Waves And Particles PDF

Chapter Waves And Particles 14 OBJECTIVES To explain the particle and wave nature (matter waves) of the quantum particles To derive the de-Broglie equation related with momentum and wavelength of the particle To discu...

Chapter Waves And Particles 14 OBJECTIVES To explain the particle and wave nature (matter waves) of the quantum particles To derive the de-Broglie equation related with momentum and wavelength of the particle To discuss the properties of matter waves and explain the experimental evidence of matter waves To derive the Schrödinger wave equation for the motion of the quantum particles To discuss the importance of the uncertainty principle To derive application of Schrödinger equation for a particle in one-dimensional box and its importance 14.1 I Classical mechanics which is based on Newton’s laws, explains the large- n scale experiments. On the other hand, Newtonian mechanics fails to explain T the different phenomena which take place at microscopic or atomic scales. Quantum theory was able to explain the nature of light waves in terms of r packets or bundles known as photons or quanta. Some of the phenomena O like interference, diffraction, polarization, etc., are explained based on the d wave nature of light. However, the phenomena like photoelectric effect, U Compton effect, Zeeman effect, etc., are explained based on the particle C nature of light. Therefore, radiation, or light, behaves like a wave as well as a particle. Thus, the wave–particle property of light radiation is known T as the dual nature of light. The concepts which prompted de Broglie to I propose the idea of matter waves are namely i) the universe is made up of O radiation and matter, and ii) nature is symmetrical everywhere. n In the present chapter, an attempt has been made to explain the dual nature of light, i.e., de Broglie waves, their experimental verification, Schrödinger wave equations and their applications in quantum mechanics. 14.2 Engineering Physics 14.2 dE BrOGLIE WAVE According to de Broglie, light waves behave sometimes as particles, and conversely particles, can have wavelike characteristic properties. Therefore, the particles like electrons, protons, neutrons, atoms or molecules will have associated waves with them known as matter waves or pilot waves or de Broglie waves. A variable quantity known as wave function is used to characterise the de Broglie waves. The wave function is denoted by the symbol y (psi). 14.3 dE BrOGLIE WAVELEnGTH Let us consider that a photon with frequency n is traveling with the velocity of light c. Then, the velocity of light c = nl (14.1) where l is the wavelength of light. According to Einstein, the energy of each photon E = hn (14.2) where E is the energy of the photon. Considering the momentum of the photon subjected to an external force, Eq.(14.1) and Eq. (14.2) can be combined as, E p= (14.3) c Substituting the value of c = u l and E = h n , we get, hn p= n l or h p= (14.4) l From the above equation, it is clear that the light photon with wavelength l and momentum p = h / l will have an associated wavelength l and also a wavelike character. In 1924, de Broglie suggested that the matter particles will have associated waves known as de Broglie waves (or) matter waves. The value of the wave function associated with a moving particle at any point (x,y,z) and time t determines the probability of finding the particle at that corresponding point and time. Consider a particle of mass m moving with a velocity n. Then, the wavelength associated with the particle is h l= (14.5) mv where p = mv is the momentum of the particle. Equation (14.5) is known as the de Broglie wavelength of the matter wave.  Waves and Particles 14.3 14.3.1 de Broglie Wavelength in terms of KE Consider a particle of mass m moving with a velocity v. Then, the kinetic energy (KE) of the particle is, E = (1 / 2) m v 2 1 p2 = m2 v 2 = 2m 2m We know that p = mv, \ The momentum p = 2mE (14.6) Substituting the value of p in Eq. (14.5), we get, h The wavelength l= (14.7) 2mE Equation (14.7) represents the de Broglie wavelength interms of kinetic energy. 14.3.2 de Broglie Wavelength in terms of v Consider an electron of mass m and charge e, accelerated through a potential difference of V volts. Then, the KE of the electron is equal to the loss in potential energy. 1 2 i.e., mv = eV (14.8) 2 or m2v2 = 2 m eV \ The momentum of the electron p = mv = 2 m eV (14.9) Substituting the value of mv in Eq. (14.5), we get, h de Broglie wavelength of electron l = (14.10) 2 m eV Substituting the value of h, e and m in Eq. (14.10), we get, 6.626 ¥10-34 De Broglie wavelength of electron l = 2 ¥ 9.1¥10-3 ¥1.6 ¥10-19 ¥ V 12.28 ¥ 10-10 = m V 12.28 l= Å (14.11) V Equation (14.11) gives the de Broglie wavelength of an electron. 14.4 Engineering Physics 14.4 PrOPErTIES OF MATTEr WAVES The following are the dominant properties of matter waves: (i) Matter waves are not physical phenomena, but a wave probability. (ii) The wavelength associated with matter waves decreases with increase in the mass of the particles, h i.e., l= (14.12) 2 m eV (iii) Matter waves are a new kind of waves and are not electromagnetic in nature. (iv) Matter waves are mainly used as a wave to pilot or guide the matter par- ticles and hence, are known as pilot waves or guide waves. The schematic representation of pilot or guide waves are shown in Fig. 14.1. (v) The velocity of matter waves depends on the velocity of the material particle. (vi) The phase velocity of a matter wave is inversely proportional to its wave- length. l Fig. 14.1 Pilot or guide wave (vii) Matter waves are not constant as electromagnetic waves and move faster than the velocity of light, i.e., u = c2/n (14.13) 14.5 MATTEr WAVES—EXPErIMEnTAL VErIFICATIOn The de Broglie hypothesis clearly states that particles like electrons will have associated wavelike characteristics and obey the properties of ordinary light like refraction, reflection, diffraction, etc. One can prove the wave nature of the particles through simple experiments, namely, Davission–Germer and Thomson’s experiments. In both the experiments, electrons have been selected to prove the wave natures mainly due to their well known characteristic properties and for the easy generation and detection. 14.5.1 davission–Germer Experiment In 1927, Davission and Germer made an attempt to prove the wave nature of atomic particles. The schematic representation of the experimental set-up used to demonstrate the dual nature of matter is shown in Fig. 14.2.  Waves and Particles 14.5 D1 D2 N L.T. F A H H.T. C S G Fig.14.2 Davission–Germer experiment—dual nature verification The electrons are produced by the hot filament F by applying the necessary potential through LT. The electrons are allowed to pass through the two thin aluminum diaphragms D1 and D2 in order to get a fine parallel beam of electrons. By applying the necessary potential V to the aluminium cylinder A, the electrons are accelerated, which are then allowed to strike the single crystal of nickel N which is fixed with the handle H. In order to get the diffraction at different points on the surface of the nickel N, one can rotate the nickel crystal through the handle H at any desired axis of the incident electron beam. When the fast-moving electron beam strikes the nickel crystal, scattering of electrons takes place according to the Braggs diffraction formula for X-rays, 2d sin q = nl (14.14) where d is the interplanar distance; q, the scattering angle; n, the order of the spectrum and λ, the wavelength of the incident beam. The scattered electron will be collected with the help of the detector known as the Faraday cylinder C. The detector moves on a graduated circular scale S. The Faraday cylinder collects the scattered electrons in all the directions by moving the same on the graduated scale. The amount of current produced by the collected electrons by C is measured through the galvanometer G. The position of the cylinder on the scale measures the angle of scattering of the electrons. Then, by measuring the peak position of the deflection in the galvanometer and the position of the cylinder on the scale, one can obtain respectively the maximum current for the scattered electrons and the maximum scattering angle. One can measure the number of electrons received on the Faraday cylinder and the scattering angle for a given accelerating voltage V. The experiment can be repeated for different accelerating voltages, V. The above observation can be plotted by taking the angle along the x-axis and the number of electrons scattered along the y-axis, as shown in Fig.14.3a. A peak in the number of scattered electrons occurs at an angle of 50˚ for an applied voltage of 54 V. This means that the diffraction effect which exhibits from the surface layer of the crystal acting as a plane grating produces the first- order spectrum at an angle of 50˚. 14.6 Engineering Physics The wave nature of the electron can be verified by determining the wavelength of the reflected waves by considering both its particle and wave nature. The wavelength of the electron, by considering the particle nature, can be determined from Bragg’s diffraction formula. The arrangements of atoms in the nickel crystal are shown in Fig.14.3b. The interatomic distance for the nickel crystal is 2.12 ¥ 10-10 m. From the top surface of the crystal, the reflecting planes in the crystal are inclined at an angle of 250. Therefore, the interplanar distance d = 2.15 ¥ 10–10 sin 25˚ = 0.909 ¥ 10–10 m (14.15) No. of scattered electrons 5° 5° 5° 65° d 50° a 0° 30° 60° q a) Arrangement of atoms b) Scattering angle versus number of electrons collected Fig.14.3 Scattering angle and arrangement of atoms in nickel crystal According to Bragg’s formula, 2 ¥ 0.909 ¥ 10–10 sin (90° – 25°) = 1 l Therefore, the de Broglie wavelength l = 1.65 Å (14.16) According to de Broglie, the electrons, which are accelerated by a potential difference of 54 V, 12.28 de Broglie wavelength l = = 1.67 Å (14.17) 54 Equations (14.16) and (14.17) confirm the close agreement between the two wavelengths of the electron obtained by considering the wave and particle nature of electrons. 14.5.2 Thomson’s Experiment Another experiment which is used to study the wave nature of the electron was developed by GP Thomson. The Thomson experiment is similar to Von Laue’s X-ray diffraction experiment. The schematic representation of the experimental set-up is shown in Fig.14.4.  Waves and Particles 14.7 A G C S Pump P Fig.14.4 Thomson experiment—dual nature verification The accelerated fast-moving narrow beam of electrons is produced by the cathode C by applying a potential of 50 kV. In order to get a fine narrow beam of electrons, it is further passed into a fine hole H as shown in Fig. 14.5. These electrons are made incident on the thin gold foil F. Thus, the diffraction of the electrons takes place at F. The diffracted electrons are made to fall either on a florescent screen S or a photographic plate P, as shown in Fig.14.4. One can use either a photographic plate (or) screen to obtain the pattern. When an electron is incident on a thin foil, both diffraction and generation of secondary X-rays takes place. The wave nature of the electrons can be verified by considering the reason for the existence of the diffracted pattern, i.e., the pattern is due to the existence of electrons or secondary X-rays. Fig.14.5 Diffraction pattern The different patterns produced by the electrons which are diffracted at the foil F is shown in Fig.14.5. The pattern consists of concentric rings with increasing diameter. This is possible due to the periodic arrangement of atoms in the gold foil and it obeys the Braggs diffraction condition to give proper reflection. The above results confirm that the pattern is only due to the existence of electrons and not due to the secondary X-rays coming out of the thin foil F. The same can be explained in another way. When the foil is removed, the entire pattern disappears which confirms that the foil is essential to obtain the diffraction pattern. On the other hand, if the electron behaves as a particle when it is incident on the foil, it gets scattered over a wide angle. The diffracted electrons are the ones that produce the diffraction pattern. From the above experiment, it is clear that electrons behave like a wave since the obtained diffraction pattern is possible only by waves. The size of the crystal unit 14.8 Engineering Physics cell can be obtained by knowing the ring diameter and the de Broglie wavelength of electrons. The size of the unit cell obtained by this method and X-rays are in close agreement. 14.6 SCHrÖdInGEr WAVE EQUATIOn In 1926, Schrödinger developed mathematical equations to explain both microscopic and macroscopic particles by predicting the wave functions at any particular point. The Schrödinger equation is one of the basic equations in quantum mechanics similar to Newton’s law of motion. Schrödinger derived two forms of wave equations, namely, time independent wave equation, and time-dependent wave equation, to explain the dual nature of matter waves. de Broglie made two assumptions to derive the equation for the wave. In his first assumption, the de Broglie wavelength concept was applied to any matter waves under any field of force. When such an external force is applied, the particles will have both potential and kinetic energy, i.e., E = PE + KE E = V + (1 /2 ) mv2 p2 The total energy E=V+ (14.18) 2m Rearranging Eq.(14.18), we get, p = [2m (E – V)] 1/2 (14.19) According to de Broglie wavelength, we know that, h or l= p h \ the de Brogile wavelength l = (14.20) [2m( E - V )]1/ 2 According to de Broglie’s second assumption, the one-dimensional wavefunction associated with the particle is y = y0 e–iwt (14.21) where y0 is the amplitude of the wave function at the point (x,y,z) at any time which is independent of time and w = 2pn, where n is the frequency of radiation. 14.6.1 Time Independent Wave Equation Consider a system of stationary waves associated with a moving particle. y is the wave function of the particle along x, y and z co-ordinate axes at any time t. Based on the Cartesian co-ordinates, one can write the differential wave equation of a progressive wave with wave velocity u as, d 2y d 2y d 2y 1 d 2y + + = ¥ (14.22) dx 2 dy 2 dz 2 u2 dt 2  Waves and Particles 14.9 The solution for Eq. (14.22) is Eq. (14.21). d 2y The value of can be obtained by differentiating Eq. (14.21) w.r.t. t twice. dt 2 dy = - iwy 0 e - iw t = - iwy dt or d 2y = - w 2y (14.23) dt 2 Substituting the value of d2y/dt2 in Eq. (14.22), we get, d 2y d 2y d 2y w2 + + =- y (14.24) dx 2 dy 2 dz 2 u2 We know that w = 2p n. (2pn ) 2 Therefore, Eq. (14.24) can be written as, =- y u2 d 2y d 2y d 2y 4p 2 n 2 + + =- y (14.25) dx 2 dy 2 dz 2 u2 Substituting the value of u ( = nl) in Eq. (14.25), we get, d 2y d 2y d 2y 4p 2n 2 2 + 2 + 2 =- y (14.26) dx dy dz l 2n 2 Substituting the value of the wavelength from Eq. (14.20) in Eq. (14.26), we get, d 2y d 2y d 2y 4p 2 2 + 2 + 2 =- ¥ 2m( E - V ) y (14.27) dx dy dz h2 We know that — 2 is the Laplacian operator and is equal to ∂2 ∂2 ∂2 — 2= 2 + 2 + (14.28) ∂x ∂y ∂z 2 The Laplacian operator operates on the wave function and reproduces the corresponding wave function and its energy. Equation (14.27) can be written as, 8p 2 —2y = - m( E - V ) y (14.29) h2 14.10 Engineering Physics 2m =- ( E - V )y 2 h where ħ= 2p 2m Therefore, —2 y + (E - V ) y = 0 (14.30) 2 The Schrödinger time independent equation can be taken either from Eq. (14.29) or (14.30) based on the requirements. 14.6.2 Time dependent Wave Equation On the other hand, for a free particle one cannot use Eq. (14.30), since for a free particle, the potential energy V = 0. Hence, a more general time-dependent equation is required. Therefore, Eq. (14.30) can be written by taking V = 0. 2m —2y + Ey = 0 (14.31) 2 According to Schrödinger’s second assumption, Eq. (14.20) can be differentiated w.r.t. t, and we get, dy = -iw y = - i 2 p n y (14.32) dt Where E = hn. We get, dy -i 2 p E E = y =-i y dt h   dy Ey = - i dt or dy Ey =i (14.33) dt According to Schrödinger’s time-independent equation, 2m 2m —2 y + 2 Ey - Vy =0  2 Substituting the value of Ey from Eq. (14.33), we get, 2m dy 2m — 2y + i ¥ - 2 Vy =0 (14.34) 2 dt  Multiplying Eq. (14.34) by  2 / 2m , we get,  Waves and Particles 14.11 2 dy — 2 y + i -V y = 0 2m dt or 2 dy - — 2 y + V y = i (14.35) 2m dt H is a Hamiltonian operator and is equal to 2 H= - — 2+V (14.36) 2m Substituting the value of H in Eq. (14.36), we get, Hy = Ey (14.37) dy where E y = i  ¥ is an energy operator. dt The Schrödinger time-dependent wave equation can be either in the form of Eq. (14.35) or (14.37) depending on the applications. 14.6.3 Physical Significance of Wave Functions The following are the significance of a wave function: (i) It gives a statistical relationship between the particle and wave nature. (ii) It is a complex quantity and hence, one cannot measure it. (iii) It is a function of position and time co-ordinate. Hence, it cannot be used to locate the position of particle. (iv) A new term known as position probability density P (r,t) is introduced to explain the wave function. It is equal to the product of the wave function y and its complex conjugate as, 2 P (r , t ) = y (r , t ) (14.38) (v) The probability of finding a particle within a volume dv is Ú 2 P = y dv (14.39) where d v = dx dy dz. (vi) When a particle definitely exists in a volume dv then its probability density is one, + Ú 2 i.e, P= y dv = 1 (14.40) - (vii) Wave function does not have any physical meaning while the probability density has physical meaning. 14.12 Engineering Physics 14.7 APPLICATIOn OF SCHrÖdInGEr’S EQUATIOn TO A PArTICLE In A BOX L V O L x Fig. 14.6 Schrödinger equation application Particle in a box Consider a one-dimensional box with width x = 0 to x = L and infinite height as shown in Fig.14.6. Let E be the total energy of the particle with a mass m moving freely in the x-direction inside the potential box. The particle is moving inside the box with limitations x = 0 and x = L, and hence, its motion is restricted by the walls. Therefore, the particle will be inside the box and hence, its potential energy is zero. Therefore, one can use the Schrödinger time-independent equation to obtain the wave function and energy of the particle inside the box. The particle is inside the box and hence, the potential energy is zero i.e., for 0 < x < L V=0 Therefore, the wave function 2 i.e., for 0 > x > L y = 0 y=0 The particle cannot be outside the box and hence, the potential energy outside the box is infinite, i.e., for 0 ≥ x ≥ L V= 2 Therefore, the wavelength y π 0 for 0< x < L According to Schrödinger, the one-dimensional time independent wave equation is d 2 y 2m + 2 ( E - V )y = 0 (14.41) dx  We know that for a freely moving particle, V = 0 in Eq. (14.41), and we get, d2y 2m 2 + Ey = 0 (14.42) dx 2 or d2y + k 2y = 0 (14.43) dx 2  Waves and Particles 14.13 2m where k2 = E (14.44) 2 Equation (14.43) is similar to the equation of harmonic motion, and the solution of the above equation gives the wave function of the particle inside the box, y = A sin kx + B cos kx (14.45) where A and Bare the amplitude of the electron’s wave function. Applying the boundary conditions, we get, x = 0 and y = 0 Therefore at this boundary condition the move function of electron is Zero. Substituting the boundary conditions no. Eq. (14.45),we get, B=0 Similarly, when x = L and y = 0, we get 0 = A sin kL We know that the electron is present inside the box and hence, A π 0 Therefore, we can write, sin kL = 0 But, sin kL = 0, only when kL = n p np Therefore, k = (14.46) L Squaring Eq. (14.46) and comparing with the value of k2 in Eq. (14.44), we get, 2m n2 p 2 E= 2 L2 Rearranging the above equation, we get, n2 p 2 2 En = ¥ L2 2m Substituting the value of  (= h / 2p ), we get, n2 p 2 h2 En = ¥ L2 8p 2 m or n2 h2 En = (14.47) 8mL2 Eq. (14.47) given the permited energy levels of an electron in one dimentional at potential box. Therefore, the wave function of the particle can be written by applying the boundary conditions to value of k from Eq. (14.46) in Eq. (14.45), and we get, np x y n = A sin (14.48) L One can determine the value of A by considering the definite existence of an electron inside the box, Ú y dx =1 2 i.e., (14.19) - 14.14 Engineering Physics L Ê np xˆ Ú A2 sin 2 Á Ë L ˜¯ dx = 1 (14.50) 0 Rearranging the above equation, we get, L Ê np xˆ 1 Ú sin 2 Á Ë L ˜¯ dx = 2 A 0 L 1 È1 - cos(2 p n x / L) ˘ or A 2 = Ú Í Î 2 ˙ dx ˚ 0 Integrating the above equation, we get, 1 L 2 = A 2 2 Therefore, A= (14.51) L Substituting the value of A in Eq. (14.48), we get, The wave function of the electron, 2 np x yn = sin (14.52) L L Equations (14.52) and (14.47) give respectively the wave function and the corresponding energy levels. For each value of n, the above equations give the corresponding wave functions and energy levels. The allowed value of the wave function is known as eigen function, and the energy value is known as eigen value. We know that the particle is inside the box and hence, it will have a discrete value of wave function and energy. These wave functions and energies are normalised and are represented respectively by y1, y2, y3, y4 etc., and E1, E2, E3, E4, etc. The normalised wave function and energy of the particle inside the box are shown respectively in Fig.14.7. y n=1 n=3 n=2 Fig.14.7 One-dimensional potential box—energy levels  Waves and Particles 14.15 14.8 HEISEnBErG UnCErTAInTY PrInCIPLE Based on the classical theory, it is clear that each and every particle will have a fixed position and momentum in a given region of space. One can determine these quantities exactly at any given point of time. During measurement, there will be an inherent uncertainty in the determination of the position and momentum. The uncertainty in the simultaneous determination of position and momentum can be explained based on the Heisenberg uncertainty principle. The Heisenberg principle states that the simultaneous determination of a pair of physical quantities like position and momentum of a particle cannot be determined with the required accuracy. Let Dx be the error in determining its position and Dp, the error in determining its momentum during the simultaneous determination. According to the Heisenberg principle, the above error quantities are related as 1 h h Dx Dp  or (14.53) 2 2p 4p Equation (14.53) states that the product of two errors is approximately of the order of Planck’s constant. Let DE and Dt be respectively the uncertainty in the measurement of energy and time. Then according to the Heisenberg principle, 1 h h Dt DE = 2 2 p or 4p (14.54) Equation (14.53) and (14.54) clearly state that it is impossible to design an experiment to prove the wave and particle nature of matter at any given instant of time. If one measures the position or momentum accurately then there will be an uncertainty in the other. Thus, the Heisenberg uncertainty principle gives the probability of determining the particle at any given instant of time instead of a certainty. 14.8.1 Illustration Consider that a beam of electrons is incident on a slit S of width Dx, which is perpendicular to the direction of the beam as shown in Fig. 14.8. The initial momentum of the electrons is p = mv. The electrons acquire momentum along OA after getting diffracted at the slit S. The angular deflection q of the diffracted electrons depend on the component of the momentum parallel to the slit, i.e., Dp = p sin q For a small angular deflection, Dp = pq (14.55) The deflection angle q0 for the first minimum of the diffracted pattern is x q0 = Dx or l Dx = (14.56) q0 14.16 Engineering Physics The product of uncertainty in position and momentum l Dx Dp = pq q0 or q Dx Dp = p l (14.57) q0 S O q q P A Fig. 14.8 Heisenberg uncertainty principle—illustration For a small value of q0, it is nearly equal to q (i.e., q ª q0 approximately) Therefore, DxDp ~ h (14.58) where p = h/l The probable deflection q of the electron is less than q0. Therefore, 1 h h D x D p≥ or (14.59) 2 2p 4p Equation (14.59) states the uncertainty relation for the position and momentum according to the Heisenberg principle. Keypoints to remember All matter particles will have an associated wave known as matter wave or de Broglie wave.  Waves and Particles 14.17 h h The de Broglie wavelength l = = mv p h The de Broglie wavelength of an electron l = 2 m eV Matter waves are known as pilot waves (or) guide waves. The Davission–Germer and GP Thomson experiments are used to verify the wave nature of the particles experimentally. Schrödinger’s time-independent and time-dependent wave equations are used to describe the dual nature of matter waves. 2m Schrödinger’s time-independent equation is — 2 y + 2 ( E - V ) y = 0  2 dy Schrödinger’s time-dependent equation is - — 2 y + V y = i 2m dt Schrödingers time–dependent equation in terms of the Hamiltonian operator Hy = Ey According to the Heisenberg uncertainty principle, it is not possible to determine simultaneously a pair of physical quantities like position and momentum to a very high precision. Solved Problems Example 14.1 Determine the de Broglie wavelength of an electron accelerated by a potential difference of (i) 150 V, (ii) 5000 V, (iii) 400 V. Given Data (i) Potential difference V = 150 V (ii) Potential difference V = 5000 V (iii) Potential difference V = 400 V Solution Case I h The de Broglie wavelength of an electron, l= (i) 2meV where h is the Planck’s constant; m, the mass of the electron; and V, the potential difference applied. Substituting the constant value in Eq. (i), we get, 14.18 Engineering Physics 6.626 ¥10-34 The de Broglie wavelength of an electron l = 2 x ¥ 9.1¥10-31 ¥ 1.6 ¥ 10-19 ¥ 150 = 1.0018 ¥ 10-10 m The de Broglie wavelength of the electron is 1.0018 ¥ 10-10 m Case II h The de Broglie wavelength of an electron, l= (ii) 2meV where h is the Planck’s constant; m, the mass of the electron; and V, the potential difference applied. Substituting the constant value in Eq. (ii), we get, 6.626 ¥10-34 The de Broglie wavelength of an electron l = 2 ¥ 9.1¥10-31 ¥1.6 ¥10-19 ¥ 5000 = 0.17353 ¥ 10-10 m The de Broglie wavelength of the electron is 0.17353 ¥ 10-10 m Case III h The de Broglie wavelength of an electron, l= (iii) 2meV where h is the Planck’s constant; m, the mass of the electron; and V, the potential difference applied. Substituting the constant value in Eq. (iii), we get, 6.626 ¥10-34 The de Broglie wavelength of an electron l = 2 ¥ 9.1¥10-31 ¥ 1.6 ¥ 10-19 ¥ 400 = 0.6136 ¥ 10-10 m The de Broglie wavelength of the electron is 61.36 ¥ 10-10 m The de Broglie wavelength of the electron (i) accelerated for a potential of 150 V is 1.0018 ¥ 10-10 m (ii) accelerated for a potential of 5000 V is 0.17353 ¥ 10-10 m (iii) accelerated for a potential of 400 V is 0.6136 ¥ 10-10 m Example 14.2 Determine the wavelength associated with an electron having an energy of 100 eV Given Data Energy of the electron, E = 100 eV Solution Energy of the electron, E = 100 ¥ 1.6 ¥ 10-19 = 1.6 ¥ 10-17 J  Waves and Particles 14.19 h The de Broglie wavelength of an electron l = 2mE 6.626 ¥10-34 Substituting the values of h, m and n, we get = 2 ¥ 9.11¥10-31 ¥1.6 ¥10-17 l = 1.227 ¥ 10-10 m The de Broglie wavelength of the electron is 1.227 ¥ 10-10 m. Example 14.3 A neuton of 1.675 ¥ 10-27 kg mass is moving with a kinetic energy of 10 KeV. Calculate the de Broglie wavelength associated with it. Given Data Kinetic energy of the neutron E = 10 keV E = 10 ¥ 103 ¥ 1.6 ¥ 10-19 = 1.6 ¥ 10-15 J Solution h The de Broglie wavelength of the neutrons l = 2mE 6.626 ¥ 10-34 = 2 ¥ 1.675 ¥ 10-27 ¥ 1.6 ¥ 10-15 = 2.861 ¥ 10-13 m The de Broglie wavelength is 2.861 ¥ 10-13 m. Example 14.4 Determine (i) the velocity of an electron, (ii) phase velocity of the electron wave, (iii) the de Broglie wavelength, (iv) momentum, and (v) the wave number of an electron wave. The electrons which are at rest are accelerated through a potential difference of 125 V. Given Data Potential difference applied, V = 125 V Solution 2eV 2 ¥ 1.6 ¥ 10-19 ¥ 125 (i) The velocity of the electron, v= = m 9.1 ¥ 10-31 = 6.6299 ¥ 106 m s-1 c2 (3 ¥ 108 ) (ii) The phase velocity of the electron, u = = v 6.6299 ¥ 106 = 1.3574 ¥ 1010 m s-1 h 6.626 ¥ 10-34 (iii)The de Broglie wavelength, l= = mv 9.1 ¥ 10-31 ¥ 6.6299 ¥ 106 14.20 Engineering Physics = 1.0982 ¥ 10-10 m l = 1.0982Å (iv) The momentum of the electron, p = mv = 9.1 ¥ 10-31 ¥ 6.6299 ¥ 106 = 6.033 ¥ 10-24 kg m s-1 1 (v) The wave number of the electron wave, n = l 1 = 1.0982 ¥ 10-10 9 -1 = 9.1058 ¥ 10 m The velocity of the electron is 6.6299 ¥ 106 m s-1. The phase velocity of the electron is 1.3574 ¥ 1010 m s-1. The de Broglie wavelength is 1.0982 Å. The momentum of the electron is 6.033 ¥ 10-24 kg m s-1. The wave number of the electron wave is 9.1058 ¥ 109 m-1 Example 14.5 Calculate the energy of the electron in the energy level immediately after the lowest energy level, confined in a cubical box of 0.1 nm side. Also, find the temperature at which the energy of the molecules of a perfect gas would be equal to the energy of the electrons in the above-said level. Given data The length of the potential well L = 0.1 ¥ 10-9m Solution For the lowest energy level n x = ny = 1 Let nz be the higher level nz = 2 and hence, the lowest level will be n x = ny = 1 The energy of the electron in the (1 1 2) state is h2 E112 = (12 + 12 + 22 ) 8 m L2 6h 2 = 8mL2 Substituting the values of h, m and L in the above equation, we get, 6 ¥ (6.62 ¥ 10 - 34) 2 E112 = 8 ¥ 9.1 ¥ 10-31 ¥ (0.1 ¥ 10-9 ) ¥ 2 = 36.12 ¥10-18 J Therefore, the energy of the electron is 36.12 ¥ 10-18 J  Waves and Particles 14.21 The energy E112 is equal to the average energy of the molecule. 3 Therefore, E112 = kT 2 2 \ The temperature T = E112 3k Substituting the values of E112 and k in the above equation, we get 2 ¥ 36.12 ¥ 10-18 T= 3 ¥ 1.38 ¥ 10-23 = 17.45 ¥ 105 K The energy of the electron is 36.12 ¥ 10-18 J The temperature of the system is 1.745 ¥ 106 K Example 14.6 An electron is trapped in a one-dimensional box of 0.1nm length. Calculates the energy required to excite it from its ground state to the fifth excited state. Given data The length of the one-dimensional box = 0.1 nm = 0.1 ¥ 10-9 m Solution We know that the energy of the electron in a one-dimensional well is n2 h2 En = 8mL2 For the ground state, n = 1 Substituting the values of h, m, and L in the above equation 12 ¥ (6.626 ¥ 10-34 ) 2 E1 = 8 ¥ 9.11 ¥ 10-31 ¥ (0.1 ¥ 10-9 ) 2 = 6.0233 J 6.0233 = = 37.64 eV 1.609 ¥ 10-19 For the fifth excited state, n = 6 62 ¥ (6.626 ¥ 10-34 ) 2 \ E6 = 8 ¥ 9.11 ¥ 10-31 ¥ (0.1 ¥ 10-9 ) 2 = 2.16805 J 2.16805 = = 1355.03 eV 1.609 ¥ 10-19 \ The energy required to excite the electron from the ground state to the fifth excited state is given by 14.22 Engineering Physics E = E6 – E1 = 1355.03 – 37.64 = 1317.93 eV \ The energy required to excite the electron from the ground state to the fifth excited state is 1317.39 eV. Example 14.7 Find the energy of the electron moving in one dimension in an infinitely by high potential box of 0.1 nm width. Given data The length of the one-dimensional potential box is 0.1 nm = 0.1 ¥ 10-9 m Solution For the lowest energy level, n= 1 The energy of the electron in a ground state is h2 E1 = (1) 2 8mL2 h2 = 8mL2 Substituting the value of h, m and L in the above equation, we get, (6.626 ¥ 10-34 ) 2 E1 = 8 ¥ 9.11 ¥ 10-31 ¥ (0.1 ¥ 10-9 ) 2 6.0233 ¥ 10-18 = 6.0233 ¥ 10-18 J = J 1.609 ¥ 10-19 = 37.64 eV The energy of the electron in the ground level is 37.64 eV Example 14.8 En +1 - En For the one-dimensional well, show that limit =0 En nÆ Given data The limit of the quantum number nÆ Solution We know that the energy of the electron in a one-dimensional well is n2 h2 En = 8mL2 Similarly, for the (n + 1)th state, the energy (n +1) 2 h 2 En+1 = 8mL2  Waves and Particles 14.23 \The difference in the energy between nth and (n + 1)th state, h2 En+1 – En = È(n + 1) 2 - n 2 ˘ 8 m L2 Î ˚ Rearranging the above equation, we get, E( n+1) - En (n + 1) 2 - n 2 = En n2 Given that n Æ \ (n + 1)2 – n2 Æ 0 Therefore, En +1 - En limit =0 En NÆ Example 14.9 If the uncertainty in position of an electron is 4 ¥ 10-10 m, calculate the uncertainty in its momentum. Given data The electron in the position Dx = 4 ¥ 10-10 m Solution We know from the Heisenberg uncertainty principle, Dx Dp ª h h Therefore, Dp = Dx Substituting the value of h and Dx in the above equation, we get, 6.6 ¥ 10-34 The error in the momentum Dp = = 1.65 ¥ 10-24 kg m s-1. 4 ¥ 10-10 The uncertainty in momentum Dp is 1.65 ¥ 10-24 kg m s-1. Example 14.10 Calculate the minimum energy that an electron can posses in an infinitively deep potential well of 4 nm width. Given data The width of the potential well L = 4 nm = 4 ¥ 10-9 m Solution n2 h2 Energy of the electron in a one-dimensional potential well = 8mL2 12 ¥ (6.626 ¥ 10-34 ) 2 E = -34 -9 2 = 3.765 ¥ 10-21 Joules 8 ¥ 9.11 ¥ 10 ¥ (4 ¥ 10 ) 14.24 Engineering Physics 3.765 ¥ 10-21 J E = = 0.235 eV 1.6 ¥ 10-19 Example 14.11 An electron is confined to a box of 10-9 m length. Calculate the minimum uncertainty in its velocity. Given data The electron in the position Dx = 10-9 m Mass of the electron m = 9.1 ¥ 10-31 kg Planck’s constant h = 6.6 ¥ 10-34 J s Solution We know that Dx Dp = h When Dx is maximum, the error in Dp, i.e., the uncertainty in momentum is minimum. i.e., (Dx)max (Dp)min = h h or ( D p ) min = ( Dx) max Substituting the values of Dx and h in the above equation, we get, 6.6 ¥ 10-34 ( D p ) min = 10-9 = 6.6 ¥ 10-25 kg m s-1 We know that minimum error in momentum ( D p ) min = m ( D v) min ( D p ) min Therefore, ( D v) min = m Substituting the values of ( D p ) min and m in the above equation, we get, The minimum error in velocity ( Dpx ) min 6.6 ¥10-25 ( D vx ) min = = = 7.3 ¥105 ms -1 m 9.1¥10-31 The minimum uncertainty in velocity is 7.3 ¥ 105 m s -1. Objectives Questions 14.1 Electrons, protons and neutrons are the examples for ______.  Waves and Particles 14.25 14.2 The wave function is denoted by ______. 14.3 Matter waves are not ______ waves. 14.4 Matter waves are used as ______ to guide the matter particles. 14.5 In Davission and Germer experiment, electron is selected as a particle due to ______ and ______. 14.6 Bragg’s diffraction formula 2d sinq = n ______. 14.7 The de Broglie wavelength of electron accelerated through a potential of 54 V is ______ A˚. 14.8 Two forms of Schrödinger’s equation are ______ and ______. 14.9 When the particle is inside the box, it’s potential energy is equal to ______. 14.10 When the electron is inside the well, the wave function associated with the electron is ______. 14.11 The wave function corresponding to each En is known as ______. 14.12 Heisenberg uncertainty principle Dx Dp = ______. KEY 14.1 Matter particles 14.2 y 14.3 Electromagnetic 14.4 Pilot 14.5 Easy generation, detection 14.6 l 14.7 1.67 14.8 Time independent, time dependent 14.9 Zero 14.10 Not equal to zero h 14.11 Eigen functions 14.12 2p Exercises 14.1 Determine the de Broglie wavelength of an electron accelerated through a potential difference of 200 eV. 14.2 Compute the de Broglie wavelength of a proton whose kinetic energy is equal to the rest energy of an electron. Mass of a proton is 1836 times than that of the electron. 14.3 Compare the de Broglie wavelength, speed and momenta of a proton and a-particle having the same kinetic energy. 14.4 Determine the energy levels and wave function for an infinitely deep two- dimensional potential well. Given that V = 0 for 0 < ¥ < d and 0 < y < d and v = a for outside of this region. 14.5 Determine the energy levels for a particle with mass m in a one-dimen- sional potential box shown in Fig. 14.10. Given that V = V1 for x £ 0, V = 0 for 0 < x < and V = V2 for x ≥ d 14.26 Engineering Physics V V1 V2 0 V Distance Fig. 14.10 Energy levels of a particle 14.6 Applying the uncertainty principle, show the non-existence of an electron inside the nucleus. 14.7 A particle is different at a certain angle and the uncertainty in the mea- surement of the angle is one second of the angle. Estimate the error in the measurement of its angular momentum. 14.8 An electron has a speed of 600 m s-1 with an accuracy of 0.005%. Determine the certainty with which one can locate the position of the electron. 14.9 Determine the uncertainty in velocity if the uncertainty in the location of a particle is equal to its de Broglie wavelength. Short Questions 14.1 What is meant by dual nature? 14.2 Explain particle wave duality. 14.3 What is meant by matter waves? 14.4 Define de Broglie wave length. 14.5 Write down the de-Broglie matter wave equation?. 14.6 What are the ideas prompted de Broglie, to propose his concept about matter waves? 14.7 Define wave function. 14.8 What is meant by Schrödinger equation? 14.9 Write the Schrödinger time independent equation and explain the various terms. 14.10 Write the Schrödinger time dependent equation and explain the various terms. 14.11 Drive an expression for the de Broglie wavelength of an electron. 14.12. What is the physical significance of wave function? 14.13 What is meant by eigen function?  Waves and Particles 14.27 14.14 Define Eigen value. 14.15 State and explain Heisenberg uncertainty principle. descriptive Questions 14.1 Explain the particle wave duality. Drive the equation for the de Broglie equation for an electron accelerated through a potential difference V. 14.2 What are the matter waves? Explain with neat sketch the existence of matter waves using Thomson’s experimental method. 14.3 Explain with neat sketch the experimental verification of matter waves using Davisson-Germer experiment. 14.4 Derive the Schrödinger time independent and time dependent wave equation. 14.5 Explain the physical significance of the wave functiony ? Explain the application of Schrödinger wave equation to a particle in a box. 14.6 Explain Heisenberg uncertainty principle with suitable illustration. 14.7 Write note on the following Matter waves Particle in a box Heisenberg uncertainty principle