Problem 1:
A negative K meson with mass m = 1000 electron masses is captured into a circular Bohr orbit around a lead nucleus (Z = 82). Assume it starts with principal quantum number n = 10 and then cascades down through n = 9, 8, 7, …, etc. What is the energy of the photon emitted in the n = 10 to n = 9 transitions?
Solution:
 | Concepts, principles, relations
that apply to the problem: The hydrogenic atom, the ground-state energy of the hydrogen atom, the Bohr radius |
| Why do they apply? The lead nucleus and the meson form a hydrogenic atom. The energy levels and the average distance of the meson from the nucleus can be determined from the ground-state binding energy of the hydrogen atom and the Bohr radius. The photon emitted in a transition has an energy equal to the energy difference between the initial and the final state. |
| How do they apply? We have a hydrogen like system with Z = 82:  .  . . |
| Details of the calculation: None |
Problem 2:
In an atom, a valence electron experiences a long range Coulomb force and the
potential well representing the interaction supports an infinite spectrum of
bound states. In contrast, the interaction between the outermost electron in a
negative ion and the neutral atomic core, which is weak and short ranged,
results in only a finite number of bound states.
For the case of s states (l = 0), the negative ion may be
approximated by a model in which the interaction between the outermost electron
of mass m and the core is represented by an attractive one-dimensional
central potential of the form
V(r) = -V0, 0 < r < a;
V(r) = 0 r ³ a.
(a) Solve the time independent Schroedinger equation and determine an
expression, in the form of a transcendental equation, relating the eigenvalues
of this system to the quantities V0, a, and m. Solve this equation graphically.
(b) Show, graphically or otherwise, that there will exist no bound states
unless
R = (2mV0a2/h2)1/2
³ p/2.
(c) Determine how many bound states exist if R = p.
(d) The condition for the existence of bound states depends on the product
of the depth and the square of the width of the potential well. Explain this in
terms of the uncertainty principle.
Solution:
| Concepts, principles, relations
that apply to the problem: Three-dimensional square potentials |
| Why do they apply? The radial equation for a particle in a central potential with l = 0 is  .Here V(r) is a square potential an m = me = m. |
| How do they apply? (a) We have to solve a square potential problem in the region 0 < r < ¥. Let region 1 extend from r = 0 to r = a and region 2 from r = a to infinity. For bound states we have E < 0. Define k2 = (2m/h2)(E + V0), r2 = (2m/h2)(-E), and k02 = (2m/h2)V0. Note: I am assuming that V0 is a positive number denoting the depth, the potential at the bottom of the well is -V0. In region 1 we have f1(r) = A sin(kr), since f1(0) = 0. In region 2 we have f2(r) = Bexp(-rr), since f2(¥) = 0. At r = a we need hat f(a) and (¶/¶x)f(x)|a are continuous. A sin(ka) = Bexp(-ra) kA cos(ka) = -rBexp(-ra) Therefore cot(ka) = -r/k. 1/sin2(ka) = 1 + cot2(ka) = (k2 + r2)/k2 = k02/k2. We can find a graphical solution by plotting |sin(ka)| and k/k0 versus k. The intersections of the two plots in regions where cot(ka) < 0 gives the values of k for which a solution exist.  |
| Details of the calculation: (b) If the slope 1/k0 > 2a/p, then no solution exists. 1/k0 > 2a/p --> k0a < p/2, or R = (2mV0a2/h2)1/2 < p/2. We need R ³ p/2 for a solution to exist. (c) If R = p, then k0a = p, 1/k0 = a/p. Only one bound state exists. (c) For a bound state to exist we need (2mV0)1/2 > ph/(2a). If a bound state barely exists, then the kinetic energy of the particle is T ~ V0. We have Dpr ~(2mT)1/2, for the uncertainty the momentum, Dr ~ a for the uncertainty in the position. Therefore DprDr ~ ph/2 for a bound state to exist. |
Problem 3:
Let r, f, z be the
cylindrical coordinates of a spinless particle. Assume the potential
energy of this particle depends only on r and not on f
and z.
(a) Write in cylindrical coordinates the differential operator associated
with the Hamiltonian. Show that H commutes with Lz and Pz.
Show from this that the wave function associated with the stationary states of
the particle can be written as
ynmk(r, f,
z) = fnm(r)exp(imf)exp(ikz),
where the values that can be taken on by the indices m and k are to be
specified.
(b) Write in cylindrical coordinates the eigenvalue equation of the
Hamiltonian of the particle. Derive from it the differential equation
which yields fnm(r).
Solution:
| Concepts, principles, relations
that apply to the problem: The Hamiltonian and Schroedinger equation in polar coordinates, the Lz and Pz operators, commuting variables |
| Why do they apply? We are asked to write the Hamiltonian H in cylindrical coordinates and show that H commutes with Lz and Pz. |
| How do they apply? (a) H = -[h2/(2m)]Ñ2 + V(r). Here m is the mass of the particle. In cylindrical coordinates H = -[h2/(2m)][¶2/¶r2 + (1/r) (¶/¶r) + (1/r2)(¶2/¶f2) + ¶2/¶z2] + V(r). Using Lz = [h/i]¶/¶f, Pz = [h/i]¶/¶z, Lz2 = -h2¶2/¶f2, Pz2 = -h2¶2/¶z2, we can write H = Pr2/(2m) + Lz2/(2mr2) + Pz2/(2m) + V(r). Here Pr2 = -h2[¶2/¶r2 + (1/r) (¶/¶r) ]. Lz depends only on f, Pz depends only on z, and Pr2 depends only on r. Therefore [Lz,Pr2] = [Lz,Pz ], = [Pr2,Pz ] = 0 and [H,Lz] = [H,Pz ] = 0. We can find a common eigenbasis of H, Lz and Pz. This means separation of variables in cylindrical coordinates is possible. {exp(imf)} = eigenbasis of Lz. {exp(ikz)} = eigenbasis of Pz. Any function y(r, f, z) can be written as a linear combination of functions f(r)exp(imf)exp(ikz), y(r, f, z) = Smexp(imf)òdkfk(r)exp(ikz), since k is a continuous index. The functions f(r)exp(imf)exp(ikz) is an eigenfunction of H if H f(r)exp(imf)exp(ikz) = E f(r)exp(imf)exp(ikz), [Pr2/(2m) + Lz2/(2mr2) + Pz2/(2m) + V(r)]f(r)exp(imf)exp(ikz) = E f(r)exp(imf)exp(ikz). The plane wave exp(ikz) represents a stream of particles moving with uniform speed along the z-axis. The energy associated with each particle is h2k2/(2m). Define E' = E - h2k2/(2m). E' is the energy associated with the motion perpendicular to the z-axis. [Pr2/(2m) + h2m2/(2mr2) + V(r)]f(r) = E' f(r). The eigenfunction f(r) does not depend on k, only on m. We label different possible solution for the same m with the index n. [Pr2/(2m) + h2m2/(2mr2) + V(r)]fnm(r) = E'nm fnm(r). |
| Details of the calculation: (b) [-(h2/(2m))[¶2/¶r2 + (1/r) (¶/¶r)] + h2m2/(2mr2) + V(r)]fnm(r) = E'nm fnm(r). [[¶2/¶r2 + (1/r) (¶/¶r)] - m2/r2 + U(r)]fnm(r) = lnm fnm(r). Here E'nm = h2lnm2/(2m) and V(r) = h2U(r)/(2m). |
Problem 4:
Consider the following non-pure state for a hydrogenic
atom:
|y> = a1|y100>
+ a2|y200> + a3|y210>
+ a4|y32-1> + a5|y432>
a1 = (3/10)1/2, a2 = (1/10)1/2,
a3 = (2/10)1/2, a4 = (1/10)1/2,
a5 = (3/10)1/2.
i) Show that |y> is normalized. What
property of the hydrogen wave functions must you exploit to show that?
ii) Calculate the probability of observing a 1s state.
iii) Calculate the probability of observing a state with n > 2.
iv) What are the possible values you would get if you measured the quantity
associated with Lz? Give the probability of measuring each of these
values.
v) Calculate the expectation value of Lz.
vi) What are the possible values you would get if you measured the quantity
associated with L2? Give the probability of measuring each of these
values.
vii) Calculate the expectation value of L2.
Solution:
| Concepts, principles, relations
that apply to the problem: The hydrogen atom, postulates of QM |
| Why do they apply? When a physical quantity described by the operator A is measured on a system in a normalized state |y>, the probability of measuring the eigenvalue an is given by P(an) = Si=0gn|<uni|y>|2, where {|uni>} (i=1,2,...,gn) is an orthonormal basis in the eigensubspace En associated with the eigenvalue an. |
| How do they apply? |y> = a1|y100> + a2|y200> + a3|y210> + a4|y32-1> + a5|y432> The |ynlm> are normalized common eigenstates of H, L2, and Lz. H|ynlm> = En|ynlm>, L2|ynlm> = l(l+1)h2|ynlm>, Lz|ynlm> = mh|ynlm>. <ynlm|ynlm> = 1. <yn’l’m’|ynlm> = 0 unless n’ = n, l’ = l, m’ = m. <ynlm|ynlm> stands for òall space| ynlm (r,q,f)|2dV. i) <y|y> = a12 + a22 + a32 + a42 + a52 = 1. The wave function is normalized because the hydrogen-atom wave functions are orthonormal. ii) The probability of observing the ith state in the expansion is |ai|2. The probability of observing the 1s state is |a1|2 = 3/10. iii) The 4th and the 5th state in the expansion have n > 2. P(n > 2) = probability of observing the 4th state + probability of observing the 5th state = 4/10. iv) The values you can obtain for Lz are mh, where m = 0, -1 or 2. P(m = 0 ) = a12 + a22 + a32 = 6/10. P(m = -1) = 1/10, P(m = 2) = 3/10. v) <Lz> = ( 0 *6/10 + -1*1/10 + 2*3/10) mh = ½mh. vi) The values you can obtain for L2 are l(l+1)h2, where l = 0, 1, 2, or 3. P(l = 0) = a12 + a22 = 4/10, P(l=1) = 2/10, P(l = 2) = 1/10, P(l = 3) = 3/10. vii) <L2> = (0 *4/10 + 2*2/10 + 6*1/10 + 12*3/10)h2 = (23/5)h2. |
| Details of the calculation: None |
Problem 5:
Find the eigenvalues and eigenfunctions for a particle in a box with sides a, b, c by solving the time independent Schroedinger equation (-h2/(2m)) Ñ2f(r+ U(r)f(r) = Ef(r). Do not just write down your answer but derive it.
Solution:
| Concepts, principles, relations that apply to the
problem: The particle in a 3-dimensional box, separation of variables |
| Why do they apply? We are asked to derive the eigenfunctions and eigenvalues of the Hamiltonian of a particle in a 3D box. |
| How do they apply? U(x,y,z) = 0 if 0 < x < a AND 0 < y < b AND 0 < z < c, U(x,y,z) = infinite otherwise. In Cartesian coordinate the 3D time-independent Schroedinger equation is (-h2/(2m))[¶2f(x,y,z)/¶x2 + ¶2f(x,y,z)/¶y2 + ¶2f(x,y,z)/¶z2] + U(x,y,z)f(x,y,z) = Ef(x,y,z).
We usually try to solve such equations by a technique called separation of
variables. The normalized solutions to the differential equations
are The eigenfunctions for the problem therefore are |
| Details of the calculation: None |
Problem 6:
A hydrogen atom is placed in a uniform electric field, E = -E.
Place the proton at the origin of your coordinate system. An electron in the
hydrogen atom then has potential energy U(r) = -kqe2/r
– qeEz. U(0,0,z) becomes increasingly positive for negative z . For
positive z the potential energy U(0,0,z) contains a “hill” and then decreases
with increasing z.
Sketch the potential energy of the electron, U(0,0,z), as function of z and
calculate the energy at the maximum at positive z. Equate this energy to the
energy of the unperturbed (zero field) hydrogen energy level and thereby
determine the value of the field required to field-ionize a hydrogen atom with
principle quantum number n (neglect tunneling).
Solution:
| Concepts, principles, relations that apply to the
problem: The energy levels of the hydrogen atom |
| Why do they apply? The perturbed potential has a local maximum. If an energy level of the unperturbed hydrogen atom lies above this local maximum, than field ionization is likely. |
| How do they apply? U(0,0,z) = - kqe2/|z| – qeEz. For positive z: dU(0,0,z)/dz = kqe2/z2 – qeE = 0. z2 = kqe/E, z = (kqe/E)1/2. Let - kqe2/(kqe/E)1/2 – qeE(kqe/E)1/2 = -(EI)/n2, with EI = 13.6 eV. Then -2qe(kqeE)1/2 = -(13.6 eV)/n2, 4kqe3E = EI2/n4, E = EI2/(4kqe3n4) = (3.21*1010 V/m)/n4. A reasonable field (kV/cm) can ionize levels with n > 24. |
| Details of the calculation: None |
Problem 7:
Two atoms of masses m1 and m2 are bound together in a diatomic molecule. The separation of their nuclei is r. What are the rotational kinetic energy levels of the molecule? How does the energy of the first excited state of 13C16O compare to that of 12C16O?
Solution:
| Concepts, principles, relations that apply to the
problem: The rigid rotator |
| Why do they apply? We model the system as a rigid rotator. |
| How do they apply? Rotation about the CM: E = L2/2I. I = m1r12 + m2 r22. r1 + r2 = r. m1r1 = m2 r2. r1 = m2r/(m1 + m2). r2 = m1r/(m1 + m2). I = r2m1m2/(m1 + m2) = mr2. m = m1m2/(m1 + m2) = reduced mass. L2 = l(l + 1)h2, l = 0, 1, 2, … . E = l(l + 1)h2(m1 + m2)/(2m1m2r2).
Ground state: E0 = 0. 13C16O: E1 = (29/208)
h2/r2 = 0.139
h2/r2. |
| Details of the calculation: None |