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1 1.1 ELEMENTS OF QUANTUM MECHANICS INTRODUC TION A revolution took place in physics between 1900 and 1930. This was the era of a new and more general scheme called quantum mechanics. This new approach was highly successful in explaining the behaviour of atoms, molecules and nuclei. Moreover, the quantum theory raduces to classical physics when applied to macroscopic systems. The basic ideas of quantum theory were first introduced by Max Planck, but most of the subsequent mathematical developments and interpretations were made by a number of distinguished physicists including Einstein, Bohr, Schrödinger, de Broglie, Heisenberg, Born and Dirac. This chapter is simply an introduction to the underlying ideas of quantum theory and the wave-particle nature of matter. We shall also discuss some simple applications of quantum theory. 1.2 WA VE-PARTICLE DU ALIT Y WAVE-PARTICLE DUALIT ALITY It is well recognised that the same light beam that can eject photoelectrons from a metallic surface can also be diffracted by a grating. In other words, the photon concept and the wave theory of light complement each other. Thus, all forms of electromagnetic radiation can be described from two points of view. In one extreme electromagnetic waves describe interference and diffraction pattern formed by a large number of photons while in other extreme, the photon description is natural when we deal with highly energetic photon of very short wavelength. Light and other electromagnetic radiation sometimes act like waves and sometimes like particles. Interference and diffraction demonstrate wave behaviour while emission and absorption of photons demonstrate the particle behaviour. Besides waves that sometimes acts like particles, quantum mechanics extend the concept of wave-particle duality to include particles that sometimes show wavelike behaviour. In these situations, a particle is modelled as an inherently spread-out entity that cannot be described as a point with a perfectly definite position and velocity. 1 2 LASER SYSTEMS AND APPLICATIONS 1.3 de BR OGLIE W AVES BROGLIE WA de Broglie postulated that a free particle with mass m moving with speed v should have a wavelength λ related to its momentum exactly the same way as for a photon. A photon of light of frequency ν has the momentum. h . p = c Since λν = c so, the wavelength of a photon is specified by its momentum according to the relation h . λ = ...(1.1) p de Broglie suggested that eqn. (1.1) is a completely general one that applies to material particles as well as to photons. The momentum of a particle of mass m and velocity v is p = mv and its de Broglie wavelength is accordingly. h . mv In eqn. (1.2) m is the relativistic mass λ = m = m0 1 v2 c2 ...(1.2) . The wave and particle aspects of moving bodies can never be observed at the same time. In certain situations a moving body exhibits wave properties and in others it exhibits particle properties. Which set of properties is most conspicuous depends upon how its de Broglie wavelength compares with its dimensions and the dimensions of whatever it interacts with. de Broglie Wave Velocity If we call the de Broglie wave velocity as W, we may write W = νλ. Where λ is the de Broglie wavelength and ν is the frequency. To find the frequency, we equate the quantum expression E = hν with the relativistic formula for total energy E = mc2 hν = mc2 mc 2 . h The de Broglie wave velocity is therefore ν = mc2 h c 2 W = νλ = . h mv v Since particle velocity v must be less than the velocity of light c, de Broglie waves always travel faster then light. In order to understand this unexpected result, we must understand distinction between phase velocity and group velocity. Phase velocity is the velocity of wave. ELEMENTS OF QUANTUM MECHANICS 1.4 3 WA VE EQU ATION WAVE EQUA The equation of the wave travelling in the direction of increasing x and whose vibrations are in the y direction can be given as x y = A cos 2π ν t W where A is the amplitude of vibrations (that is, their maximum displacement on either side of the x-axis) and ν their frequency. The wave has some speed W and travels the distance x in time t. Since the wave speed W is given by W = νλ. We have wave equation x y = A cos 2π t . ...(1.3) Equation (1.3) is more convenient to use. The most widely used description of a wave, however, is still another form with angular frequency and wave number. Angular frequency ω and wave number k can be defined by the formulas 2 . W The unit of ω is the radian per second and that of k is the radian per meter. In terms of ω and k eqn. (1.3) becomes y = A cos (ωt – kx). ...(1.4) ω = 2πν, k = 1.5 PHASE AND GR OUP VEL OCITIES GROUP VELOCITIES The wave representation of a moving particle corresponds to a wave packet or wave group as shown in Fig. 1.1. Mathematically a wave group can be described in terms of the superposition of individual waves of different wavelengths whose interference with one another results in the variation in amplitude. If the velocities of the Fig. 1.1. Wave group waves are the same, the velocity with which the wave group travels is the common wave velocity. If the wave velocity varies with wavelength, the different individual waves do not proceed together and the wave group has a velocity different from those of the waves that compose it. Let us suppose that a wave group arises from the combination of the following two waves that have the same amplitude A but differ in angular frequency and wave number. 4 LASER SYSTEMS AND APPLICATIONS y1 = A cos (ωt – kx) y2 = A cos [(ω + dω)t – (k + dk)x]. The resultant displacement y at any time t and any position x is the sum of y1 and y2. y = y1 + y2 = 2A cos cos 1 [(2ω + dω)t – (2k + dk)x] 2 1 (dωt – dkx). 2 Since dω and dk are small compared with ω and k respectively. 2ω + dω ≈ 2ω 2k + dk ≈ 2k d dk t x ...(1.5) y = 2A cos (ωt – kx) cos 2 2 Equation (1.5) represents a wave of angular frequency ω and wave number k which has superimposed upon it a wave (the process is known as d dk modulation) of angular frequency and wave number . The effect 2 2 of this modulation is to produce successive ‘wave groups’. The ‘phase velocity’ or wave velocity W is given by k while the group velocity u of the ‘group of waves’ is given by W = νλ = 2 2 d . dk The group velocity may be greater or less than the phase velocity. It depends on the manner in which phase velocity varies with wave number. For light waves in vacuum, the group and phase velocities are the same. The angular frequency and wave number of the de Broglie waves associated with a particle of rest mass m0 moving with the velocity v are Angular frequency of de Broglie waves u = 2 m0 c 2 2 mc 2 . h h 1 v2 c2 Wave number of de Broglie waves 2 m0 v 2 2 mv . k = h h 1 v2 c2 ω = 2πν = de Broglie phase velocity c2 k v which exceeds both the velocity of the particle v and the velocity of light c, since v < c. W = ELEMENTS OF QUANTUM MECHANICS 5 The group velocity u of the de Broglie waves associated with the particle is u = d d dv dk dk dv 2 m0 v d = dk h(1 v 2 c 2 )3/2 2 m0 dk = dv h(1 v 2 c 2 )3/2 and so u = v. The de Broglie wave group associated with a moving particle travels with the same velocity as the particle. The wave velocity W of de Broglie waves has no simple physical significance. The wave velocity is a function of wavelength even in free space and here it differs from the velocity of light which may be shown as under p = mv and total energy E = mc2 E v and now using the relativistic expression for the mass of a c2 particle, the equation for its momentum can be given as So, p = m0 p = v v2 1 2 c where m0 is the rest mass of the particle. Eliminating v between the last two equation, we get 2 E2 p c4 c2 h Since we known that p = and E = hν. m02 = So, m0 = h 2 1 c c2 2 For de Broglie wave W = Hence m0 = νλ. h W2 1 c c 2 2 2 W = c 1 m02c2 2 h2 This equation shows that for a particle of rest mass m0 > 0, the wave velocity W is always greater than c and also the wave velocity of de Broglie waves is a function of λ even in free space. From which 6 LASER SYSTEMS AND APPLICATIONS Now consider electromagnetic wave as a special case of de Broglie wave and so, these de Broglie waves are propagated with a velocity W = c. The velocity of the associated particle i.e., photon is also equal to c. If W = c is substituted in the last equation given above, we find that rest mass of the photon m0 = 0 i.e., there is no such thing as a photon at rest, photon always moves with the velocity c. 1.6 DA VISSON AND GERMER EXPERIMENT DAVISSON In 1927, Davisson and Germer predicted experimentally the electron waves predicted by de Broglie. Davisson and Germer were studying the scattering of electrons from a nickel target using an apparatus like that sketched in Fig. 1.2. The energy of the electrons in the primary beam, the angle at which they are incident upon the target and the position of the detector could all be varied. The nickel target was subjected to such a high temperature treatment that the crystal was transformed into a group Fig. 1.2 of crystals. In this case the reflection became anomalous and the reflected intensity showed striking maxima and minima instead of a continuous variation of scattered electron intensity with angle. The position of the maxima and minima observed depended upon the electron energy. Then, they suspected that the beam of electron might be diffracted from the crystals like X-rays. This shows that electrons behave like waves under certain circumstances. Typical polar graphs of electron intensity after the heat treatment are shown in Fig. 1.3. Fig. 1.3 ELEMENTS OF QUANTUM MECHANICS 7 The method of plotting is such that the intensity at any angle is proportional to the distance of the curve at that angle from the point of scattering. If the intensity were the same at all scattering angles, the curves would be circles centered on the point of scattering. de Broglie hypothesis suggested the interpretation that the electron waves were being diffracted by the target, much as X-rays are diffracted by the planes of atoms in a crystal. This idea was confirmed when it was realised that the effect of heating a block of nickel at high temperature is to cause many small individual crystals of which it is normally composed to form into a single large crystal all of whose atoms are arranged in a regular lattice. To verify whether de Broglie waves are responsible for the findings of Davisson and Germer, an analysis of the observation should be made. For the beam of electrons falling normally on the surface of the crystal, the current observed in detector is a measure of the intensity of the diffracted beam. Several curves were obtained for different voltage electrons when graphs were plotted between the colatitude (angle between the incident beam and the beam entering the detector) which are shown in Fig. 1.4. It is observed that a bump begins to appear in the curve at 44 volt electrons. This bump moves upward for 54 volts at colatiFig. 1.4 tude of 50°. Above 54 volts the bump again diminishes. The bump at 54 volts offers the evidence for the existence of electron waves. The angles of incidence and scattering relative to the family of Bragg plane shown in Fig. 1.4 are both 65°. The spacing of the planes in this family, which can be measured by X-ray diffraction is 0.091 nm. The Bragg equation for maxima in the diffraction pattern is nλ = 2d sin θ. Here d = 0.091 nm, θ = 65°. For n = 1, the de Broglie wavelength λ of the diffracted electrons is λ = (2) (0.091) (sin 65°) = 0.165 nm, We use de Broglie formula to calculate the expected wavelength of the electrons. The electron kinetic energy of 54 eV is small compared with its rest energy m0c2 of 0.51 MeV so we can ignore relativistic considerations. Since K = 1 mv 2 . 2 8 LASER SYSTEMS AND APPLICATIONS The electron momentum mv is mv = 2mk (2) (9.1 10 31 ) (54) (1.6 10 19 ) = 4.0 × 10–24 kg m/s. The electron wavelength is therefore λ = 6.63 1034 h = 1.66 × 10–10 m = 0.166 nm mv 4.0 10 24 is in excellent agreement with the observed wavelength. The Davisson-Germer experiment thus provides direct verification of de Broglie’s hypothesis of the wave nature of moving bodies. SOL VED EXAMPLES SOLVED EXAMPLE 1: Calculate the de Broglie wavelength associated with a proton mov1 th of the velocity of light. 20 SOLUTION: Velocity of proton ing with a velocity equal to 3 10 8 = 1.5 × 107 m/s 20 Mass of the protonm = 1.67 × 10–27 kg v = λ = 6.6 10 34 h = 2.634 × 10–14 m. mv 1.67 10 27 1.5 107 EXAMPLE 2: Calculate the de Broglie wavelength of neutron of energy 12.8 MeV. Given h = 6.62 × 10–34 J.sec, m = 1.67 × 10–27 kg. SOLUTION: m0c2 = 1.507 1010 = 941.87 MeV 1.6 1019 Since 12.8 MeV is very small compared to rest mass energy hence relativistic consideration may be ignored λ = λ = h mv h E = Ve where V is voltage in volts 2mE 6.62 10 34 2 1.67 10 27 12.8 106 1.6 1019 = 8.0 × 10–5 Å. 9 ELEMENTS OF QUANTUM MECHANICS EXAMPLE 3: Show that the de Broglie wavelength for a material particle of rest mass m0 and charge q accelerated from rest through a potential difference of V volts relativistically is given by h λ = qV 2m0 qV 1 2m0 c 2 . SOLUTION: Kinetic energy Ek = Vq. Ek ≠ mv2 because v is relativistic velocity and so, we cannot find momentum directly from Ek. Now, we use relativistic formula E2 = p2c2 + m02c4 E = Ek + m0c2 = Vq + m0c2 P2c2 = E2 – m02c4 = (Vq + m0c2)2 – m02c4 P2c2 = V2q2 + 2Vq m0c2 P2 = V 2 q2 Vq 2Vq m0 2m0 Vq 1 2 c 2m0 c 2 Vq 2m0 Vq 1 2m0 c2 P = ∴ de Broglie wavelength λ = 1 Vq 2m0 Vq 1 2m0 c 2 EXAMPLE 4: Calculate the wavelength associated with an electron accelerated to a potential difference of 1.25 keV. SOLUTION: If E is the K.E. of the electron, the de Broglie wavelength of the wave associated with the electron is λ = h 2mE 6.6 10 34 2 9.1 10 31 1.25 10 3 1.6 10 19 = 3.46 × 10–11. EXAMPLE 5: What will be the kinetic energy of an electron if its de Broglie wavelength equals the wavelength of the yellow line of sodium (5896 Å). The rest mass of electron is m0 = 9.1 × 10–31 kg and h = 6.63 × 10–34 J. sec. SOLUTION: de Broglie wavelength λ = h mv or v= h . m 10 LASER SYSTEMS AND APPLICATIONS In the absence of relativistic consideration m = m0 λ = h 1 , Kinetic energy K = m0 v 2 m0 v 2 K = h 1 h2 . m0 2 2m0 2 m0 2 Putting the values of h, m0 and λ K = (6.63 10 34 )2 6.95 10 25 2 9.1 1031 (5896 10 10 )2 1.6 1019 = 4.34 × 10–6 eV. EXAMPLE 6: A particle of rest mass m0 has a kinetic energy K. Show that its de Broglie wavelength is given by hc . λ = K (K 2m0 c 2 ) Hence, calculate the wavelength of an electron of kinetic energy 1 MeV. What will be the value of λ if K << m0c2? SOLUTION: According to de Broglie’s concept of matter wave, the wavelength λ = 1/2 v2 1 2 c 1 m02 m2 = m0 m = v2 c2 v2 = c2 or m2v2 = or mv = Substituting this value or λ = or h mv λ = m= m0 (1 v 2 c 2 )1/2 m2 m02 m2 (m – m02)c2 c(m2 – m02)1/2. of mv in equation of wavelength, we get 2 h c m m02 2 hc c 2 m2 m02 c2(m2 – m02)1/2 = [c4(m – m0) (m + m0)]1/2 = [c2(m – m0) {(m + m0)c2}]1/2 = [(m – m0)c2 {(m – m0)c2 + 2m0c2}]1/2 or c2(m2 – m02)1/2 = [K(K + 2m0c2)]1/2. 11 ELEMENTS OF QUANTUM MECHANICS λ = Therefore hc . [K (K 2m0 c 2 ]1/2 For an electron m0c2 = 9.1 × 10–31 × (3 × 108)2 joule = For K = 1 MeV, 81.9 1015 eV = 0.51 × 106 eV = 0.51 MeV 1.6 1019 λ = hc 1(1 (2 0.51)) hc 2.02 λ = 6.62 10 34 3 108 m 2.02 1.6 10 19 106 = 8.78 × 10–13 m = 8.78 × 10–3 Å. If K << m0c then K + 2m0c2 = 2m0c2 2 λ = hc 2m0 Kc 2 h . 2m0 K EXAMPLE 7: What is the de Broglie wavelengths of any electron which has been accelerated from rest through a potential difference of 100 V. SOLUTION: ∴ 12.25 Å V V = 100 volts λ = λ = 12.25 = 1.225 Å. 100 EXAMPLE 8: Can a photon and an electron of the same momentum have the same wavelength? Compare their wavelengths if the two have the same energy. S OLUTION : Using de Broglie concept of matter wave, momentum of the electron may be written as pe = as pph = h h . c p h and momentum of photon e So, if the electron and photon have same momentum, then we have λe = λp. Thus, the photon and electron of the same momentum have the same wavelength. The de Broglie wavelength of electron is given by h and mv where E is the energy of the electron λe = So, λe = 1 2mE 1 mv 2 = E or mv = 2 2mE 12 LASER SYSTEMS AND APPLICATIONS The de Broglie wavelength of photon is given by h hc but E = hν = = pc p hc = E λph = λph ph 2mc 2 hc 2mE 2m c . E h E e E EXAMPLE 9: Find the energy of the neutron in units of electron volt whose de Broglie wavelength is 1 Å. Now = SOLUTION: Given mass of the neutron = 1.674 × 10–27 kg Planck’s constant h = 6.60 × 10–34 joule/sec We know that or where λ = h mv h 2mE h2 2m 2 m = 1.674 × 10–27 kg λ = 1 Å = 10–10 m h = 6.60 × 10–34 J.s. E = E = = (6.60 1034 )2 2 1.674 1027 (10 10 )2 = 13.01 × 10–21 joules 13.01 10 21 = 8.13 × 10–2 eV. 1.6 10 19 EXAMPLE 10(a): What would be the wavelength of quantum of radiant energy emitted, if an electron transmitted into radiation and converted into one quantum? SOLUTION: When the energy of an electron is transmitted into radiation, we use the following relations to get the value of λ E = mc2 and So λ = E = hν = hc 6.6 10 34 h mc 9.1 10 31 3 108 = 0.0242 × 10–10 m = 0.0242 Å. EXAMPLE 10(b): A lamp of 150 W is emitting 8% of its energy as blue light having a mean wavelength of 4500 Å. How many photons are being emitted by the lamp per second. SOLUTION: Power emitted by the lamp = 150 8 W 12 W 100 The energy emitted per second = 12 J 13 ELEMENTS OF QUANTUM MECHANICS Energy contained in one photon 6.625 10 34 3 108 4500 1010 –20 = 44.2 × 10 J = hc/λ = No. of photons emitted per second = 1.7 12 27.15 1018. 44.2 10 20 UNCERT AINT Y PRINCIPLE UNCERTAINT AINTY It is impossible to know both the exact position and exact momentum of an object at the same time. To regard a moving particle as a wave group implies that there are some fundamental limits to the accuracy with which we can measure position and momentum of a particle. Fig. 1.5 The narrower the wave group of the particle, its position can be precisely determined (Fig. 1.5a). The wavelength and hence particle’s momentum cannot be established h because λ = and it is not well defined in a narrow packet. mv On the other hand, a wide wave group such as in Fig. 1.5b, the wavelength can be precisely determined but not the position of the particle. The relationship between the distance Δx and the wave number spread Δk depends upon the shape of the wave group and upon how Δx and Δk are defined. If ψ(x) representes a wave group and the function g(k) describes how the amplitudes of the waves that contribute to ψ(x) vary with wave number k and Δx and Δk are taken as the standard deviations of the respective func1 tions ψ(x) and g(k), then the minimum value of Δx Δk = . 2 It is more realistic to express the relationship between Δx and Δk as 1 . Δx Δk ≥ 2 The de Broglie wavelength of a particle of momentum p is h . λ = p The wave number corresponding to this wavelength is 2 2p . k = h 14 LASER SYSTEMS AND APPLICATIONS Hence an uncertainty Δk in the wave number of the de Broglie waves associated with the particle results in an uncertainty Δp in the particle’s momentum according to the formula hk . Δp = 2 1 1 Since Δx Δk ≥ , Δk ≥ 2(x) 2 h . Δx Δp ≥ ...(1.6) 4 Equation (1.6) states that the product of the uncertainty Δx in the position of an object at some instant and the uncertainty Δp in its momentum compoh . For nent in the x direction at the same instant is equal to or greater than 4 a narrow wave group Δx is small and then Δp will be large and in a broad wave group Δp is small and then Δx will be large. h is abbreviated as (h bar) Generally 2 h = = 1.054 × 10–34 J.s. 2 Thus, in terms of , the uncertainty principle becomes Δx Δp ≥ . 2 In the above discussion we have considered wave nature of particle while uncertainty principle can be arrived considering particle nature of waves. It is worth mentioning that the lower limit of for Δx Δp is rearely 2 attained; more usually Δx Δp ≈ or even Δx Δp ≈ h. Another form of the uncertainty principle is sometimes useful. Let energy emitted E be measured during the time interval Δt in an atomic process. The limited time available imposes restriction on the accuracy with which we can determine the frequency ν of the waves. The uncertainty Δν in our frequency measurement is 1 Δν ≥ . t Because the minimum uncertainty in the number of waves we count in a wave group is one wave and to get uncertainty Δν, it is divided by the time inverval. The corresponding energy uncertainty is ΔE = hΔν h and so ΔE ≥ or ΔE Δt ≥ h. t A more precise calculation changes this result to ΔE Δt ≥ . 2 15 ELEMENTS OF QUANTUM MECHANICS SOL VED EXAMPLES SOLVED EXAMPLE 11: Calculate the smallest possible uncertainty in the position of an electron moving with velocity 3 × 107 m/s. SOLUTION: (Δx)min (Δp)max = h 2 (Δp)max = p (momentum of the electron) mv = (Δx)min = m0 v (1 v 2 c 2 ) h 1 v2 c2 2 m0 v h = 1.05 × 10–34 J.s., v = 3 × 107 m/s, m0 = 9 × 10–31 kg, c = 3 × 108 m/s 2 (3 107 )2 1 (3 108 )2 9 10 31 3 107 1.05 10 34 (Δx)min = = 3.8 × 10–12 m = 0.038 Å. EXAMPLE 12: An electron is confined to a box of length 10–8 meter; calculate the minimum uncertainty in its velocity. Given m = 9 × 10–31 kg; = 1.05 × 10–34 joule second. SOLUTION: (Δx)max = 10–8 meter (Δp)min = 1.05 10 34 kg-m/s ( x )max 10 8 = 1.05 × 10–26 kg-m/s (Δp)min = m(Δv)min = 1.05 × 10–26 (Δv)min = 1.05 10 26 1.05 10 26 = 1.17 × 104 m/s. m 9 10 31 EXAMPLE 13: Find the uncertainty in the momentum of a particle when its position is determined within 0.01 cm. Find also the uncertainty in the velocity of an electron and an α-particle respectively when they are located within 5 × 10–8 cm. SOLUTION: According to uncertainty principle h 2 h Δp = 2 x (Δx) (Δp) = or 16 LASER SYSTEMS AND APPLICATIONS h = 1.05 × 10–34 J.s 2 Δp = Δx = 0.01 × 10–2 meter 1.05 10 34 = 1.05 × 10–30 kg m/s 0.01 102 Δp = mΔv ∴ Δv = h . 2m x Uncertainty in the velocity of electron m = 9 × 10–31 kg Δx = 5 × 10–10 m Δv = 1.05 10 34 9 1031 5 10 10 = 2.33 × 105 m/sec. Uncertainty in the velocity of α-particle, mass of α-particle = 4 × mass of proton = 4 × 1.67 × 10–27 = 6.68 × 10–27 kg Δx = 5 × 10–10 m Δv = 1.05 10 34 6.68 1027 5 10 10 = 31.4 m/sec. EXAMPLE 14: An electron has a speed 4 × 105 m/s within the accuracy of 0.01 per cent. Calculate the uncertainty in the position of the electron. Given h = 6.625 × 10–34 J.s. Mass of electron = 9.11 × 10–31 kg. SOLUTION: Uncertainty in velocity = 0.01 × 4 × 105 = 40 m/s 100 h (Δx) (Δp) = 2 (Δx) = = 1.055 10 34 h m v 2 ( p) 1.055 10 34 = 2.895 × 10–6 m. 9.11 10 31 40 EXAMPLE 15: An excited atom has an average lifetime of 10–8 sec. That is during this period it emits a photon and returns to the ground state. What is the minimum uncertainty in the frequency of this photon? SOLUTION: h 2 E = hν ΔE Δt ≥ 17 ELEMENTS OF QUANTUM MECHANICS or ΔE = hΔν h 2 1 Δν ≥ 2 t hΔν Δt ≥ or Δt = 10–8 sec Δν ≥ 1 = 15.92 × 106 sec–1. 2 3.14 10 8 EXAMPLE 16: If an excited state of hydrogen atom has a lifetime of 2.5 × 10–14 sec. What is the minimum error with which the energy of this state can be measured? SOLUTION: The uncertainty in the energy of the photon is equal to the uncertainty in the energy of the excited state of the atom due to energy conservation. According to uncertainty principle ΔE Δt ≥ ΔE ≥ = h 2 6.63 10 34 h = 4.22 × 10–21 J 2 t 2 3.14 (2.5 10 14 ) 4.22 10 21 = 0.026 eV. 1.6 1019 EXAMPLE 17: Using the uncertainty relation ΔE.Δt = h , calculate the time 2 required for the atomic system to retain rotation energy for a line of wavelength 6000 Å and width 10– 4 Å. SOLUTION: We know that E = hν = hc hc 2 where Δλ is the width of the spectral lines. Here Δλ = 10– 4 × 10–10 m. Using uncertainty relation ΔE = ΔE . Δt = Δt = h , we have 2 h 2 E Putting the value we get Δt = 2 h 2 c hc 2 2 (6 107 )2 2 3.14 3 108 1014 = 1.9 × 10– 8 second. 18 LASER SYSTEMS AND APPLICATIONS EXAMPLE 18: A nucleon is confined to a nucleus of diameter 5 × 10–14 m. Calculate the minimum uncertainty in the momentum of the nucleon. Also calculate the minimum kinetic energy of the nucleon. SOLUTION: We have the relation (Δp)min (Δx)max = (Δp)min = h 2 6.626 1034 h 2(x)max 5 10 14 2 3.14 = 0.21 × 10–20 kg m/sec. Since p cannot be less than (Δp)min Emin = ( Pmin )2 2m Putting the values of Pmin = 0.21 × 10–20 kg . m/sec and m = 1.675 × 10–27 kg. Emin = = (0.21 1020 )2 = 0.063 × 10–13 J 2 1.675 1027 0.063 10 13 = 0.039 MeV. 1.6 10 19 EXAMPLE 19: An electron is confined to a box of length 1.1 × 10–8 m. Calculate the minimum uncertainty in its velocity. Given m = 9.1 × 10–31 kg and × 10–34 J.s. h = 1.05 2 SOLUTION: We know from uncertainty principle that Δx . Δp ≥ h 2 Δp = h 2 x Let Δv be the uncertainty in the velocity of a particle of mass m, so we have Δp = mΔv or Δv = Δx = 1.1 × 10–8 m, Δv = p m h = 1.05 × 10–34 J.s. and m = 9.1 × 10–31 kg 2 1.05 1034 h 2 m x 9.1 10 31 1.1 10 8 = 1.06 × 104 m/sec. 19 ELEMENTS OF QUANTUM MECHANICS EXAMPLE 20(a): What is the minimum uncertainty in the frequency of a photon whose life time is about 10–8 sec? SOLUTION: From uncertainty principle, we have h 2 h h Δν Δt ≥ 2 ΔE Δt ≥ But E = hν or Δν ≥ or ΔE = h Δν 1 2 t 1 = 15.92 × 106/sec. 2 3.14 108 EXAMPLE 20(b): A typical atomic nucleus is about 5 × 10–15 m in radius. Use the uncertainty principle to place a lower limit on the energy an electron must have if it is to be part of a nucleus. Putting Δt = 10–8 sec, Δν = SOLUTION: Let Δx = 5 × 10–15 m 1.054 10 34 2x 2 5 10 15 ≥ 1.1 × 10–20 kg . m/s If this is the uncertainty in a nuclear electron’s momentum, the momentum p itself must be at least comparable in magnitude. An electron with such a momentum has kinetic energy many times greater than its rest energy m0c2. Thus we have E = pc E = pc ≥ (1.1 × 10–20) (3 × 108) ≥ 3.3 × 10–12 J Since 1 eV = 1.6 × 10–19 J, the energy of an electron must exceed 20 MeV if it is to be on nuclear constituent. Experiments indicate that even the electrons associated with unstable atoms never have more than a fraction of this energy and we conclude that electrons are not present within nuclear. Δp ≥ 1.8 WA VE FUNCTION AND WA VE EQU ATION WAVE WAVE EQUA The quantity with which quantum mechanics is concerned is the wave function Ψ of a body. The square of its absolute magnitude | Ψ |2 evaluated at a particular place at a particular time is proportional to the probability of finding the body there at that time. The momentum, angular momentum and energy of the body are other quantities that can be established from Ψ. In quantum mechanics we determine Ψ for a body when its freedom of motions is limited by the action of external forces. If the wave function Ψ is complex with both real and imaginary parts, the probability density | Ψ |2 is written as ΨΨ*. Where Ψ* is complex conjugate of Ψ. A complex wave function Ψ can be written in the form Ψ = A + iB where A and B are real functions Ψ* = A – iB Ψ*Ψ = A2 + B2 [i2 = –1]. 20 LASER SYSTEMS AND APPLICATIONS Thus ΨΨ* is always a positive real quantity. The condition for an acceptable wave function is that the integral of | Ψ |2 over all space must be finite, if | Ψ |2 is equal to P, then it must be true that 2 PdV dV = 1 since =1 is the mathematical statement that the particle exists somewhere at all times. A wave function that obeys above equation is said to be normalised. Every acceptable wave function can be normalised by multiplying by an appropriate constant. Besides being normalisable, Ψ must be single valued, since P can have only one value at a particular place and time and continuous. Momentum , , be finite conx y z tinuous and single valued. The wave functions with all these properties can yield physically meaningful results when used in calculations. considerations require that the partial derivatives For a particle restricted to motion in the x direction, the probability of finding it between x1 and x2 is given by Probability = x2 x1 2 dx Schrödinger’s equation which is the fundamental equation of quantum mechanics is a wave equation in the variable Ψ. 1.9 SC HR ÖDINGER’S EQU ATION: TIME-DEPENDENT SCHR HRÖDINGER’S EQUA The wave function of a freely moving particle moving in x direction can be represented by a wave function Ψ as Ψ = Ae x i t ...(1.15) Replacing ω in the above formula by 2πν and v = νλ, we get Ψ = Ae–2πi(νt – x/λ) ...(1.16) Which is convenient since we know ν in terms of the total energy E and momentum p of the particle being described by Ψ. Since E = hν = 2π ν and λ = h 2 p p we have Ψ = Ae–(i/)(Et – px). ...(1.17) Equation (1.17) is a mathematical description of the wave equivalent of an unrestricted particle of total energy E and momentum p moving in the + x direction. The expression for the wave function Ψ given by eqn. (1.17) is correct only for freely moving particles. ELEMENTS OF QUANTUM MECHANICS 21 Schrödinger’s equation which is the fundamental equation of quantum mechanics is a wave equation in the variable Ψ. Differentiating eqn. (1.17) twice with respect to x, we get p2 2 = 2 x 2 and differentiating eqn. (1.17) once w.r.t. t, we get ...(1.18) iE . = ...(1.19) x At speeds small compared with that of light, the total energy E of a particle is the sum of its kinetic energy p2/2m and its potential energy V, where V is in general a function of position x and time t. E = p2 V. 2m ...(1.20) Multiplying both sides of eqn. (6) by the wave function Ψ gives p2 V . 2m From eqns. (1.18) and (1.19), we see that EΨ = EΨ = and ...(1.21) i t 2 ...(1.22) x 2 Substituting these expressions for E Ψ and p2 Ψ into eqn. (1.21), we get p2 Ψ = 2 2 2 = ...(1.23) V . t 2m x 2 Equation (1.23) is the time dependent form of Schrödinger’s equation. In three dimensions the time-dependent form of Schrödinger’s equation is i i 2 2 2 2 = V . t 2m x 2 y 2 z 2 ...(1.24) Where the particle’s potential energy V is some function of x, y, z and t. The restrictions on particle’s motion affect the potential energy funciton V. Once V is known Schrödinger’s equation may be solved for the wave function Ψ of particle from which its probability density | Ψ |2 may be determined for a specified x, y, z, t. The extension of Schrödinger’s equation from the special case of freely moving particle (V = constant) to the general case of a particle subject to arbitrary forces that vary in time and space [V = V(x, y, z, t)] is entirely plausible. 22 LASER SYSTEMS AND APPLICATIONS Thus, Schrödinger’s equation is postulated, solve it for a variety of physical situations and compare the results of the calculations with the results of experiments. If they agree, the postulate is valid and if disagree it may be discarded and some other approach would have to be explored. Equation (1.24) can be used for non-relativistic problems and in practice it has turned out to be accurate in predicting the results of experiments. 1.10 SC HR ÖDINGER’S EQU ATION: STEAD Y-ST ATE SCHR HRÖDINGER’S EQUA STEADY -STA FORM It has been observed that in many situations, the potential energy of a particle does not depend on time and vary with the position of the particle only. When this is true, Schrödinger’s equation may be simplified by removing time. The one-dimensional wave function Ψ of an unrestricted particles may be written as Ψ = Ae–(i/)(Et – px) = Ae–(iE/)t e+(ip/)x ...(1.25) –(iE/)t Thus, Ψ is the product of a time-dependent function e and a position dependent function Ψ. (Ψ = Ae(ip/)x) Substituting the Ψ of eqn. (1.25) into the time-dependent form of Schrödinger’s equation, we obtain 2 2 (iE/ )t e V e (iE/ )t 2 2m x Dividing throughout by the common exponential factor E ψ e–(iE/)t = 2 2m 2 (E V ) = 0. ...(1.26) x 2 Equation (1.26) is the steady-state form of Schrödinger’s equation. In three-dimensions it is 2 2 2 2m 2 2 2 (E V ) = 0. x 2 y z ...(1.27) Schrödinger’s steady-state can be solved only for certain values of the energy E. To solve Schrödinger’s equation we obtain a wave function ψ which fulfils the requirements for an acceptable wave function, i.e., its derivatives be continuous, finite and single valued. If there is no such wave function, the system cannot exist in a steady-state. The energy quantisation in wave mechanics is a natural element of the theory and is a characteristic of all stable systems. Particle in a Box The simplest quantum-mechanical problem is that of a particle trapped in a box with infinitely hard walls. We may specify the particle’s motion along the ELEMENTS OF QUANTUM MECHANICS 23 x-axis between x = 0 and x = L. A particle does not lose energy when collides with infinitely hard walls so that its total energy remains constant. From the formal point of view of quantum mechanics, the potential energy V of the particle is infinite on both sides of the box while V is constant say 0 for convenience on the inside. The particle cannot have infinite amount of energy. So, the wave function Ψ is zero for x ≤ 0 and x ≥ L. Thus, our aim is to find Ψ between x = 0 and x = L (Fig. 1.6). Schrödinger’s equation for the condition specified (x = 0 and x = L) becomes 2 2m 2 E = 0 [V = 0] ...(1.28) x 2 Equation (1.28) has the solution 2mE 2mE x B cos x where A and B are constants to be evaluated. Applying the boundary conditions that Ψ = 0 for x = 0 and for x = L. Since cos θ = Fig. 1.6. Potential well which 1, the second term cannot describe the parcorresponds to a box with ticle because it does not vanish at x = 0. infinitely hard walls Hence, B = 0. Since sin θ = 0, the sine term always yields ψ = 0 at x = 0 but ψ will be 0 at x = L only when ψ = A sin 2mE L = nπ, n = 1, 2, 3 ... ...(1.29) This is true because the sines of the angles π, 2π, 3π, ... are all 0. From eqn. (1.29) it is clear that the energy of the particle can have only certain values known as eigenvalues. These eigenvalues constituting the energy levels of the system are n2 2 2 where n = 1, 2, 3 ...(1.30) 2mL2 The wave function of a particle in a box whose energy is En is En = 2mEn x Substituting eqn. (1.30) for En, we get ψn = A sin ...(1.31) nx . ...(1.32) L where ψn is the eigenfunctions corresponding to the energy eigenvalues En. ψn = A sin It can be verified that these eigenfunctions meet all the requirements. For each quantum number n, ψn is a single valued function of x and ψn and ∂ψn/ ∂x are continuous. The integral of | ψn |2 over all space is finite. By integrating | ψn |2 dx from x = 0 and x = L. 24 LASER SYSTEMS AND APPLICATIONS n 2 dx = L 0 n 2 dx A2 L 0 sin 2 nx dx L L . ...(1.33) 2 2 To normalise ψ we assign a value of A such that | ψn | dx is equal to the probability P dx of finding the particle between x and x + dx, rather than merely proportional to P. If | ψn |2 dx is equal to P dx, then it must be true that = A2 n 2 dx = 1 since P dx = 1. ...(1.34) is the mathematical way of stating that the particle exists somewhere at all times. Comparing eqns. (1.33) and (1.34), we find that the wave function of a particle in a box are normalised if 2 L . = 1, A = A2 L 2 The normalised wave functions of the particle are therefore 2 nx sin ψn = n = 1, 2, 3. L L The normalised wave functions ψ1, ψ2 and ψ3 together with the probability densities | ψ1 |2, | ψ2 |2 and | ψ3 |2 are plotted in Fig. 1.6(a). While ψn may be negative as well as positive, | ψn |2 is always positive and, since ψn is normalised, its value at a given x is equal to the probability density P of finding the particle there. In every case | ψn |2 = 0 at x = 0 and x = L, the boundaries of the box. Fig. 1.6(a). Wave functions and probability densities of a particle confined to a box with rigid walls At a particular place in the box the probability of the particle being present may be very different for different quantum numbers. For instance | ψ1 |2 has its maximum value of 2/L in the middle of the box while | ψ2 |2 = 0. A particle in the lowest energy level of n = 1 is most likely to be in the middle of the box, while a particle in the next higher state of n = 2 is never there. Classical physics suggests the same probability for the particle being anywhere in the box. 25 ELEMENTS OF QUANTUM MECHANICS SOL VED EXAMPLES SOLVED EXAMPLE 21: Show that ψ(x) = eicx where c is some finite constant is acceptable eigenfunctions. Also normalise it over the region – a ≤ x ≤ a. SOLUTION: The wave function ψ(x) can be written as ψ(x) = eicx = cos cx + i sin cx. Its derivative is given by d ( x ) = ic eicx = ic (cos cx + i sin cx) dx = – c sin cx + ic cos cx. The following points may be observed: (i) sin cx and cos cx are periodic functions with maximum value 1 and d ( x) are finite for all values of c is finite constant. Thus ψ(x) and dx x. d ( x) are single-valued because cos cx and dx sin cx are also continuous for all values of x. Hence ψ(x) is an acceptable form of the eigenfunction. To normalise, the wave function we may write ψ(x) as ψ(x) = Aeicx. Now we have to determine the value of A and we may write (ii) The function ψ(x) and a a A2 * ( x )( x ) dx = 1 a a e icx e icx dx = 1 a A2 x a = 1 A2(2a) = 1 or A= . 2a Hence normalised wave function is ψ(x) = icx e . 2a EXAMPLE 22: A particle is moving in one-dimensional potential box (of infinite height) of width 25 Å. Calculate the probability of finding the particle within an interval of 5 Å at the centres of the box when it is in its state of least energy. SOLUTION: The wave function of the particle can be written as ψ(x) = 2 nx sin a a 26 LASER SYSTEMS AND APPLICATIONS For the particle in the least energy state n = 1 and hence 2 sin x a a ψ(x) = At the centre of the box x = a , the probability of finding the particle in 2 the interval Δx is given as P = | ψ(x) |2 Δx 2 2 ( a 2) 2 2 2 | ψ(x) |2 = sin a sin 2 a a a P = 2 5 10 10 2 x a 25 10 10 a 25 Å = 0.4 x 5 Å EXAMPLE 23: A particle is in motion along a line between x = 0 and x = a with zero potential energy. At points for which x < 0 and x > a, the potential energy is infinite. The wave function for the particle in the nth state is given by nx . a Find the expression for the normalised wave funciton. ψn = A sin SOLUTION: The probability of the particle between x and x + dx for the nth state is given as nx dx. a Since the particle is found in the region x = 0 and x = a for all times, we have | ψn(x) |2 dx = A2 sin 2 A2 a 0 a 0 a 0 n A2 sin 2 2 dx = 1 nx dx = 1 a 1 2nx 1 cos dx = 1 2 a a A2 a 2nx sin x = 1 2 2n a 0 A2 a = 1 or 2 Now the normalised wave funciton is ψn(x) = 2 nx sin . a a A= 2 . a 27 ELEMENTS OF QUANTUM MECHANICS EXAMPLE 24: Find the energy of an electron moving in one dimension in an infinitely high potential box of width 1 Å, given mass of the electron 9.11 × 10–31 kg and h = 6.63 × 10–34 Js. n2 h 2 (n = 1, 2, 3, ...) 8ma2 For least energy of the particle n = 1. SOLUTION: Since we know that E = Now (6.63 10 34 )2 h2 = joules 2 8ma 8 9.11 10 31 (10 10 )2 = 9.1 × 10–19 joules = 9.1 10 19 eV = 5.68 eV. 1.602 10 19 EXAMPLE 25: Show that Ψ = ψe–iωt is a wave function of a stationary state. SOLUTION: If ψ is the wave functions of a stationary state, then the value 2 2 of at each point must be constant, independent of time. To find , we first take the complex conjugate of Ψ = ψe–iωt which is Ψ* = ψ* e+iωt. Then 2 = Ψ*Ψ = (ψ* e+iωt) (ψ e–iωt) = ψ*ψ e0 = 2 where ψ is not a function of time, so 2 it has been shown that Ψ = ψe –iωt is also independent of time. Now, 2 2 2 = , so is independent of time and is a wave function of a stationary state. EXAMPLE 26: Find the probability that a particle trapped in a box L wide can be found between 0.45 L and 0.55 L for the ground and first excited states. SOLUTION: This part of the box is one tenth of the box’s width and is centered on the middle of the box. Classically, we could expect the particle to be in this region 10 per cent of the time. The quantum mechanics gives quite different predictions that depend on the quantum number of the particle’s state. The probability of finding the particle between x1 and x2 when it is in the nth state is P = x2 x1 2 n dx 2 L x2 x1 sin 2 x 1 2nx 2 x sin = L x1 L 2n x1 = 0.45 L and x2 = 0.55 L For ground state which corresponds to n = 1, we have P = 0.198 = 19.8% nx dx L 28 LASER SYSTEMS AND APPLICATIONS For the first excited state n = 2 P = 0.0065 = 065% 2 This low figure is consistent with the probability density of n = 0 at x = 0.50 L. EXAMPLE 27: (a) Show that y = Ax + B where A and B are constants, is a solution to the Schrödinger equation for an E = 0 energy level of a particle in a box. (b) Show, however, that the probability of finding a particle with this wave function is zero. SOLUTION: (a) The Schrödinger equation for a particle in a box is d2 dx 2 2m 2 E = 0 Differentiating ψ = Ax + B twice with respect to x gives d2 = 0 for dx 2 the left side of the Schrödinger equation. Also E = 0 gives zero of the right side. Since 0 = 0, ψ = Ax + B is a solution to this Schrödinger equation for E = 0. (b) The wave function ψ equals zero outside the box (x < 0 and x > L). In order that ψ = Ax + B may be continuous at x = 0, it must be true that y = 0 at x = 0. So A(0) + B = 0 or B = 0. Similarly, in order that ψ may be continuous at x = L, it must be true that ψ = 0 at x = L, so A (L) + 0 = 0 or A = 0. With both A and B equal to zero, ψ = Ax + B = 0. Thus the wave function equals zero inside the box as well as outside the box and the probability of finding the particle anywhere with this wave function is zero. EXER CISES EXERCISES 1. 2. 3. 4. 5. 6. 7. 8. Discuss the dual nature of matter and waves. Find de Broglie wavelength of an electron of energy MeV. What are de Broglie matter waves? Calculate the de Broglie wavelength associated with a proton moving with a velocity equal to 1/20 velocity of light. Derive time dependent and time-independent Schrödinger wave equation. What is uncertainty principle? Apply this to prove the non-existence of the electron in the nucleus. Calculate the smallest possible uncertainty in the position of electron moving with velocity 3 × 107 m/sec. [Ans. 0.038 Å] An electron has a speed of 1.05 × 104 m/sec within the accuracy of 0.01 per cent. Calculate the uncertainty in the position of the electron. [Ans. 1.10–4 m] ELEMENTS OF QUANTUM MECHANICS 29 9. Show that the de Broglie wavelength for a material particle of rest mass m0 and charge q, accelerated from rest through a potential difference of V volts relativistically is given by λ = h/{2m0qV [1 + qV/ 2m0c2]}1/2. 10. Find the uncertainty in the momentum of a particle when its position is determined within 0.01 cm. Find also the uncertainty in the velocity of an electron and α-particle respectively when they are located within 5 × 10–8 cm. [Ans. 1.05 × 10–30 kg m/s, 2.33 × 105 m/s, 31.4 m/s] 11. An electron is confined to a box of length 1.1 × 10–8 meter. Calculate the minimum uncertainty in its velocity. Given m = 9.1 × 10–31 kg, [Ans. 1.06 × 104 m/s] = 1.05 × 10–34 J. sec. 12. Calculate the kinetic energy of an electron if the wavelength of electron equals the yellow line of Sodium. [Ans. 4.34 × 10–6 eV] 13. Calculate the de Broglie wavelength of neutron of energy 28.8 eV. Given h = 6.62 × 10–34 J.sec, m = 1.67 × 10–27 kg. [Ans. 5.3 × 10–2 Å] 14. Derive the formula for de Broglie wavelength of particle in terms of kinetic energy and its rest energy m0c2. 15. Describe Davisson-Germer experiment to demonstrate the wave nature of particle. 16. Deduce an expression for de Broglie wavelength of helium atom having energy at temperature T°K. 17. What are de Broglie waves and how do they help in the interpretation of the Bohr’s quantisation rule? 18. Show that the phase velocity of de Broglie wave is greater than the velocity of light; but the group velocity is equal to the velocity of the particle with which the waves are associated. 19. Calculate de Broglie wavelength associated with nitrogen at 3.0 atmospheric pressure and 27°C temperature. 20. Can a photon and an electron of the same momentum have the same wavelength? Compare their wavelengths if the two have the same energy. 21. X-rays of wavelength 0.91 Å fall on a metal plate having work function 2.0 eV. Find the wavelength associated with the emitted photoelectrons. [Ans. 8.7 Å] 22. Calculate kinetic energy of a neutron having de Broglie wavelength 1 Å. [Ans. 8.13 × 10–2 eV] 23. What is the minimum uncertainty in the frequency of a photon whose lifetime is about 10–8 sec? [Ans. 15.92 × 106 per sec.] 24. A certain excited state of hydrogen atom is known to have a lifetime of 2.5 × 10–4 sec. What is the minimum error, with which energy of the excited state can be measured. [Ans. 0.026 eV] 30 LASER SYSTEMS AND APPLICATIONS 25. What is de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 volts. [Ans. 1.23 Å] 26. Find the expression for the energy state of a particle in one-dimensional box. OBJECTIVE QUES T IONS QUEST 1. A particle possesses a kinetic energy E and mass m, then its de Broglie wavelength is h (a) (b) h 2mE 2mE 1 2mE (d) h 2mE h 2. An electron is accelerated through a potential difference of V volts. The de Broglie wavelength of the electron is 12.27 12.27 (a) (b) V V (c) (c) 12.27 V (d) 12.27 V 3. Read the statements (A and B) and choose the answer: (A) The de Broglie wavelength of a moving particle is inversely proportional to its momentum (B) Only a charged particle in motion is associated with matter waves (a) A and B are correct (b) A and B are wrong (c) A is wrong and B is correct (d) A is correct and B is wrong 4. If the momentum of a particle is doubled then its de Broglie wavelength (a) halves (b) doubles (c) quadruples (d) remains the same 5. The masses of neutron and electron are mn and me respectively. If they have the same de Broglie wavelength, then their velocities should be in the ratio (a) 1 : 1 (c) mn me (b) (d) me mn me2 mn2 6. An α-particle has a mass 4 mp and a proton mp. If they possess the same kinetic energy, then the ratio of the de Broglie wavelengths is 31 ELEMENTS OF QUANTUM MECHANICS (a) (c) 7. The (a) (b) (c) (d) 8. The 1:4 (b) 4 : 1 2:1 (d) 1 : 2 characteristic of wave function ψ are real function, finite and discontinuous complex, single valued, finite and continuous function complex, infinite and discontinuous function complex single valued and infinite wave function for the motion of a particle in a potential well of nx width a is given as ψn = B sin , then B is a (a) 1 a (b) 1 a (c) 2 a (d) a 2 9. When the wave function is normalized then (a) ψ ψ* = 1 (b) ψ ψ* dx = a 1 (c) 1 * dx 1 (d) * d x 1 a 1 10. Increasing the potential difference between anode and filament in Davisson-Germer experiment (a) causes an increase in the wavelength of the electron-waves (b) causes a decrease in the wave velocity (c) causes a decrease in the wavelength of the electron-waves (d) causes a decrease in the momentum of the electron 11. For a particle of mass m confined to a cubical box of side L, the allowed values of energy E are given as (a) (c) n2 h 2 2 mL2 h 2 L2 , n = 1, 2, 3 (b) , n = 1, 2, 3 (d) n2 h2 8mL2 , n = 1, 2, 3 n2 h2 , n = 1, 2, 3 mL 2mn 2 12. An electron is confined to a potential box of infinite height and width 10 Å. The probability of finding the particle in a small interval Δx at the centre of the box for the energy state immediately above the ground state is (a) zero (b) 0.5 (c) 0.9 (d) 1 32 LASER SYSTEMS AND APPLICATIONS 13. If a charged particle of mass m is accelerated through a potential difference of V volts, the de Broglie wavelength is proportional to (a) 1 14. 15. 16. 17. 18. 19. 20. 21. 22. (b) V V2 1 (d) V2 (c) V 2 The de Broglie hypothesis is associated with (a) wave nature of α-particles only (b) wave nature of radiations (c) wave nature of electrons only (d) wave nature of all material particles Which of the following phenomena cannot be expressed by wave nature of light? (a) Interference (b) Diffraction (c) Polarization (d) Photoelectric effect Matter waves (a) are latitudinal (b) show diffraction (c) are electromagnetic (d) always travel with speed of light Of the following moving with the same velocity, the one which has largest wavelength is (a) a photon (b) an electron (c) an α-particle (d) a neutron The uncertainty principle holds for (a) macroscopic particles only (b) microscopic particles only (c) macroscopic and microscopic particles both (d) neither macroscopic nor microscopic particles The uncertainty principle cannot hold for the following pairs: (a) Energy and time (b) Position and momentum (c) Angular momentum and angle (d) Linear momentum and angle Matter waves were first experimentally observed by (a) Frank-Hertz (b) Davisson and Germer (c) Ster-Gerlach (d) de Broglie If the momentum of a particle is increased to four times, then the de Broglie wavelength will become (b) four times (a) half (c) twice (d) one-fourth The Schrödinger time independent wave equation for free particle is (a) 2 2 (E v ) 0 2m (b) 2 2 ( v E) 0 2m 33 ELEMENTS OF QUANTUM MECHANICS 2 2 2 2 E 0 E 0 (d) 2m 2m 23. Schrödinger time dependent wave equation for free particle is (c) (a) 2 2 i t 2m x 2 2 2 i t 2m x 2 i 2m 2 i (d) 2 2 t t x x 24. An α-particle and a proton have the same kinetic energy. The ratio of their wavelength is (m∝ = 4mp) (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 25. In Davisson-Germer experiment, nickel crystal acts as (a) perfect absorber (b) perfect reflector (c) two dimensional grating (d) three dimensional diffraction grating 26. Momentum of a photon of energy hν is (a) hν (b) hνc (c) hν/c (d) has no momentum (c) 2m 2 (b) 2 2 ANSWERS 1. 6. 11. 16. 21. 26. (a) (d) (b) (b) (d) (c) 2. 7. 12. 17. 22. (b) (b) (a) (b) (a) 3. 8. 13. 18. 23. (d) (c) (c) (b) (b) 4. 9. 14. 19. 24. (a) (d) (d) (d) (c) 5. 10. 15. 20. 25. (b) (c) (d) (b) (d)