Quantum Mechanics PDF

Dr M CHANDRA SHEKHAR REDDY Dr M Chandra Shekhar Reddy Black body radiation A black body is a perfect absorber and emitter. It absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence and emits electromagnetic radiation at thermal equilibrium....

Dr M CHANDRA SHEKHAR REDDY Dr M Chandra Shekhar Reddy Black body radiation A black body is a perfect absorber and emitter. It absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence and emits electromagnetic radiation at thermal equilibrium. This radiation is called as Black body radiation. Dr M CHANDRA SHEKHAR REDDY A black body in thermal equilibrium Emits as much as more energy at every frequency than any other body at the same temperature. The energy is radiated isotropically, independent of direction. Dr M CHANDRA SHEKHAR REDDY Stefan-Boltzmann law The total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature. The radiant heat energy emitted from a unit area in one second (that is, the power from a unit area) is given by 𝜎 = 5.670374419 × 10−8 𝑊𝑚−2 𝐾 −4 , Dr M CHANDRA SHEKHAR REDDY 4 𝑃/𝐴 = 𝜎𝑇 𝑐𝑎𝑙𝑙𝑒𝑑 𝑎𝑠𝑆𝑡𝑒𝑓𝑎𝑛 − 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Rayleigh–Jeans law Electromagnetic radiation emitted by a black body at a given temperature for wavelength λ is 8KT E (T ) = 4 The Rayleigh–Jeans law agrees with experimental results at large wavelengths (or, equivalently, low frequencies) but strongly disagrees at short wavelengths (or high frequencies). Wien’s displacement law The maximum wavelength emitted by a black body is b  max = m. inversely proportional to Temperature T b is a constant equal to 3×10-3 mK. The hotter an object is, the shorter the wavelength at which it will emit most of its radiation. Dr M CHANDRA SHEKHAR REDDY The Wien’s displacement law agrees with experimental results at shorter wavelengths (or, equivalently, high frequencies) but strongly disagrees at longer wavelengths (or low frequencies). Planck's law of blackbody radiation (1901): Plank derived a theoretical expression for energy distribution on the basis of quantum theory. He made the following assumptions in deriving the equation. An oscillator can absorb or radiate energy only in discrete amounts proportional to its frequency i.e. E=hν Dr M CHANDRA SHEKHAR REDDY The energy of oscillator increases by absorption and decreases by emission of radiation by a minimum amount of hν. In other words the energy of oscillator is quantized. It can have only energies like hν, 2hν, 3hν….n hν. Therefore the energy of oscillator E=n hν where n=1,2,3… The energy per unit volume per unit 8πh𝜐 3 1 frequency can be written as Eυ = c 3 𝑒 ℎ𝜐Τ𝐾𝑇 − 1 Or the energy per unit volume per 8πhc 1 unit wavelength can be written as Eλ = 𝜆5 𝑒 ℎ𝐶 Τ𝜆𝐾𝑇 − 1 Photo electric effect (Heinrich Hertz in 1887) It is the emission of electrons from the metal plate when light of suitable wavelength impinges The emitted electrons are called as photoelectrons Dr M CHANDRA SHEKHAR REDDY Properties of Photo electrons The kinetic energy of photoelectrons increases with increase in the frequency of light but independent of intensity of light. The Photo current increases with the increase in intensity of light but independent of the frequency of light. The kinetic energy of photoelectrons remains constant as light amplitude increases. 1 𝑇ℎ𝑒 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑝ℎ𝑜𝑡𝑜𝑒𝑙𝑒𝑐𝑡𝑜𝑛 𝑖𝑠, 𝐸 = 𝜙 + 𝑚𝑣 2 2 𝜙 = ℎ𝜐0 is called as the work function of the metal and 𝜐0 is known as the threshold frequency Predictions of Classical theory about photo electric effect The classical theory predicts that the oscillating electric field of light wave heats the electrons, causing them to vibrate, freeing them from the metal surface. According to classical theory, the intensity of wave is equal to the square of amplitude of wave. Hence, the kinetic energy of emitted photoelectrons should increase with the light amplitude. The rate of electron emission, which is proportional to the measured electric current, should increase as the light frequency is increased Failure of Classical theory in explaining the photo electric effect Experiments observations of photo electric effect are The kinetic energy of photoelectrons increases with light frequency Electric current remains constant as light frequency increases. Electric current increases with light amplitude. The kinetic energy of photoelectrons remains constant as light amplitude increases. All the above observations are in contradicts with the classical theory predictions Work functions of few metals Classical mechanics Developed basing on the Newton’s laws of motion Applicable for macroscopic particles Dr M CHANDRA SHEKHAR REDDY Failed to explain properties of microscopic particles Quantum mechanics Developed by Max Plank It is the study of matter and its interactions with energy on the scale of atomic and subatomic particles. Particles will have wavelike properties, and governed by the Schrodinger equation. Applicable for microscopic as well as macroscopic particles. Successfully explained the phenomenon like black body radiation, photo electric effect by proposing that the energy on an oscillator is quantized.  NDRA SHEKHAR REDDY Waves and particle nature of light Light is a wave interference, diffraction and polarization A wave is specified by frequency, wavelength, phase velocity, amplitude and intensity. A wave is a spread-out disturbance and occupies a large space. Light is a Particle photo electric effect, Compton effect and black body radiation The particle characterized by the mass, velocity acceleration etc A particle occupies a definite position in space. Hence the light has dual nature Matter waves de Broglie extended the wave particle dualism of light for the matter Einstein’s mass energy relation, we have 𝐸 = 𝑚𝑐 2 ℎ𝑐 From Planks radiation law we have 𝐸 = ℎ𝜐 = 𝜆 ℎ𝑐 ℎ From the above equations we can write 𝐸 = 𝑚𝑐 2 = (or) 𝜆 = 𝜆 𝑚𝑐 ℎ A material particle travels with a velocity v, hence we can write 𝜆 = 𝑚𝑣 Hence the de Broglie principle can be stated as, Every moving particle is Here m is 𝑚0 ℎ 𝑚= associated with a wave whose 𝜆= relativistic mass 𝑣2 1− 2 wavelength is, 𝑚𝑣 and given by 𝑐 The wave nature of the particle is noticeable only when the associated wavelength is comparable with the dimensions of the particle. Dr M CHANDRA SHEKHAR REDDY 1 1 𝑚2 𝑣 2 𝑃2 If E is the energy of particle, 𝐸 = = 𝑚𝑣 2 = 𝑜𝑟 𝑝 = 2𝑚𝐸 2 2 𝑚 2𝑚 ℎ 𝑜𝑟 𝜆 = 2𝑚𝐸 If the considered particle is electron accelerated with a potential V, then E=eV ℎ Hence wavelength associated with the electron, 𝜆= 2𝑚𝑒𝑉 Ignoring the relativistic condition, if we substitute all the constants, 6.63  10 −34 12.27 × 10−10 = 𝑜𝑟 𝜆 = − 31 −19 𝑉 2  9.1  10  1.6  10 V Characteristics of Matter waves 1. Lighter the particle, greater is the wavelength associated with it. 2. Lesser the velocity of the particle, longer the wavelength associated with it. 3. For v = 0, λ = ∞. Hence the matter waves are associated with only moving particles. 4. Whether the particle is charged or not, matter waves are associated with it. Dr M CHANDRA SHEKHAR REDDY 5. No single phenomena exhibit both particle nature and wave nature simultaneously. 6. The wave nature of matter introduces an uncertainty in the location of the particle & the momentum of the particle when both are determined simultaneously. 2𝑚𝑒𝑉 Wave number : Reciprocal of the wavelength 𝑊𝑎𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟, 𝜈ҧ = ℎ Wave velocity (u) : Product of the wavelength and frequency 𝐸 𝑃2 1 ℎ 𝑃2 ℎ 1 From 𝐸 = ℎ𝜐 we can write, 𝜐 = = × = × = × ℎ 2𝑚 ℎ 2𝑚 ℎ2 2𝑚 𝜆2 \\ by multiplying the numerator and denominator with h. ℎ 1 ℎ ∴ Wave velocity u = 𝜐𝜆 = × ×𝜆 = 2𝑚 𝜆2 2𝑚𝜆 Experimental evidences for the existence of matter waves Davisson and Germer’s Experiment Dr M CHANDRA SHEKHAR REDDY Experimental Arrangement: The electrons were produced by heating the filament F with a low tension battery. These electrons were accelerated with the help of high tension battery and collimated into fine pencil of beam with the help of slits S1 and S2. Dr M CHANDRA SHEKHAR REDDY This beam of electron is made to fall on a single large nickel crystal. These electrons acting like the waves are diffracted by the different atomic planes. The angular distribution of the electrons is measured by the electron detector. This electron detector can move on a circular scale between 20° to 90° and connected to a galvanometer. The pronounced scattering directions was found to be 50° and for a voltage of 54 volts. Explanation Interatomic distance of Ni a = 2.14×10-10 m Inter planar spacing d = a sin φ = 2.14×10-10 sin25◦ = 0.909 ×10-10 m From Braggs law we have 2d sinθ= nλ For first order diffraction we can write 2×0.909 ×10-10 sin (90-25) = 1×λ (or) λ =we 1.165 -10 Similarly from de Broglie principle, have×10 m 12.27×10−10 12.27×10−10 𝜆= = 1.67 ×10-10 m 𝑉 54 This value is in agreement with the experimental value. Hence the experiment proves the wave nature of matter. G.P. Thomson’s Experiment A high speed electron beam is produced by the cathode. This beam is accelerated with Dr M CHANDRA SHEKHAR REDDY high potential differences up to 50,000 V and made to fine pencil using slits. This beam of electrons is incident on the thin gold film of thickness 10-8 m. To avoid the loss of energy of electron beam due to the collisions with air molecules, the total chamber is evacuated. The electron beam transmitted through the metal foil gets scattered producing diffraction pattern consisting of concentric circulars rings around a central spot. To make sure that the pattern is due to electron beam but not due to x rays, the cathode rays were subjected to magnetic field. This resulted in the shift of the entire fringe pattern. Hence it is confirmed that the pattern is only due to electrons as x rays are not affected by the magnetic field.  Dr M CHANDRA SHEKHAR REDDY 𝑟 𝑟 From the fig tan 2𝜃 = 2𝜃 = // For small angles tan 2𝜃 = 2𝜃 = 𝑙 𝑙 From Braggs law, 2𝑑 sin𝜃 = (2𝑑)𝜃 = 𝑛𝜆 r rd d = n (or )  = l nl 12.27×10−10 Similarly from de Broglie equation, we have 𝑜𝑟 𝜆 = 𝑚 𝑉 The wavelengths calculated from the above two methods are in good coordination hence the wave nature of the particle is proved.  Heisenberg Uncertainty Principle “It is impossible to measure a particular pair of physical variables precisely and simultaneously. The uncertainty in measuring such pairs is on the order of Planks constant” Dr M CHANDRA SHEKHAR REDDY In quantum mechanics, a particle is described in terms of wave packet. According to Max Born probability interpretation the particle may be found anywhere in the wave packet. When the wave packet is small, the position of the particle may be fixed but the velocity there by the moment become indeterminate as the particle spread rapidly. On the other hand, when wave packet is large, the velocity can be fixed but the position become indeterminate. If ∆P and ∆x are uncertainties in the momentum and position ℎ measurements, according to uncertainty principle ∆𝑃 × ∆𝑋 ≥ 4𝜋 Similarly, If ∆E is uncertainty in measuring the energy of an atomic ℎ process and ∆t is uncertainty in measuring the time interval ∆𝐸 × ∆𝑡 ≥ 4𝜋 If ∆J is uncertainty in measuring the angular momentum and ∆θ ℎ ∆𝐽 × ∆𝜃 ≥ is uncertainty in measuring the angle. 4𝜋 Schrödinger’s Time Independent Wave Equation Schrödinger equation (1926) can be considered as the fundamental equation in quantum mechanics just like F=ma in case of classical mechanics. Dr M CHANDRA SHEKHAR REDDY wave function is represented by Ψ(x,y,z,t) (or) Ψ Schrödinger’s Time Independent Wave Equation From classical mechanics, the differential 𝜕2𝜓 2 𝜕2𝜓 𝜕2𝜓 𝜕2𝜓 =v + + equation of wave motion is given by 𝜕𝑡 2 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜕2 𝜓 2 2 𝜕2 𝜓 𝜕𝑡 2 =v 𝛻 𝜓 (or) − v 2 𝛻 2 𝜓 = 0 − −(1) 𝜕𝑡 2 𝜕2 𝜕2 𝜕2 𝛻2 = 2 + 2 + 2 → 𝐿𝑎𝑝𝑙𝑎𝑐𝑖𝑎𝑛 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟 𝜕𝑥 𝜕𝑦 𝜕𝑧 Equation (1) is a second order differential equation. Let the trial solution is 𝜓 = 𝜓0 sin 𝜔𝑡 𝜕𝜓 𝜕2𝜓 2 𝜓 sin 𝜔𝑡 = −𝜔 2 𝜓 = 𝜔𝜓0 cos 𝜔𝑡 = −𝜔 0 𝜕𝑡 𝜕𝑡 2 𝐵𝑢𝑡 𝑤𝑒 𝑘𝑛𝑜𝑤 𝜔 = 2𝜋𝜐  𝜕2𝜓 2 4𝜋 2 v 2 = − 2𝜋𝜐 𝜓 = − 𝜓 𝜕𝑡 2 𝜆2 ℎ Dr M CHANDRA SHEKHAR REDDY From de Broglie’s equation we have 𝜆 = 𝑚𝑣 𝜕2 𝜓 2𝜓 4𝜋2 v2 = − 2𝜋𝜐 = − 2 𝑚v 2 𝜓----(2) 𝜕𝑡 2 ℎ 4𝜋 2 From (1) & (2) 𝛻 𝜓 + 2 𝑚v 2 𝜓 = 0 2 ℎ We know the total energy of the system E= Kinetic Energy + Potential Energy 1 𝐸= 𝑚v 2 + 𝑉 𝑜𝑟 𝑚v 2 = 2𝑚(𝐸 − 𝑉) 2 2 8𝑚𝜋 𝛻 2𝜓 + (𝐸 − 𝑉)𝜓 = 0 ℎ2 ℎ 2𝑚 Let ℏ = 𝛻2𝜓 + 2 (𝐸 − 𝑉)𝜓 = 0 2𝜋 ℏ The above two equations are called as Schrödinger time independent wave equations Schrödinger’s Time Dependent Wave Equation Can be obtained by eliminating “E” from Schrödinger time independent wave equation. Dr M CHANDRA SHEKHAR REDDY 𝜕𝜓 𝐿𝑒𝑡 𝜓 = 𝜓0 𝑒 −𝑖𝜔𝑡 = −𝑖𝜔𝑒 −𝑖𝜔𝑡 = −𝑖𝜔𝜓 𝜕𝑡 2𝜋𝐸 𝐸 𝜕𝜓 −𝑖𝐸𝜓 𝐸𝜓 𝜕𝜓 𝐵𝑢𝑡 𝜔 = 2𝜋𝜐 = = ∴ = = 𝑜𝑟 𝐸𝜓 = 𝑖ℏ ℎ ℏ 𝜕𝑡 ℏ 𝑖ℏ 𝜕𝑡 Substituting in Schrodinger time independent wave equation 2𝑚 𝜕𝜓 𝜕𝜓 ℏ2 2 𝛻 2𝜓 + 2 (𝑖ℏ − 𝑉𝜓) = 0 (or) 𝑖ℏ − 𝑉𝜓 = − 𝛻 𝜓 ℏ 𝜕𝑡 𝜕𝑡 2𝑚 𝜕𝜓 ℏ2 2 𝑖ℏ = − 𝛻 +𝑉 𝜓 𝜕𝑡 2𝑚 ෠ = 𝐻𝜓 𝑜𝑟 𝐸𝜓 ෡ ෠ 𝜕 ℏ2 2 𝐸 = 𝑖ℏ → 𝐸𝑛𝑒𝑟𝑔𝑦 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟 ෡=− 𝐻 𝛻 + 𝑉 → 𝐻𝑎𝑚𝑖𝑙𝑡𝑜𝑛𝑖𝑎𝑛 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟 𝜕𝑡 2𝑚 Physical Significance of wave function According to Max Born, the product of the wave function (ψ) with its complex conjugative (ψ*) gives the probability density i.e. the probability of finding particle. Dr M CHANDRA SHEKHAR REDDY The probability of finding particle in a volume dxdydz is 𝜓 2 𝑑𝑥𝑑𝑦𝑑𝑧 The total probability of finding particle is unity ම 𝜓 2 𝑑𝑥𝑑𝑦𝑑𝑧 = 1 normalization condition Limitations of Wave function It must be finite for all values of x,y,z. It must be single valued i.e. for each set of x,y,z, ψ must have only one value. It must be continuous in all regions except where the potential energy is infinite. Ψ is analytic i.e. it possess continuous first order derivatives. Ψ vanishes at boundaries. Applications of Schrodinger wave equation Particle in one dimensional box: Consider a particle of mass m is bouncing back and forth between the walls of a one Dr M CHANDRA SHEKHAR REDDY dimensional box of length a. Let the motion of particle is restricted only along the x axis i.e. in between x=0 and x=a. Let the potential energy of the walls of the box is infinity and potential energy inside the box is uniform. For simplicity, this uniform potential energy can be considered as zero. Hence we can write the potential function as 𝑉(𝑥) = 0 𝑓𝑜𝑟 0 < 𝑥 < 𝑎, inside the box 𝑉 𝑥 =∝ 𝑓𝑜𝑟 𝑥 < 0& 𝑥 > 𝑎, outside the potential well. The Schrodinger wave equation inside the potential well is 𝑑2 𝜓 2𝑚 + 𝐸𝜓 = 0 --------(1) 𝑑𝑥 2 ℏ2 𝑑2𝜓 2 𝜓 = 0 − −(2) 𝑤ℎ𝑒𝑟𝑒 𝑘 2 = 2𝑚𝐸 ---------(3) 𝑜𝑟 + 𝑘 ℏ2 𝑑𝑥 2 The above equation is a second order differential equation. Let the trail solution is 𝜓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘𝑥 + 𝐵 𝑐𝑜𝑠 𝑘𝑥 The values of A & B can be obtained by applying the boundary conditions. Boundary condition -01 As the particle cannot penetrate the potential well, ψ(x) = 0 at x =0 Hence, we can write 0= 𝐴 𝑠𝑖𝑛 𝑘(0) + 𝐵 𝑐𝑜𝑠 𝑘(0) (or) B=0 The trial solution reduces to 𝜓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘𝑥 Dr M CHANDRA SHEKHAR REDDY Boundary condition -02 𝑛𝜋 𝑜𝑟 0 = 𝑠𝑖𝑛 𝑘𝑎 𝑜𝑟 𝑘 = ± -----------(4) 𝑎 𝑛𝜋 ∴ 𝜓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑎 𝑥 --------(5) 1 2𝑚𝐸 2 𝑛𝜋 From equation (3) & (4) we can write 𝑘= =± ℏ2 𝑎 𝑛2 𝜋 2 ℏ2 𝑛2 ℎ2 𝑜𝑟 𝐸 = = 2𝑚𝑎 2 8𝑚𝑎2 𝜋 2 ℏ2 𝑓𝑜𝑟 𝑛 = 1, 𝐸1 = − 𝑍𝑒𝑟𝑜 𝑝𝑜𝑖𝑛𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 2𝑚𝑎2 𝜋 2 ℏ2 𝑓𝑜𝑟 𝑛 = 2, 𝐸2 = 4 × 2𝑚𝑎2 = 4𝐸1 𝜋 2 ℏ2 𝑓𝑜𝑟 𝑛 = 3, 𝐸3 = 9 × = 9𝐸1 2𝑚𝑎2 Hence we can write the equation as 𝐸𝑛 = 𝑛2 𝐸1 This shows that the energy of the particle in one dimensional box is quantized The spacing between nth level and (n+1)th level is (𝑛 + 1)2 𝐸1 −𝑛2 𝐸1 = 2𝑛 + 1 𝐸1 Dr M CHANDRA SHEKHAR REDDY Here the integers corresponding to n=1,2,3 are called as quantum numbers for corresponding En. The wave function corresponding to is called as eigen function of the particles. To calculate the value of A 𝑎 2 The normalization condition in one dimension can be written as ‫׬‬0 𝜓 𝑑𝑥 = 1 −− −(6) 𝑎 2 2 𝑛𝜋 Substituting (5) in (6) න 𝐴 𝑠𝑖𝑛 𝑥 𝑑𝑥 = 1 0 𝑎 𝐴2 𝑎 2𝑛𝜋 = න 1 − 𝑐𝑜𝑠 𝑥 𝑑𝑥 = 1 // cos 2𝜃 = 1 − 2 sin2 𝜃 2 0 𝑎 𝑎 𝐴2 𝑎 2𝜋𝑛 𝐴2 𝑎 2 = 𝑥− 𝑠𝑖𝑛 𝑥 =1= (or)𝐴 = 2 2𝜋𝑛 𝑎 0 2 𝑎 Therefore the wave equation in one 2 𝑛𝜋 dimensional potential box will become 𝜓 𝑥 = 𝑠𝑖𝑛 𝑥 𝑎 𝑎 Probability of finding particle 2 𝑛𝜋 Probability of finding particle is given by 𝑃 𝑥 = 𝜓 × 𝜓 ∗ = ψ 2 = 𝑎 𝑠𝑖𝑛2 𝑥 𝑎 𝑛𝜋 𝜋 3𝜋 5𝜋 𝑎 3𝑎 5𝑎 (or) Probability is maximum for 𝑥 = 2, , −− − 𝑜𝑟 𝑥 = , , −− − 𝑎 2 2 2n 2n 2n Dr M CHANDRA SHEKHAR REDDY 𝑎 For n=1, the most probable position of finding particle is 𝑥 = 2 𝑎 3𝑎 For n=2, the most probable position of finding particle is 𝑥 = 4 , 4 𝑎 3𝑎 5𝑎 For n=3, the most probable position of finding particle is 𝑥 = , , 6 6 6 Particle in there dimensional box: Similarly if we consider three dimensional rectangular box with sides a,b,c, we can write the wave equation and energy of particles as Dr M CHANDRA SHEKHAR REDDY 2 𝑛𝜋 2 𝑛𝜋 2 𝑛𝜋 𝜓 𝑥, 𝑦, 𝑧 = 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 𝑦 𝑠𝑖𝑛 𝑧 𝑎 𝑎 𝑏 𝑏 𝑐 𝑐 𝜋 2 ℏ2 𝑛𝑥2 𝑛𝑦2 𝑛𝑧2 𝐸(𝑛𝑥, 𝑛𝑦, 𝑛𝑧 ) = + + 2𝑚 𝑎2 𝑏2 𝑐 2 Here nx, ny, nz are quantum numbers along x, y, z directions Similarly for cubic box i.e. a=b=c, Hence we can write 8 𝑛𝜋 𝑛𝜋 𝑛𝜋 𝜓 𝑥, 𝑦, 𝑧 = 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 𝑦 𝑠𝑖𝑛 𝑧 𝑎3 𝑎 𝑏 𝑎 𝜋 2 ℏ2 2 𝐸(𝑛𝑥, 𝑛𝑦, 𝑛𝑧 ) = 2 (𝑛𝑥 + 𝑛𝑦2 + 𝑛𝑧2 ) 2𝑚𝑎 PROBLEMS 1. Calculate the de Broglie wavelength of a dust particle of mass 0.002 mg and moving with velocity of 3.50×104 m/s. Dr M CHANDRA SHEKHAR REDDY 2. Calculate the velocity and kinetic energy of an electron having wavelength of 0.21 nm 3. Calculate the wavelength associated with a neutron whose kinetic energy is 1.5 times the rest mass of electron.(Given that Mass of neutron = 1.676 × 10−27 kg, Mass of electron = 9.1 × 10−31 kg, Planck’s constant = 6.62 × 10−34 J-sec, Velocity of light = 3 × 108 m/s). 4. Electrons are accelerated by 344volts and reflected from a crystal. The first order maximum occurs when the glancing angle is 60o. Determine the spacing of crystal 5. Calculate the wavelength associated with an electron having the kinetic energy equal to 1 MeV. 6. Calculate the minimum energy of an electron moving in a one dimensional infinite high potential well of width 1.23 nm. 7. Calculate the lowest energy possessed by neutron confined to a nucleus of size 10-14 m. 8. An electron is confined to an one-dimensional infinite potential well of width of 0.2 nm. It is found that when the energy of the particle is 230 eV, the eigen function has 5 antinodes. Find the mass of the particle and show that it can never have the energy equal to 1K eV. 9. Find the energies of six lowest energy levels of a particle in a cubical box of width 1 nm. Dr M CHANDRA SHEKHAR REDDY Which of these levels are degenerate? 10. If the electron having the de Broglie wavelength as 1.21×10-10 m is confined in one dimensional box, how far apart must be the walls of the box when five loops of the de Broglie wave span are formed from one wall to other. 11. Consider an electron is confined to move between two rigid walls separated by 1 nm. It require 1.36 fs to cross the gap. What quantum numbers describe this motion? Consider the quantum numbers with nearest integer. 12. An electron in a hydrogen atom is known to be somewhere between 0.050 nm and 0.10 nm from a proton. What is the minimum uncertainty in the speed of that electron? 13. Compare the uncertainty in the velocities of electrons and protons confined to a one dimensional box 14. A nucleon is confined to a nucleus of diameter 5×10-4 meter. Calculate the minimum uncertainty in the momentum of the nucleon. Also calculate the minimum kinetic energy of the nucleon 15. Show that if the uncertainty in the location of a particle is equal to the de Broglie wavelength, then the magnitude of uncertainty in its velocity is equal to the magnitude of the velocity of the particle.