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Prepare a sketch! The four interior nodes divide the wire into five equal length segments, node-to-node spacings. Thus, 5 (/2) = 15 m or = 30 m/5 = 6 m. From the BWP, f = v/ = 12/6 s -1 = 2 Hz. C




(17) Kinetic Theory of Gases

Gas molecules have an average translational kinetic energy of 3/2 k T (three-halves times Boltzmann’s constant times the absolute temperature). The pressure arises due to the many molecular collisions per second with the walls so it increases for higher speeds v (v increases if temperature increases) and for higher density. The molecules have a cross sectional area for interaction  and so sweep out a volume  v t in time t. The mean time between collisions is found as the time for a molecule to sweep out the average volume per molecule.

v tcollision free = V/N so tfree = V/Nv

The mean free path free is the distance traveled to sweep out V/N.  free = V/N.



Ex. 17) Cubical tanks X and Y have the same volumes and share a common wall. There is 1 gram of helium in tank X and 2 grams of helium in tank Y, and both samples are held at the same temperature. Which of the following is the same for both samples?

(A) the number of molecular collisions per second on the

common wall

(B) the average speed of the molecules

(C) the pressure exerted by the helium

(D) the density of the helium

(E) the mean free path of the molecules




The number of collisions per second, pressure and density for sample Y are double the X values. The mean free path for sample Y is half that for sample X because there are twice as many things to hit in the same volume. The average speed depends on the temperature and molecular mass. It is identical for samples X and Y. B
(18) Ideal Gas Law

isothermal: process at constant temperature

isobaric: process at constant pressure

iso-entropic: process at constant entropy

adiabatic: process in which there is no heat transfer; insulated from the environment

isochoric, isometric, iso-volumetric -------- process at constant volume


The Ideal Gas Law: P V = n R T P V = N k T

number of moles: n

ideal gas constant R = 8.31 (J/ mol-K)

number of molecules: N

Boltzmann’s constant: k = 1.38 x 10-23 J/K
LaChatlier’s Principle: A stable system acts to counter any change in its parameters. If a the volume of a sample of gas is reduced, its pressure increases to fight additional decreases in volume.
(Ex. 18) If one mole of an ideal gas doubles its volume as it undergoes an isothermal expansion, it pressure is

(A) quadrupled

(B) doubled

(C) unchanged

(D) halved

(E) quartered

The process is isothermal (same temperature). P1 V1 = n R T = P2 V2

D

Use ratios whenever possible.


(19) Thermodynamics and efficiency

Entropy Change: Ideal Heat Engine Efficiency:

Use absolute temperatures. 0 0C = 273 K
Ex 19.) A power plant takes in steam at 527 0C to power turbines and then exhausts the steam at 127 0C. In any given time, it consumes 100 megawatts of heat energy from the steam. The maximum output power of the plane is

(A) 10 MW (B) 20 MW (C) 50 MW (D) 75 MW (E) 100 MW


The problem statement is a little fuzzy with regard to its use of the terms power and energy!

Using 0 0C = 273 K, Thigh = 800 K and Tlow = 400 K. The ideal thermal efficiency for a reversible engine operating between these temperatures is 50% so the maximum possible mechanical/electrical power out is 50 MW. C


QUANTUM MECHANICS AND ATOMIC PHYSICS:
Particles have associated wave properties. As a crude characterization, particles propagate like waves and interact as (point) particles. It may be helpful to assume that each particle has a guide wave that feels out the space to generate the probability distribution for the particle’s location. Waves such as optical radiation also display particle-like character. When they interact, energy transferred into and out of the wave in quanta (photons) each with energy hv (and momentum hv/c.)

The information that can be known about a system is encoded in a wavefunction (x,t) that satisfies the Schrödinger equation. The probability to find the particle between x and x + dx is (x,t)(x,tdx (Born interpretation). Additional information can be extracted from the wavefunction using the operator for the dynamical quantity of interest. The operator for x is x. The operator for px is - i x. Dynamic quantities in Hamiltonian mechanics are functions of position, momentum and time. The quantum operator for the quantity is formed by keeping the coordinates and the time while replacing each momentum pq with –i/q. px  - i x; p- i ; …..



Q(x,y,z, px,py,pz,t) (x,y,z, i x,i y,i z, t); K;

Lz = xpy – ypx

Shrödinger’s equation identifies two operators for energy, the Hamiltonian and i t. In other words, the Hamiltonian represents the total energy (kinetic + potential), and it is the time-development operator for wavefunction (multiplied by i/).



(Use separation.)

The values En are the energy eigenvalues, the total energies, a set of real values that are characteristic of the problem and the associated temporal dependence is: Tn(t) = .

A general solution has the form, .

Operators operate on wavefunctions to return other (or perhaps the same) wavefunctions.

Operators operate on their eigenfunctions to return a scalar multiplying the same wavefunction.

Eigenvalue equation: px u(x) = –i x u(x) = (momentum eigenvalue) u(x) = p u(x)

i x u(x) = p u(x)  u(x) = A

A e i (kx - t) is a momentum eigenfunction with eigenvalue



Hn(x,t) = En n (x,t)

Operators for physical observables are Hermitian (or self-adjoint) an, as such, they have real eigenvalues and expectation values.


Wavefunction Behavior: In a classically allowed region, the kinetic energy K, E ­– ­V, is greater than zero so the net curvature of  has a sign opposite to that of . The function value oscillates back and forth across zero. In the case that V > E or K < 0 (which is forbidden classically; regions for which V > E are classically forbidden), the curvature and function have the same sign  behavior like growing and decaying exponentials.

All the functions oscillate in the classically allowed region and decay in the forbidden regions. The decay is more rapid for larger energy deficits. Each higher state has an additional node.

Notice that the spatial rate of oscillation increases as the kinetic energy increases (or the ‘local wavelength decreases).

The n = 36 QHO wavefunction is plotted for t = 0. Note the more rapid oscillation around x = 0 where the kinetic energy is greatest.


Free Particle States for piecewise constant potentials are of the form where . At any abrupt change in V(x), the wave is partially transmitted and partially reflected. The wave function ca tunnel through classically forbidden regions (V > Ek = i and the wave is like ex).

Fails to show the reflected wave!

Energy eigenstates are stationary  the associated probability density * is time-independent.





Energy Level Spacing: Energy levels in an infinite well with width a have the form so the n to n + 1 spacing is ; the level to level spacing grows with n and is larger for smaller a (tighter confinement). The finite-well states penetrate into the forbidden region and so are less tightly confined leading to lower energies as compared to the infinite well with the same inside width. For the hydrogen atom problem, the electron’s range approaches infinitely wide as the energy approaches zero from below. The level spacing approaches zero in this limit. For positive energies, the particle is not confined and the allowed energies run continuously (zero spacing).
Transitions between levels: A quantum system absorbs or emits energy in chunks or quanta equal to the energy difference between the initial and final state of the system.

A hydrogen atom in an n = 4 state can make a spontaneous transition to an n =2 state emitting a photon of energy 2.55 eV.

- 13.6 eV/42 – (- 13.6 eV/22) = 2.25 eV   = 486 nm a fantastic shade of blue or blue-green!
Commutation: If the operators for two quantities do not commute, then those quantities have a minimum uncertainty product. [x,px] = i   xpx  ½ .

General Uncertainty Relation:

If operators commute, they can have simultaneous eigenvalues. For the hydrogen problem, H, L2 and Lz (plus the operator for electron spin) commute. The hydrogen atom states are labeled by nm representing the eigenvalues of energy ( - 13.6 eV/n2), orbital angular momentum squared ([+1]2) and z component of orbital angular momentum (m) (plus spin – up or down).
Generalized Uncertainty Principle. Experimental measurements are performed on systems in order to extract average values and the corresponding uncertainties. In quantum systems, constraints can arise between two uncertainties due to the nature of the corresponding operators. Given two operators A and B, and knowledge of the commutator [A,B] = iC, then there exists a hard lower limit on the product of uncertainties A B >= ½ ||. Example: [x,px] = i   xpx  ½ .

If two operators commute, then there is no lower-level uncertainty constraint on knowledge of the simultaneous values of operators (the outcomes of measurements), and the expectation values of these operators serve as a very useful label to identify and characterize a system. For the bare Coulomb potential (simple H-atom problem), the operators H, L2, and Lz commute, so the values of the three can be known exactly at the same time. Therefore we identify individual states by the eigenvalues (energy, a.m. magnitude, a.m. z-comp) or their corresponding labels n,,m (quantum numbers).


Expectation value:

Mixed State: more than one an  0. Here, an is the amplitude to be found in state n and | an |2 is the probability for the system to be found in state n. As energy eigenstates have been used above, E. A precise measurement of the energy will yield one of the energy eigenvalues. For a set of identically prepared systems, the eigenvalue En will be found with probability | an |2. Making a precise measurement and finding the value En places the system in a state that has that energy. If a measurement on the state:

returns the energy, - 3.40 eV, the energy for the n = 2 states, then the atom is left in a state of the form by the measurement: immediately after the measurement. (All the n = 2 parts remain; the parts for other n’s are removed.) The act of making a precise measurement of a physical observable collapses the state to a state that is consistent with the measured value. (Removes the parts that are not consistent; preserves remainder with relative phases.)

After the collapse, the state will time develop as directed by the Hamiltonian, and the state may not be consistent with the measured eigenvalue after some time has passed. If the operator for the eigenvalue commutes with Hamiltonian, the expectation value for the quantity will be time independent.

time development of expectation values

for operators Q without explicit time dependence

Time independent operators that commute with the hamiltonian have time independent expectation values.

Example with non-zero commutator:

The time rate of change is momentum is the ‘force’. Ehrenfest’s Theorem: Quantum mechanical expectation values obey the corresponding classical equations of motion.


Every Hermitian operator has a complete set of eigenstates (Qm = qmm) which can be used as an basis set for representing a general solution. . It follows that:

. The value |ck|2 is the probability that the measurement returns the eigenvalue qk for the state k. A precise measurement of Q will return an eigenvalue and place the system in a state with that eigenvalue immediately after the measurement. The eigenvalues of the measurement operator are the only possible outcomes of a good measurement.


(20) Ionization and the Periodic Chart

According to L. Pauling, “the power of an atom in a molecule to attract electrons to itself.” Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. Electro-negativity increases from bottom to top of a column.



Electro-negativity increases from left to right across a row.

Atomic radius decreases → Ionization energy increases → Electronegativity increases →



The inner electrons screen the nuclear charge seen by the outer electrons. The apparent unscreened charge increases as you move across the table filling an electron shell. It is a maximum for group VIII and a minimum for the single electron outside a closed shell in group I. The effective unscreened charge increases as you move down the table to the higher Z atoms. Group I has one electron outside a closed shell and are the easiest to ionize (easiest at top after H; lowest screened charge). In group VII, the effective charge gets higher as you move down the table  most electron hungry (highest electron affinity). Electron affinity increases as you move down and across from I to VII. VIII is the closed shell nobles; they wish neither to give nor receive electrons. Outer valence shell has two S plus 6 P states.


Electron Shell Filling Sequence: Valence electrons appear rightmost

Z

Atom

Configuration




Comments

1

H

(1s)

2S½

Filling the 1s shell ; single active electron

2

He

(1s)2

1S0

Shell filled  spherical, zero angular momentum 1S0

3

Li

(2s)

2S½

Filling 2s; single active electron outside closed shell

4

Be

(2s)2

1S0

Sub-shell filled  spherical, zero angular momentum 1S0

5

B

(He)(2s)2(3p)1

2P½

Filling the 2p sub-shell

6

C

(He)(2s)2(3p)2

3P0

Maximize unpaired electrons to minimize exchange energy

7

N

(He)(2s)2(3p)3

4S3/2

Half filled sub-shell; max net spin J = |L – S|

8

O

(He)(2s)2(3p)4

3P2

Spin decreases as electrons added pairing previous ones

9

F

(He)(2s)2(3p)5

2P3/2

J = |L + S|

10

Ne

(He)(2s)2(3p)6

1S0

Sub-shell filled  spherical, zero angular momentum 1S0

11

Na

(Ne)(3s)1

2S½

single active electron outside closed shell

12

Mg

(Ne)(3s)2

1S0

Shell filled  spherical, zero angular momentum 1S0

13

Al

(Ne)(3s)2(3p)1

2P½

Filling the 2p sub-shell

14

Si

(Ne)(3s)2(3p)2

3P0

Maximize unpaired electrons to minimize exchange energy

15

P

(Ne)(3s)2(3p)3

4S3/2

Half filled sub-shell; max net spin J = |L – S|

16

S

(Ne)(3s)2(3p)4

3P2

Spin decreases as electrons added pairing previous ones

17

Cl

(Ne)(3s)2(3p)5

2P3/2

J = |L + S|

18

Ar

(Ne)(3s)2(3p)6

1S0

Sub-shell filled  spherical, zero angular momentum 1S0

19

K

(Ar)(4s)1

2S½

single active electron outside closed shell

20

Ca

(Ar)(4s)2

1S0

Shell filled  spherical, zero angular momentum 1S0

21

Sc

(Ar)(4s)2(3d)1

2D3/2

Filling inner sub-shell hidden inside the 4s electrons

22

Ti

(Ar)(4s)2(3d)2

3F2

The full set has similar chemical character.

23

V

(Ar)(4s)2(3d)3

4F3/2


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