Solution for the Homework 4

Transcription

1 Solution for te Homework 4 Problem 6.5: In tis section we computed te single-particle translational partition function, tr, by summing over all definite-energy wavefunctions. An alternative approac, owever, is to sum over all possible position and momentum vectors, as we did in section 2.5. Because position and momentum are continuous variables, te sums are really integrals, and we need to slip in a vector of / 3 to get a unitless number tat actually counts te independent wave functions. Tus, we migt guess te formula tr 3 d 3 rd 3 pe Etr/kT, () were te single integral sign actually represents six integrals, tree over te position components (denoted d 3 r) and tree over te momentum components (denoted d 3 p). Te region of integration includes all momentum vectors, but only tose position vectors tat lie witin a box of volume. By evaluation te integrals explicitly, sow tat tis expression yields te same result for te translational partition function as tat obtained in te text. (Te only time tis formula would not be valid would be wen te box is so small tat we could not justify converting te sum in equation 6.78 to an integral.) Solution: From te Eq. 6.8, E tr p2 x 2m + p2 y 2m + p2 z 2m. (2) Substitute tis to te above equation and recall tat v Q (/ 2πmkT ) 3. Ten, tr 3 d 3 r d 3 p e Etr/kT (3) ( ) ( ) 3 d 3 r dp x e p2 x /2mkT dp y e p2 y /2mkT dp z e p2 z /2mkT (4) ( 2πmkT ) 3 3. (5) v Q Tis result is agree wit te Eq Problem 6.52: Consider an ideal gas of igly relativistic particles (suc as potons or fast-moving electrons), wose energy-momentum relation is E pc instead of E p 2 /2m. Assume tat tese particles live in a one-dimensional universe. By following te same logic as above, derive a formula for te single-particle partition function,, for one particle in tis gas.

2 Solution: Since te energy sould be a positive, let E tr p c. Consider te partition function wit te one-dimensional case and put te given energy. Ten, dr dp e Etr/kT (6) ( ) dr dp e p c/kt (7) 2L dp e pc/kt 2kT L L, (8) c l Q were l Q c/2kt. 0 Problem 7.2: In a real emoglobin molecule, te tendency of oxygen to bind to a eme site increases as te oter tree eme sites become occupied. To model tis effect in a simple way, imagine tat a emoglobin molecule as just two sites, eiter or bot of wic can be occupied. Tis system as four possible states (wit only oxygen present). Take te energy of te unoccupied state to be zero, te energies of te two singly occupied states to be -0.55e, and te energy of te doubly occupied state to be -.3e (so te cange in energy upon binding te second oxygen is -0.75e). As in te previous problem, calculate and plot te fraction of occupied sites as a function of te effective partial pressure of oxygen. Compare to te grap from te previous problem (for independent sites). Can you tink of wy tis beavior is preferable for te function of emoglobin? Solution: Tere are four states and te grand partition function of tis system is + 2e (0.55e +µ)/kt + e (.3e +2µ)/kT. (9) Ten te occupancy of tis system is n N/N N(s)P(s) 2 s ( e (0.55e +µ)/kt + e (.3e +2µ)/kT ) /, (0) and te cemical potential of ideal gas is µ kt ln ( int /Nv Q ). n P bar 2

3 Te internal partition function of diatomic molecule is rotational partition function and ten int kt/η were η e (tis is not a constant for te temperature and pressure, so I just approximate). Ten te occupancy vs. pressure is like below figure. Wen te pressure of oxygen is ig like in te lung, eme sites prefer to occupied by oxygen and te pressure of oxygen is lower tan before like in te cell, eme sites prefer to unoccupied. Tis process moves te oxygen. Problem 7.5: Consider a system consisting of a single impurity atom/ion in a semiconductor. Suppose tat te impurity atom as one extra electron compared to te neigboring atoms, as would a posporus atom occupying a lattice site in a silicon crystal. Te extra electron is ten easily removed, leaving beind a positively carged ion. Te ionized electron is called a conduction electron, because it is free to move troug te material; te impurity atom is called a donor, because it can donate a conduction electron. Tis system is analogous to te ydrogen atom considered in te previous two problems except tat te ionization energy is muc less, mainly due to te screening of te ionic carge by te dielectric beavior of te medium. (a) Write down a formula for te probability of a single donor atom being ionized. Do not neglect te fact tat te electron, if present, can ave two independent spin states. Express your formula in terms of te temperature, te ionization energy I, and te cemical potential of te gas of ionized electrons. Solution: Tere are tree states, one wit energy 0 is ionized, and te oters wit energy -I are not. In tis case, te grand partition function is So, te probability of a single donor atom being ionized is + 2e (I+µ)/kT. () P ion. (2) + 2e (I+µ)/kT (b) Assuming tat te conduction electrons beave like an ordinary ideal gas (wit two spin states per particle), write teir cemical potential in terms of te number of conduction electrons per unit volume, N c /. Solution: From te Eq. 7.0 wit int e 2, ( ) 2 µ kt ln N c v Q (c) Now assume tat every conduction electron comes from an ionized donor atom. In tis case te number of conduction electrons is equal to te number of donors tat are ionized. Use tis condition to derive a quadratic equation for N c in terms of te number of donor atoms (N d ), eliminating µ. Solve for N c using te quadratic formula. (Hint: It s elpful to introduce some abbreviations for dimensionless quantities. Try x N c /N d, t kt/i, and so on.) Solution: Te answer of (a) is exactly same as te ratio of conduction electrons and donor ions. And a cemical potential is given by te result of (b), N c N d + 2e (I+µ)/kT + e I/kT N cv Q (3) (4) 3

4 Tis is just quadratic equation, so we can easily get te result wit te condition tat te number of conduction electron sould be positive: [ N c 2v Q e I/kT + + 4v Q e I/kT N d. (5) (d) For posporus in silicon, te ionization energy is 0.044e. Suppose tat tere are 0 7 P atoms per cubic centimeter. Using tese numbers calculate and plot te fraction of ionized donors as a function of temperature. Discuss te results. Solution: Put te values and ten we can plot te N c /N d as a function of T. N C N d T Problem 7.6: Sow tat wen a system is in termal and diffusive equilibrium wit a reservoir, te average number of particles in te system is N kt, (6) were te partial derivative is taken at fixed temperature and volume. Sow also tat te mean square number of particles is N 2 (kt ) (7) Use tese results to sow tat te standard deviation of N is σ N kt ( N/), (8) in analogy wit Problem 6.8. Finally, apply tis formula to an ideal gas, to obtain a simple expression for σ N in terms of N. Discuss your result briefly. 4

5 Solution: In te previous section, we are done tis for energy. Similarly, starts from te grand partition function: s e (E(s) µn(s))/kt. (9) Ten, N s N 2 s [ e (E(s) µn(s))/kt N(s) kt [ N 2 e (E(s) µn(s))/kt (s) (kt )2 (20) 2 2 µ. (2) So, σ N (kt ) 2 2 ( kt 2 µ ) 2 (kt ) µ + (kt ) 2 kt N. (22) Let s consider te ideal gas wit te number of particle is N N. From te Eq. 6.93, we can calculate te σ N for ideal gas: N ( N ) N kt σ N N. (23) Tis means tat te deviation of particles is same wit te average of particles and tis case is called Poisson distribution. 5