Physics (RGPV) - Quantum Mechanics PDF

# Physics (RGPV) - Quantum Mechanics ## 4-19 **Questions:** - What is De-Broglie Hypothesis? (2 Marks) - What do you understand by matter waves? (2 Marks) - Derive De-Broglie wavelength of matter waves. (6 Marks) **De-Broglie Hypothesis:** According to De-Broglie's concept, a moving particle al...

# Physics (RGPV) - Quantum Mechanics ## 4-19 **Questions:** - What is De-Broglie Hypothesis? (2 Marks) - What do you understand by matter waves? (2 Marks) - Derive De-Broglie wavelength of matter waves. (6 Marks) **De-Broglie Hypothesis:** According to De-Broglie's concept, a moving particle always has a wave associated with it and the motion of the particle is guided by that wave in a similar manner as a photon is controlled by a wave. **Matter Waves:** - Like electromagnetic waves, matter waves can also be propagated in a vacuum. - They are not mechanical waves like sound waves and unlike electromagnetic waves, they are also associated with the motion of electrically neutral bodies. - They are not electromagnetic waves but they are a new kind of wave. **Derivation of De-Broglie Wavelength:** - The expression for the wavelength of a matter wave or de-Broglie wave can be obtained by the analogy of the behavior of radiation with matter. - The wavelength of a matter wave or de-Broglie’s wave should be the same as that for electromagnetic radiation. - A light wave of frequency *v* is associated with a photon of energy *E*, given by Planck's relation as, E = hv ---- (4.5.1) - According to Einstein’s theory of relativity, a particle of mass *m* is equivalent to an amount of energy *E* as: E = mc^2 ---- (4.5.2) - de-Broglie assumed *m* to be the mass of the photon of energy *hv*, which moves with the velocity of light *c*. - Comparing the above two relations of energy, we get for photon: hv = mc^2 ---- (4.5.3) - If *p* is the momentum of a photon, then: p = mc - So, from Equation (4.5.3), we can write: hv = pc hv = pνλ ---- (c = νλ) - or, λ = h/p ---- (4.5.4) - This is the de-Broglie’s wavelength for photons. - de-Broglie further suggested that the same expression may also be used for the wavelength of material particles or matter waves. ## 4-20 - In the case of a particle of mass *m* moving with velocity *v* (which is comparable with the velocity of light, but always less than *c*) and has a momentum *p = mv*, the expression for the wavelength of a wave that guided the motion of the material particles or matter waves, is obtained by substituting: *p = mv* in Equation (4.5.4). - Hence, we get the wavelength of matter waves is: λ = h/mv = h/p ---- (4.5.5) - Therefore, we conclude that any particle in motion is associated with matter waves whose wavelength is given by the Equation (4.5.5). **Question:** Derive De-Broglie wavelength of an electron. (4 Marks) - Consider an electron subjected to a potential of *V* volts. Let *me* be the rest mass of electron and *v* be the velocity attained by the electron. - The kinetic energy to the electron is provided by electrical energy. - **Step I:** Kinetic energy = 1/2 * mv² = eV or 1/2 * 2me * v^2 = eV ---- (4.5.6) - **Step II:** The de-Broglie wavelength λ of matter waves associated with the electron is: λ = h/p = h/me * v ---- (4.5.7) - **Step III:** Putting for *p = me * v* from Equation (4.5.6) into Equation (4.5.7), we have de-Broglie wavelength: λ = h/√2 * me * eV ---- (4.5.8) - Now h = 6.625 × 10^-34 Js me = 9.1 × 10^-31 kg e = 1.6 × 10^-19 coulomb - Hence putting these values in Equation (4.5.8), we have: λ = 12.26/√V A° ---- (4.5.9) **Significance:** **Question:** What is the significance of De-Broglie wavelength of electron? (2 Marks) - Higher the accelerating potential corresponds to lower the wavelength associated with the electron. So, the frequency of the electron is more, i.e., energy of electron increases. **Question:** Write down the properties of matter waves. (4 Marks) - The wavelength of matter waves is given by λ = h/mv. - The properties of matter waves are: - λ ∝ 1/m, hence lighter the particle greater is the wavelength associated with it. - λ ∝ 1/v, hance greater the velocity of the particle, smaller is the associated wavelength. - As *v* → ∞, λ → 0. But the wave becomes indeterminate when *v* → 0 and λ → ∞. This shows that matter waves are associated with the particles in motion. These waves do not depend on the charge of the particle and hence matter waves are not electromagnetic in nature. The waves are not mechanical in nature. - The velocity of matter waves is not a constant like electromagnetic waves but depends on the velocity of the particle generating them: υp = c^2/v, hence matter waves travel faster than light. (v < c) - The wave velocity of matter waves depends inversely on the wavelength λ. This is the basic difference between matter waves and light waves (light waves have the same velocity for all wavelengths). - Matter wave representation is a symbolic representation. It is a wave of probability indicating the likelihood of locating the particle. - The wave and particle duality of matter is not exhibited simultaneously. ## 4-21 **Question:** Prove that velocity of matter waves greater than the speed of light. (4 Marks) - According to de-Broglie hypotheses, a moving particle is associated with matter waves. Assume velocity of matter waves: υp = νλ ---- (4.5.10) - We know that: E = hv E = mc^2 ---- (4.5.11) - So, we get: ν = mc^2/h ---- (4.5.11) - According to de-Broglie Hypothesis, the wavelength associated with a matter wave is given by: λ = h/mv ---- (4.5.12) - Put Equations (4.5.11) and (4.5.12) in (4.5.10), we get: υp = mc^2/h x h/mv = c^2/v or υp * v = c^2 - If the velocity of the particle is always less than *c* (i.e. *v* < *c*), then the velocity of matter waves is greater than *c* (i.e. υp > *c*). - So, the velocity of matter waves is always greater than light. ## 4-22 **Ex. 4.5.1:** The energy of photon is 5.28 × 10^-19 J. Calculate frequency and wavelength. (2 Marks) **Soln:** - E = hv - v = E/h = 5.28 x 10^-19 J / 6.625 x 10^-34 J.s = 7.96 × 10^14 Hz - Again ν = c/λ - λ = c/ν = 3 × 10^8 m/s / 7.96 × 10^14 = 0.3768 × 10^-6 m = 3768 × 10^-10 m = 3768 A° **Ex. 4.5.2:** Electrons moving with a speed of 7.3 × 10^7 m/s have a wavelength of 0.1A°. Calculate Planck's constant. (2 Marks) **Soln:** - Given: λ = 0.1 A° = 0.1 × 10^-10 m, v = 7.3 × 10^7 m/s - Formula required: λ = h/mv - h = λ. mv = 0.1 × 10^-10 × 9.1 × 10^-31 × 7.3 × 10^7 = 6.643 × 10^-34 J.s **Ex. 4.5.3:** At what velocity the de-Broglie wavelength of an alpha particle is equal to the wavelength of 1 KeV X-ray photons? Given mass of alpha particle is four times the mass of a proton. Mass of proton = 1.67 x 10^-27 kg. (4 Marks) **Soln:** - Given: E = 1 KeV, m = 1.67 x 10^-27 kg - Formula required: λ = h/mv = hc/E; λ = h/mv - For a photon, the energy E = hv = hc/λ ---- (1) - The de-Broglie wavelength of the alpha particle of mass m and velocity v is: λ = h/mv ---- (2) - As per the data, Equation (1) = Equation (2) - h/E = h/mv - velocity of alpha particle having same energy as 1 KeV X-ray photon is: v = E/m **Ex. 4.5.4:** Find the de-Broglie wavelength of 10 KeV electrons. (2 Marks) **Soln:** - Given: E = 10 KeV = 10 × 10^3 eV = 10 × 10^3 × 1.6 × 10^-19 J - Formula required: λ = h/√2mE = 6.625 × 10^-34 / √2 × 9.1 × 10^-31 × 10^4 × 1.6 × 10^-19 J = 1.227 × 10^-11 m = 0.1227 A° **Ex. 4.5.5:** Find the K.E. of a neutron which has a wavelength of 3A°. At what angle will such a neutron undergo first order Bragg reflection from a calcite crystal for which the grating space is 3.036 A°? Mass of neutron = 1. 67 × 10^-27 kg. (4 Marks) **Soln:** - Given: λ = 3 A° = 3 × 10^-10 m; d = 3.036 A = 3.036 × 10^-10 m - m = 1.67 × 10^-27 kg. - Formulae required: Wavelength of neutron is λ = h/√2mE - 2d sin θ = ηλ - Neutron energy E = h^2/2mλ^2 = (6.63 x 10^-34)^2 / 2 x 1.67 x 10^-27 x (3 x 10^-10)^2 = 1.449 x 10^-21 J - For Bragg reflection, 2 d sin θ = nλ - n = 1 - sin θ = λ/2d = 3 × 10^-10 / 2 x 3.036 × 10^-10 = 0.4940 - θ = sin-1 (0.4940) = 29° 36' **Ex. 4.5.6:** Determine the velocity and kinetic energy of a neutron having de-Broglie wavelength 1.0A° (mass of neutron is 1.67 x 10^-27 kg). (2 Marks) **Soln:** - We have, λ = h/mv - v = h/mλ = 6.63 × 10^-34 / 1.67 × 10^-27 × 10^-10 = 3.97 × 10^3 m/s - Again we can write, λ = h/√2mE - E = h^2/2mλ^2 ## 4-26 - V = 4.356 x 10^-67 /4.66 x 10^-70 = 934.76 Volts. - V = 12.27/0.3^2 = 1672.81 volts **Ex. 4.5.13:** What potential difference must be applied to an electron microscope to obtain electrons of wavelength 0.3 A°? (2 Marks) **Soln:** - λ = 12.27/√V A° but λ = 0.3 A° - √V = 12.27/0.3 - V = (12.27/0.3)^2 = 1672.81 volts **Ex. 4.5.14:** Calculate the velocity and de-Broglie wavelength of an α-particle of energy 1 Kev. (2 Marks) **Soln:** - λα = h/√2mE = 6.63 x 10^-34 / √2 x 4 x 1.67 x 10^-27 x 1 x 10^3 x 1.6 x 10^-19 = 45.34 x 10^-14 m = 0.0045 A° - λα = h/ma * Va - Va = h/ma * λα = 6.63 x 10^-34 / 4 x 1.67 x 10^-27 x 4.5 x 10^-13 = 0.22 x 10^6 m/s = 2.2 x 10^5 m/s **Ex. 4.5.15:** An electron has kinetic energy equal to its rest mass energy. Calculate de-Broglie’s wavelength associated with it. (2 Marks) **Soln:** - Data: h = 6.63 x 10^-34 Js, m = 9.1 x 10^-31 kg, c = 3 x 10^8 m/sec, λ = ?, E = mc^2 - Formula: λ = h/√2mE = h/√2mc² = (6.63 x 10^-34)/√2 × 9.1 × 10^-31 × 3 × 10^8 = 0.1715 × 10^-11 m **Ex. 4.5.16:** In a T.V. Set, electrons are accelerated by a p.d. of 10 KV. What is the wavelength associated with these electrons? (2 Marks) **Soln:** - Given formula: λ = h/√2meV = 12.3/√V A° = 12.3/√10 A° = 0.12 A° ## 4-29 **Question:** Define phase velocity. (2 Marks) - Phase velocity, also known as the wave velocity, υp, of a monochromatic wave is the velocity with which a definite phase (e.g. crest, trough etc.) of the wave propagates in the medium. **Question:** Define group Velocity. (2 Marks) - A wave group can be described in terms of a series of individual waves of slightly different wavelength, whose interference with one another results in variation in amplitude that defines the group shape. - The velocity with which the hump i.e. wave group travels is called the group velocity υg. If the speeds of the waves are the same, the speed of the wavegroup is the same as the speed of wave. But if the wavespeed varies with wavelength, the wave group has a speed different from that of the individual waves. **Question:** Derive the mathematical expression for phase velocity and group velocity. (4 Marks) - Consider the formation of a wave group by the superposition of two waves of the same amplitude A, but slightly different angular frequencies and propagation constants. - Let the two waves be expressed by, y1 = A sin (ωt - kx)---- (4.5.14) and y2 = A sin [ (ω + dω) t – (k + dk) • x] ---- (4.5.15) - The resultant displacement y at any time t and position x is given by, y = y1 + y2 - C + D sin C + sin D = 2 sin (C - D)/2 * cos (C + D)/2 - y = 2A cos (dωt/2 - dk/2 * x) * sin (ωt – kx) ---- (4.5.16) - dω and dk are very small as compared to ω and k respectively. - 2ω + dω ≈ 2ω - 2k + dk ≈ 2k - dk/2 ≈ dω/2 - y = 2A cos (dωt/2 - dk/2 * x) * sin (ωt – kx) represents a wave of angular frequency ω and propagation constant k. - The amplitude varies as per the cosine term with angular frequency dω/2 and propagation constant dk/2. - The effect of this modulation (amplitude variation) is to produce the successive wave groups. - The wave velocity or the phase velocity is υp = ω/k ---- (4.5.17) ## 4-34 - From equations (4.5.24) and (4.5.25), we observe that the theoretical and experimental values of λ are in good agreement with each other. This confirms the de-Broglie hypothesis of matter waves. ## 4.6 Uncertainty Principle - In classical physics, the dynamical variables like momentum and co- ordinates of a particle (i.e. its position) can be measured accurately at any instant. But such measurements cannot be performed beyond a certain limit in quantum mechanics. - According to the principles of wave mechanics, a moving material particle is associated with a wave which describes all about the particle. - A particle is always localized in space and hence a wave packet or wave group represents the moving particle. The particle lies somewhere in the wave packet and the probability of finding the particle at a given point is proportional to the wave amplitude at that point. - Thus although the particle is somewhere within the wave packet moving with the group velocity, it is impossible to determine the exact position and exact velocity at any given moment. There is always a certain uncertainty in the experimental measurement of position and momentum of a particle of very small size, like an electron, in motion. - This uncertainty is not due to the imperfection of the measuring instrument but is something inherent in the wave nature of the moving particle. The Newtonian mechanics is applicable to the motion of macroscopic bodies but fails in case of microscopic particles. **Question:** State the situation where uncertainty found in experimental measurement. (2 Marks) - Infinitely small wave packet. - Wide wave packet. **Question:** Discuss the situation where uncertainty found in experimental measurement. (4 Marks) - **(A) Infinitely Small Wave Packet:** - In the case of infinitely small wave packet, the amplitude is large over a very small region of the space and is negligible elsewhere, as shown in Fig. 4.6.1. - This small region of space can be associated with the position of the particle. Hence the position of the particle can be fixed with minimum error or uncertainty. - But at the same time the wavelength λ and hence the momentum (p = h/λ) cannot be measured accurately and hence is uncertain. - **(B) Wide Wave Packet:** - If the wave packet is sufficiently wide as shown in Fig. 4.6.2, then the wavelength λ and hence the momentum p can be determined with more accuracy. But at the same time the position of the particle becomes very uncertain. - Thus consideration of infinitely small and infinitely large wave packets shows that certainty about momentum involves complete uncertainty about the position and vice versa. Hence it is impossible to determine simultaneously both the momentum and position of a moving particle. - Experiments never tell us that the particle occupies a particular position at a particular instant and has a particular velocity at that instant. They only give us the limits within which the position and momentum can be determined i.e. they only give the probability. - Heisenberg was the first to realize such consequences and expressed them mathematically by means of a relation known as the uncertainty principle. **Question:** State and explain Heisenberg’s uncertainty principle. (2 Marks) - As per this principle, 'it is impossible to determine accurately and simultaneously the values of both the members of particular pairs of physical variables that describe the behaviour of atomic system’. ## 4-36 **Question:** Derive mathematical expression for uncertainty principle. (4 Marks) - The uncertainties in the knowledge of the coordinate x and the momentum px of a particle is expressed by writing x + Δx and px + Δpx, respectively. Thus, a particle is located somewhere between x + Δx and x; the momentum of a particle is located between px - Δpx and px + Δpx. As per the uncertainty relation, the product Δx. Δpx is always greater than Planck’s constant *h*. - Therefore, Δx * Δpx ≥ *h* - Thus the uncertainty principle states that in any simultaneous determination of position and momentum of a particle of atomic size, the product of uncertainties is equal to or greater than Planck’s constant *h*. - The pair of variables related by the uncertainty principle are called canonically conjugate variables. Hence, we also have: Δy * Δpy ≥ *h* Δz * Δpz ≥ *h* - The smallness of *h* makes the uncertainty of interest primarily in connection with systems of atomic size. - It should also be noted that the uncertainty principle does not apply to a single measurement on a single particle. It is a statement about a statistical average over a large number of measurements of position and momenta. - More precisely, the uncertainty principle is expressed by the relation Δx * Δpx ≥ ħ - where ħ = h/2π **Heisenberg’s Principle Applied to the Pair of Variables:** - The kinetic energy and time form a pair of canonically conjugate variables. The K.E. of a particle of mass *m* moving with velocity *v* is: E = 1/2 * mv^2 ---- (4.6.1) - To find the uncertainty ΔE in the measurement of energy, we differentiate Equation (4.6.1) assuming the mass as constant and get: ΔE = 1/2 * m * 2 * v * Δv = v * (m x Δv) = v * Δp ---- (v = Δx/Δt = Δx/Δt) - Therefore, ΔE * Δt = Δx * Δp - But Δx * Δp ≥ *h* by Heisenberg's relation - Therefore, ΔE * Δt ≥ *h* ---- (4.6.2)