linearly with intensity. #### Experiment 2. Study of variation of photoelectric current with applied voltage. Now keep the intensity of light fixed at I₁ and select a frequency v such that photoelectric current flows. Slowly vary the voltage from - V to + V and note the corresponding photoelectric current. The behaviour of the current as a function of potential is shown by curve 1 in Fig. 1.4. We repeat the experiment with increased intensity I2. We will now get the curve 2. #### Conclusions. 1. There exists a minimum potential Vo below which there is no photoelectric current. This potential which is just sufficient to stop the photoelectric current is called stopping potential or the cut off potential. * 2. The stopping potential Vo is independent of the intensity of incident light. * Explanation. When a light of suitable frequency falls on a photocathode, the electrons are emitted. The velocities of emitted electrons have values between zero and a certain maximum. Let umax be the velocity of the fastest electron. Then kinetic energy of the fastest electron * 1 2 * T= mv * 2 * mu max ### Origin of Quantum Mechanics #### Experiment 3. Study of variation of stopping potential with frequency. Introduce a light filter between source of light and the photocathode. This will allow a light of certain frequency to fall on the photocathode. Make sure that photoelectric current flows in the circuit. If no photoelectric current flows, change the filter. Now go on reducing the accelerating potential from positive value to negative value till the photoelectric current becomes zero. Note the value of the stopping potential. Repeat the above experiment with a series of different light frequencies. Plot a graph between the stopping potential Vo and frequency v of incident light. It will be a straight line as shown in Fig. 1.5. #### Conclusions. 1. There exists a minimum frequency vo at which the stopping potential is zero. This means that no electrons are emitted if the frequency of incident light is below vo The minimum frequency vo below which there is no photoelectric current is called threshold frequency. * 2. The stopping potential increases linearly with the frequency. Since eVo = 1/2 mumax, this means that the maximum velocity of emitted electrons increases linearly with frequency. ### 1.3 Laws of Photoelectric Emission The results of the above experiments may be summed up as the laws of photoelectric emission. * 1. The photoelectric current increases linearly with intensity of light. * 2. The maximum velocity of photoelectrons depends only on the frequency of incident light and is independent of its intensity. * 3. The photoelectric effect does not occur below a certain frequency called the threshold frequency. This frequency depends on the metal used in photocathode. * 4. The emission of photoelectrons is an instantaneous process. ### 1.4 Classical Explanation of Photoelectric Effect Even though it is well know that light is a form of wave motion, yet we can't explain the various features of photoelectric effect on the basis of wave theory. * 1. According to classical theory, the energy of a wave is only dependent on its intensity and has nothing to do with frequency. Therefore the photoelectric effect should occur for any frequency of light. This is in contradiction with the existence of the threshold frequency. * 2. When a light wave falls on the electron, it will start giving energy to the electron. The electron will come out only when it has acquired sufficient energy to overcome binding from the metal surface. This will take some time. Thus there should be some time lag between the switching of light and the ejection of electron. But the electrons are emitted instantaneously even when the intensity is low. * 3. More intense light will have more energy. If all this energy is given to the electron, then a part of it will be used in overcoming the metallic binding. The rest will appear as the kinetic energy of the electron. This would mean that the stopping potential should depend upon the intensity of light. This is certainly not observed experimentally. * 4. According to classical theory, the frequency of light has nothing to do with its energy. Therefore, there should be no influence of frequency on the photoelectric current. This also contradicts the experimental results. ### 1.5 Work Function A metal contains a large number of free electrons. These electrons are not tightly bound to particular atoms but are bound to the whole metal. In order to remove an electron from the metal, a certain amount of energy is required. This is called the binding energy of an electron and is denoted by w. Thus the binding energy w is the energy required to remove the electron from the metal. All electrons are not equally bound. So they do not require the same energy to escape from the metal. The minimum energy required to release an electron from the metal is called work function and is denoted by wo. The work function is characteristic of the material. ### 1.6 Einstein's Explanation of Photoelectric Effect In the year 1905 Einstein offered an explanation of photoelectric effect on the basis of quantum theory of light. #### Assumptions. (a) A light wave of frequency v consists of tiny bundles of energy called quanta or photons. Each photon has energy hv and travels with the speed of light. * (b) In the photoelectric effect one photon is completely absorbed by one electron in the phtocathode. #### (1) Einstein's photoelectric equation. When light of frequency v shines on a metal, an electron instantly absorbs a photon of energy hv. If hv is greater than binding energy, the absorbed photon will eject the electron from the metal surface and will appear as a photoelectron. The energy of a photon is used in two ways. * (i) The part of absorbed photon energy is used in releasing the electron from metal surface. * (ii) The balance of energy appears as the kinetic energy of the electron. Thus we can write Energy of incident photon = Binding energy of electron + K.E. of electron * or * hv = w + 1/2 mv² * or * mv² = hv * 2m - w * When w = wo, the kinetic energy of emitted electrons will be maximum. Thus * 1 * 2 * mu² max * = hv - wo * ...(1.2) * Suppose a photon of frequency v = vo has just sufficient energy to remove the least bound electron from the metal. Then the maximum kinetic energy of emitted electron is zero. Using this fact we have * 0 = hvo - wo * hvo = wo * ...(1.3) * The frequency vo determines the threshold frequency. It will depend on the work function of a particular material. * Substitute the value of wo from (1.3) in eq. (1.2) to obtain * 1 * 2 * mu² max * = hv - hvo * This equation is called Einstein's photoelectric equation. *...(1.4) * (2) Explanation of the laws of photoelectric emission. According to quantum theory of light the intensity of the light beam is defined as the number of photons striking per unit area per unit time. When increase the intensity of incident light, this means that more photons will strike the photocathode. Since each photoelectron is due to a single photon, the electrons emitted will increase. Thus an increase in intensity causes an increase in photoelectric current. This is the first law of photoelectric emission. * 2. It is evident from eq. (1.4) that the maximum velocity vmax of photoelectrons depends only on the frequency of incident light and is independent of its intensity. This is precisely the second law of photoelectric emission. * 3. Eq. (1.4) tells us that if frequency of light vis less than vo the kinetic energy becomes negative. So electron emission can occur. This is nothing but the third law of photoelectric emission. * 4. According to Einstein's theory the light, energy incident on the photocathode is supplied in concentrated bundles or photons. These photons are immediately absorbed by some atoms. As a result of the immediate emision of photoelectron starts. This is exactly the fourth law of photoelectric emission. ### PROPERTIES OF PHOTONS The photons have the following properties. * (1) Energy of photon. The energy of photon associated with light of frequency v is hv. * (2) Speed of photon. Each photon travels with a speed of light c = 3 x 10⁸ m/s. * (3) Charge on photon. Photons are electrically neutral. So they can not be deflected by electric or magnetic fields. * (4) Mass of photon. The rest mass of photon is always zero. * (5) Momentum of photon. According to Einstein theory of relativity the mass and energy are equivalent and related by * where m is mass of particle. * E = mc² * ...(1.5) * Since the photon travels with the speed of light, the above relation can be applied to it. Thus mass of the photon is * m = E/c² * hv * m = * c² * The momentum of photon is given by * p=mc * hv * P=c= * c * hv * ...(1.7) ### 1.8 The Compton Effect The scattering of monochromatic X-rays from targets was first studied by Compton in the year 1923. The experimental arrangement for studying the compton effect is shown in fig. 1.6. #### Experimental arrangement. A monochromatic beam of X-rays of wavelength is incident on a carbon target. The target scatters X-rays in all directions. Compton measured the wavelength and intensity of scattered radiations with a Bragg spectrometer for various angles of scattering. Fig. (1.7) shows the results of Compton's experiment. This figure shows the variation of intensity of scattered X-rays with wavelength at various angles of scattering. #### Result. (i) At each scattering angle there are two peaks in the spectrum. This means that scattered radiations consist of two components or lines. One component U has the same wavelength as the incident radiation. This component of unchanged wavelength in the scattered beam is called unmodified radiation or line. The second component M has a slightly longer wavelength a'. The radiation of longer wavelength is known as modified radiation or line. (ii) The wavelength a' of modified line depends on the angle of scattering. It increases with the increasing scattering angle 4. Thus we conclude that when a monochromatic beam of high frequency radiations is scattered by a substance, the scattered radiations has two components. One component has the same wavelength as the incident radiation whereas the second component has wavelength longer than that of incident radiation. This phenomenon is known as compton effect. The scattering of X-rays with the change in wavelength is called compton scattering. ### 1.9 Classical Explanation of Compton Effect According to classical theory, the incident X-rays is an electromagnetic wave of wavelength 1. When this wave falls on the carb-on target, the electric field of the wave exerts force on the electrons in the target. As a result the electrons start vibrating with the frequency of incident radiations. The oscillating electrons produce electromagnetic waves in all directions. These waves have the same wavelength as the incident radiation. Thus in classical picture the scattered wave should have the same wavelength as the incident wave. Hence classical theory successfully explains the presence of unmodified radition or line in spectrum of scattered radiation. However it fails to account for the presence of modified radiation or line. ### 1.10 Explanation of Compton Effect by Quantum Theory Compton offered an explantion of compton effect on the basis of quantum theory. According to this theory X-rays of frequency v consists of tiny packets of energy called photons. Each photon has energy h v and travels with the speed of light. The phenomenon of compton scattering can be explained by treating it as an elastic collision between the incident X-ray photon and a free electron in the scattering target. When a photon of energy E collides with an electron, it transfers some of its energy to the electron. Therefore the scattered photon has less energy than the incident photon. Since energy of photon is inversely proportional to its wavelength, so that wavelength' of the scattered photon will be greater than the wavelength A of incident photon. The change in wavelength of the scattered radiation is known as compton effect. Thus quantum theory successfully accounts for the appearance of modified radiation by regarding X-rays as particles rather than as waves. #### (a) Expression for wavelength of modified radiation. Consider a photon of frequency v incident on an electron at rest. (Fig. 1.8). When this photon collides with the electron, it gives up some of its energy to the electron. As a result of collision the electron acquires some velocity say v and recoils at an angle o to the direction of the incident beam. Suppose the photon is scattered through an angle & after the collision. Letv 'be the frequency of scattered photon. Since the collision between photon and electron is elastic, therefore total energy and total momentum of the system must remain conserved. #### (A) Before Collision * (i) Energy of incident photon E = hv * ...(1.8) * (ii) Momentum of incident photon p = hv/c * ...(1.9) * (iii) Rest energy of the electron E = mc² * ...(1.10) * (iv) Momentum of electron at rest = 0 * ...(1.11) #### After Collision * (i) Energy of scattered photon E' = hv' * ...(1.12) * (ii) Momentum of scattered photonp' = hv'/c * ...(1.13) * (iii) energy of the electron E', = mc² * ...(1.14) * where m is relativistic mass of electron moving with velocity to * (iv) Momentum of recoil electron pe = mv * ...(1.15) * (a) Application of law of conservation of energy. The law of conservation of energy requires that total energy of system before collision = total energy of system after collsion * E + E = E' + E'c * Using eq. (1.8), eq. (1.10) eq. (1.12) and eq. (1.14) it becomes * hv+ mo c² = hv' + mc² * Using the relations * hv-hv'+ mo c² = mc² * and * v = x * m = mo (1 - v²/c²) * Eq. (1.17) can be written as * 1/λ' + moc/hc = 1/λ + moc/hc * Divide both sides by c to obtain * (1/λ') + mo c/hc = (1/λ) + mo c/hc * Squaring both sides we obtain * (1/λ')² + m² c²/h² + 2 moc/h (1/λ') = (1/λ)² + m² c²/h² + 2 moc/h (1/λ) * (1/λ')² - (1/λ)² + 2 moc/h (1/λ' - 1/λ) = m² c²/h² * h²/c² (1/λ')² + 2 moc/h (1/λ' - 1/λ) = 0 * h² (1/λ')² + 2 moch (λ' - λ)/λλ' = 0 * h² + 2 moch (λ' - λ)/λλ' = 0 * h² + 2 moch (λ' - λ)/λλ' = 0 * λ² + 2 moch (λ' - λ)/ λλ' = 0 * λ² + 2 moch (λ' - λ)/λλ' = 1 * (ii) Application of law of conservation of momentum. The law of conservation of momentum requies Momentum of system before collision = Momentum of system after collisions * → * P+0=p+P * P-P=P * p-p * 0 = 1257 10 novoslo to * The magnitude of momentum of electron after collision is * p² = p² + p² - 2pp cos * (where is the angle between pand p * 22 2ppcos bata to milit * ...(1.29 * Substituting the values of P., P and p' from eq. (1.15), eq. (1.9) and (1.13) respectively we obtain * (BLK)... * m² v² = * h²v² / c² + h²v'² / c² - 2h²vv'/ c² cos * Since v = c/λ, we may write it as * h² * λ² + λ'² - 2λλ' cos φ = m² v² * Substituting the value of m from eq. (1.19) we obtain * h² * λ² + λ'² - 2λλ' cos φ = m² v² * λ² + λ'² - 2λλ' cos φ = m² v² * λ² + λ'² - 2λλ' cos φ = m² v² * (1-v²/c²) * h² λ² + 2 moch (λ' - λ)/λλ' = 1 * λ² + 2h² λ² / λλ' + 2 moch (λ' - λ)/ λλ' = 1 * λ² + 2h² λ² / λλ' + 2 moch (λ' - λ) = λ² + 2h² λ² / λλ' * h² * λ² + λ'² - 2λλ' cos φ = m² v² * h² * λ² + λ'² - 2λλ' cos φ = m² v² * h² * λ² + λ'² - 2λλ' cos φ = m² v² * h² * λ² + λ'² - 2λλ' cos φ = m² v² * h² * λ² + λ'² - 2λλ' cos φ = m² v² * h² * λ² + λ'² - 2λλ' cos φ = m²v² * h² * λ² + λ'² - 2λλ' cos φ = m² v² * h² * λ² + λ'² - 2λλ' cos φ = m² v² * h² * λ² + λ'² - 2λλ' cos φ = m² v² * h² * λ² + λ'² - 2λλ' cos φ = m² v² * h² * λ² + λ'² - 2λλ' cos φ = m² v² * (1-v²/c²) * h² (1/λ')² + 2 moc/h (1/λ' - 1/λ) = 0 * h² (1/λ')² + 2 moch (λ' - λ)/λλ' = 0 * h² + 2 moch (λ' - λ)/λλ' = 0 * h² + 2 moch (λ' - λ)/λλ' = 0 * λ² + 2 moch (λ' - λ) / λλ' = 0 * λ² + 2 moch (λ' - λ)/λλ' = 1 * (1-v²/c²) * λ² + 2 moch (λ' - λ) / λλ' = 1 * λ² + 2h² λ² / λλ' + 2 moch (λ' - λ) / λλ' = 1 * λ² + 2h² λ² / λλ' + 2 moch (λ' - λ) = λ² + 2h² λ² / λλ' * h² * λ² + λ'² - 2λλ' cos φ = m² v² * Subtract eq. (1.28) from eq. (1.22) This gives * 2h² 2 moch * + * λλ' * 2h² * (2 - 1) + * h² * COS = 0 * 2h² (1- cos 4) = * λλ. (λ-λ) * h * λ' - λ = * moc * (1-cosp) * h * λ' = λ + (1 − cos $) * moc * ...(1.29) * ...(1.30) * This equation gives the wavelength of modified radiation. It shows that wavelength of modified radiation * (i) depends on wavelength of incident radiation * (ii) depends on the angle of scattering φ * (iii) is always greater than wavelength of incident radiation * (iv) is independent of the nature of the scatter. * (b) Variation of wavelength of modified radiation with angle of scattering. * Case 1. Suppose the angle of scattering of radiation is zero. That is * φ = 0 * Substitute this value in eq. (1.30). This gives * λ' = λ * Thus there is no modified radiation along the direction of incidence. * Case 2. Suppose the angle of scattering of radiation is 180°. That is * φ = 180° * Substitute this value of φ in eq. (1.30) to obtain * 2h / moc * λ' = λ + * moc * Thus we observe that as φ varies from 0 to 180°, the wavelength of scattered radiation changes from λ to λ + 2h / moc. * (c) Compton shift. The quantity h / moc has dimensions of length and is known as compton wavelength. The difference in wavelength between the scattered X-rays and the incident X-rays is called compton shift. It is denoted by A λ. Thus * In the discussion of compton effect it has been assumed that the electron is free. This will be approximately true for the outer electrons of atoms in the carbon. The electrons close to the nucleus are tightly bound. When such bound electrons scatter X-rays, the collision is not considered taking place between the photon and the whole atom. In such collisions the whole atom is rather the individual electron. The compton shift in wavelength is still given by eq. (1.31) except that mass of electron is replaced by mass of the recoiling atom. If M is the mass of atom then compton shift is * h * Δλ = (1 − cos φ) * Mc * ...(1.3 * The mass of atom is about 10⁴ times greater than mass of electron. Therefore the compton shift is so small as to be unobservable in X-ray scattering experiment. Thus the scattered photon has the same wavelength as the incident photon. This process gives rise to the unmodified line observed by compton. * We observe that the phenomenon of compton scattering gives support for the particle picture of electromagnetic radiation. Similar result was obtained in the photoelectric effect. ### 1.12 Kinetic Energy of Recoil Electron The kinetic energy T of recoil electron is given by * E = E - Emsiani * From eq. (1.16) we have * e * ph * E' - E = E - EM * Using eq. (1.8) and (1.12) it becomes * E' - E = hv -hv' * Substitute this result in eq. (1.33) to obtain * T=hv-hv' * T= * hc hc * 0= * gaid (0.1) pa ni sulavaidsstede * boonbibas beijibon on ti yet nối * 1 * 1 * =-=c(-) * T=(1-2) * 01 * The wavlength of modified line is given by eq. (1.30). That is. * mort esguardo notteibar borsa 1 + * λ' * T * moroni al bas * λ' * h * (1- * moc * ... * Cosmoday 2016(1.30) * =1+(1-cos) * oumib and * hv * COS) * 13 * tintup Tiida notama * pwd Agroloniw * where * λ' * = 1 + x(1-cos φ) * x = * hv * C2 * a notady to ymon тос * λ' * Substitute this value of in eq. (1.34). This gives * T=[1-1+x(1-cos)] * T=hv * Using eq. (1.36) it becomes * [ * - * x (1 − cos φ) * 1+x 1 + x (1-cos). * h²² * T = * 1+ * c2 * hv * (1-cos * cos $) * тосг(1-cos) * 1.11 * ...(1.35) * ...(1.36) * seid ni a zi motodo lo baagaadisonia 11 * atoil mio suley sdt gaihustada? * १ * TO * idgallo digestov odtahoved * froa * q mutremoni galud r...(1.37) * This gives the kinetic energy of the recoil electron in terms of the energy of incident photon and scattering angle φ. ### 1.13 Dual Nature of Electromagnetic Radiations Both the photoelectric and compton effects show that electromagnetic radiations possess particle properties. The photon concept successfully explains the results of both the experiments. The results of these experiments indicate that transfer of energy by radiation occurs in discrete quanta of energy known as photons. The interference and diffraction phenomena can be explained only on the hypothesis that electromagnetic radiations propagate as a wave motion. Thus electromagnetic radiations such as visible light, infrared and ultraviolet radiations, X-rays can show both wave and particle properties depending on the experimental situation. Hence we conclude that electromagnetic radiations has a dual nature. It means that electromagnetic radiations behave as wave in some processes and as particles in others. It never shows both characteristics in any one experiment. ### 1.14 Dual Nature of Matter De-Broglie has the feeling that nature must be symmetrical. Therefore he suggested that the wave-particle duality of light should also apply to matter. This means that material particles such as electron, proton etc. must show particle as well as wave nature. Just as a photon has a light wave associated with it, a material particle has matter wave associated with it. The matter waves associated with material particles are called De-Broglie waves. It may be noted that matter waves are not electromagnetic waves. They are also entirely different from other waves such as sound waves, waves on a string etc. ### 1.15 De Broglie Relation #### (a) Wavelength of a photon. The energy of a photon of frequency v is * E = hv * ...(1.38) * If m is the mass of the photon corresponding to energy E, then according to Einstein's theory of relativity * E = mc² * Equating (1.38) and (1.39), we have * ...(1.39) * hv=mc² * hv * m = * c² * Since the speed of photon is c in free space, therefore the momentum p of photon is * p=mc * Substituting the value of m from eq. (1.40) it becomes * hv * p = * C * h * or * or * P=x * h * λ= * P * -(1.41) * where a is the wavelength of light wave associated with a photon. Eq. (1.41) gives the wavelength of a photon having momentum p. * (ii) de-Broglie wavelength. De-Broglie postulated that the wavelength of de-Broglie waves associated with a mateial particle is also determined by the relation (1.41). If m is the mass of the material partick moving with velocity v then the wavelength A of the de-Broglie ware * Fig. 1.9 * associated with it is * h * λ = * mv * This is called the de-Broglie wavelength. A typical pictorial representation of the matter wave associated with a material particle is shown in Fig. (1.9). ### 1.16 Properties of Matter Waves De-Broglie wavelength is given by * λ * λ= * It is evident from eq. (1.41) that * (1) the wavelength of de-Broglie wave is independent of charge on the particle * (2) the faster the particle of a given mass moves, smaller is its de-broglie wavelength. * (3) Heavier the partile, lesser is its de-broglie wavelength. * (4) When v = 0 then A = ∞. This means that wave becomes indeterminate. On the other hand if v = ∞ then a = 0. This shows that matter waves are produced by the motion of particles. ### 1.17 De Broglie Wavelength Associated with An Electron The de-Broglie wavelength associated with any particle of mass m isГАЛЯ З * h * h * λ = * mv P * where p is the momentum of the particle. * The kinetic energy of particle is * E * = * 1 * mv² * = * 2 * P * p² * 2m * amba motoqadi to zastada * sado (92.1) bas (821) * or * p = √2m E * 1.13 * ...(1.42) * Substitute this value of p in eq. (1.41). Thus the wavelength of de Broglie wave associated with any particle is * h * λ = * √2mE * ...(1.43) * Supose now that the particle is an electron. If V is the potential difference through which an electron is accelerated, then its kinetic energy is * E = eV * Substitute this value of E in eq. (1.43) to obtain * This gives the de Broglie wavelength associated with an electron * Substituting pacem = 9 × 10-31 * and * h = 6.63 × 10-34 J.s, e = 1.6 × 10-19 C * we get * 12-3 * λ = * × 1010m081) * 12-3 * 1 = * For * Ada (2.1) pa ni bash to aaravaadaturitada * V = 100 volts * λ = 1.23 Å * x100x * This wavelength is of the order of wavelength for X-rays. It should therefore be possible to demonstrate wave like properties of electrons by diffracting an electron beam from a crystal. ### 1.18 Davisson and Germer Experiment The first experiment to test the wave nature of material paritcles was done accidently by Davisson and Germer. Their apparatus is shown in Fig. (1.10). #### Experimental arrangement. G is an electron gun which provides incident electrons. It consists of a directly heated filament F which acts as a source of electrons. P is a set of plates having pin holes. Electrons from heated filament are accelerated by a potential difference V between F and P. A narrow pencil of electrons emerges from the hole H with kinetic energy * V * GOT * P * INCIDENT * Η ΒΕΑΜΘ * ATTERED * AM * DETECTOR CURRENT * 1.14 * 0 = 50° * 35 40 45 50 55 60 65 70 * POTENTIAL DIFFERENCE V * АЗИАНКIM MODERN PHYSICS * of electron from experiment. * (a) Wavelength * is maximum when * that the detector current * tion is satisfied. The Bragg condition is * 2d sin 0 = ηλ * n = 1, 2... * It was om * the Braggond * مم * where d is the spacing between successive crystal pla * as shown in fig. (1.10) and e is the angle * reflection occurs. * is * at which s * For nickel crystal, the distance d between atomic p * nisido at (8.1) ps at a lo aules * d = 0.91 Å *