Solution 01. Sut 25268

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Marvin Richard
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1 Solution. Since this is an estimate, more than one solution is possible depending on the approximations made. One solution is given The object of this section is to estimate the the size of an atom. While the Schrodinger equation can be solved for the hydrogen atom, only the uncertainty relation will be used here. This approach is mathematically simpler and provides some insight into what determines the size of an atom. Hydrogen consists of a positively charged proton and a negatively charged electron. The force of attraction between the proton and the electron is given by Coulomb's law, The potential energy of the electron can be calculated by integrating the force as a function of the position. The potential energy is, = 4 The potential energy is negative. The electron lowers its energy as it approaches the proton. t would seem that the electron would want to get as close to the proton as possible. However, there is another energy that prevents the electron from getting too close to the proton. This is the quantum confinement energy. The quantum confinement energy can be estimated using the Heisenberg uncertainty relation, ΔxΔpx h/(4π). For a hydrogen atom, the uncertainty in the position of the electron is about the radius of the atom Δx = r. This means that the then Δpx h/(4πr). The definition of the uncertainty in momentum is, =. Here the angular bracets < > signify that the average value of the quantity in the braets should be used. Since the electron is confined to a region near the proton, the average momentum of the electron is zero, <px> =. This means that the average inetic energy of the electron in the x-direction is, 2 h 32 There is an analogous increase in the inetic energy in the y and z directions so the total increase in inetic energy is three times the increase calculated above. = h 32 The electron decreases its potential energy by getting close to the proton but increases its confinement energy by concentrating itself too much around the proton. The radius of an atom can be estimated by determining the radius that minimizes the total energy of the electron. The radius that minimizes + is = 3h 4 =4. [] This is ¾ of the Bohr radius, the true radius of an electron in hydrogen that can be calculated using a full quantum mechanical analysis. The important result of this calculation is that the inetic energy of a particle increases as the particle is confined to a smaller region. t is the balance between this confinement energy and the electrostatic energy that is largely responsible for determining the size of atoms. This discussion also shows that quantum effects have to be considered when discussing phenomena at the nanometer scale. 2. DeBroglie Wavelength is as λ=h where is momentum and for a particle with zero velocity mass can be written as = 2 and for photon (a particle with no zero mass) can be written as = where is the velocity of light. Knowing that: h=6.626 / =9. ~ =.67 = / =.6

2 Solution -4-5 λ (m) -6-7 photon -8-9 electron - - proton & neutron -2 E (ev) a) Normalization: b) = = = c) = = = = = = hence = = = = = 2 = 4 2 = = = d) ~, :, :, ~, :, :, = 4. a) Before = : = + where = 2hence = 8 After = : = +2 where = 4 or = hence = 32 or = 4. The wave function of the particle at times > is,= where =,= 2 = cos 4,=,,2, Hence, for all even s (corresponding to odd eigenfunctions), will be zero. The probability of finding particle in even eigenfunctions will be as = = =,3,5, b) expectation value of the energy can be written as:

3 Solution = = 6 2+ [2+ 4] (Summation performed only on the odd values of ) which shows energy is conserved! c) the characteristic velocity of the system is =2 = 2and results are valid if = 5. a) = = = + Note that is the operator of the location so did not apply on it! 2 += 2 + = 2 Replacing 2 in, we will get = 2 + = 2 + Knowing that = = 2 [ + ] = Knowing the fact that ± = or = = = Similarly = = + Using 2 = = = Which is the quantum counterpart of = b) Considering = and nowing that = + = + = 2 = =., comparing with + = 6. < 2 = = +,h = 2 > = 2=,h = 2

4 Solution << 2 + = = +,h =2 Using conditions for continuity of and at = and = we will get = = + 4 sinh Call =/ and =2, can be written as = see plot for =2 and =5 lim = = + 4 sinh T β= β=2.2. E / V bv bv bv bv -a a 2a Periodic -function potentials (a simplified model to the periodic potential in a one-dimensional lattice). We can express mathematically this potential energy term as: V ( x) = bv δ ( x na) n= n region (), the wavefunction is given by: = Aexp[ iβ x] + B exp [ iβ x], β = 2mE Further relations between the coefficients in different regions can be obtained from Bloch s theorem. Each stationary solution of the Schrödinger equation for a periodical potential can be written in the form x = u x exp ix [ ] [ ] [ ] where u [x] has the same periodicity as the potential. From equation (6), it also follows that

5 Solution [ x + a] = exp[ ia] [ x] which means that a translation by a only results in a phase shift of exp[il]. The same translation symmetry is valid for the derivative of the wave function. Therefore, we can write the wavefunction in region () in terms of the one in region () using Bloch theorem, to get: ia iβ ( x a) iβ ( x a) ia = ( x a) e = Ae + Be e We also now that for a wavefunction to be a proper function, it must satisfy the continuity requirement, i.e. ( a) = ( a) Which gives ia iβ a i a ia ( ) ( β ) A e e = B e e () The continuity of the first derivative is not satisfied when is a -function. This can be shown directly from the D TSE, d d 2m + V ( x) E ( V ( x) E) 2 = = 2 2 2m d x d x f this equation is integrated in the neighborhood of =, we get: d d d 2m 2m dx = = ( bv δ ( x) E) dx = bv () d x d x + d x x = x= Using the expression for, we arrive at a second equation that is relating coefficients A and B: ia i a 2mbV ia ia i a 2mbV ia A β β ie iβe e = B ie iβe + e 2 2 (2) Dividing equations () and (2) and rearranging the terms leads to the following final expression: sin[ β a] cos[ a] = cos [ β a] + P. (3) β a Where β = 2mE and P = 2mbV 2 The right hand side of the (3) is plotted in the next figure in blue, the left hand side could be any number between - and (as shown in red lines) (plotted for P=6)

Problems and Multiple Choice Questions

Problems and Multiple Choice Questions 1. A momentum operator in one dimension is 2. A position operator in 3 dimensions is 3. A kinetic energy operator in 1 dimension is 4. If two operator commute, a)