CHM 3411 - Physical Chemistry 2
First Hour Exam
February 4, 2011
There are six problems on the exam. Do all of the problems. Show your work.
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NA = 6.022 x 1023me = 9.109 x 10-31 kg 1 eV = 1.602 x 10-19 J
c = 2.998 x 108 m/sk = 1.381 x 10-23 J/K1 cm-1 = 1.986 x 10-23 J
h = 6.626 x 10-34 J.s
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1. (16 points) Rigel is the brightest star in the constellation Orion. It is a bright blue color and can easily be seen in the winter in the night sky over Miami.
a) Close examination reveals that the peak intensity of light emitted by Rigel occurs at = 265. nm. Based on this observation estimate the value for the surface temperature of Rigel.
b) What is the speed and kinetic energy of an electron whose deBroglie wavelength is equal to 265. nm? Give your answer in units of m/s (for speed) and J (for energy).
2. (15 points) In a photoelectric effect experiment light of wavelength = 254. nm was used to illuminate a grounded piece of bismuth metal. The maximum speed of the electrons that were ejected was v = 4.8 x 105 m/s. Based on this information, find 0, the critical wavelength for bismuth. Give your answer in units of nm.
3. (16 points) For each of the following operators and functions, determine under what conditions (if any) the function is an eigenfunction of the operator. For those functions that are eigefunctions, find the corresponding eigenvalues.
a) Ô = d2/dx2 , f(x) = exp(ax) , where a is a positive constant
b) Ô = [ (1/x) d/dx ] , f(x) = exp(ax2) , where a is a positive constant
c) Ô = [ x2 – (d2/dx2) ] , f(x) = exp(-ax2) , where a is a positive constant
4. (16 points) Consider the following function as a possible solution to the one dimensional particle in a box problem discussed in class
(x) = Ax cos(bx)0 < x < L
(x) = 0x 0 or x L
where A and b are positive constants.
a) For what value(s) of b, if any, will (x) satisfy the boundary conditions that apply to the one dimensional particle in a box?
b) Is (x) an eigenfunction of the Hamiltonian operator for the particle in a box inside the box (0 < x < L)? Justify your answer.
5. (22 points) The normalized solution to the particle in a box for n = 4 is, for 0 < x < L
(x) = (2/L)1/2 sin(4x/L)
a) What value(s) of x correspond to the places where it is most likely to find a particle in the n = 4 state?
b) What is the probability of finding the particle in the region 0 < x < L/3?
c) What is <x2> for a particle in the n = 4 state?
6. (15 points) To a first approximation the pi electrons in a C=C double bond in an alkene can be modeled using the one dimensional particle in a box. The size of the “box” containing the pi electrons can be taken as equal to twice the bond length, or 0.270 nm. Assuming that both pi electrons are initially in the n = 1 energy level of the box, what is the longest wavelength of light that an alkene will absorb? Give your answer in units of nm.
Exam solutions.
1) a) If we assume the star can be approximated as an ideal blackbody, then
maxT = AW , where AW is the Wein’s law constant.
SoT = AW/max = (2.898 x 106 nm.K)/(265. nm) = 10900. K
b) dB = h/mv
Sov = h/mdB = (6.626 x 10-34 J.s)/(9.109 x 10-31 kg)(265. x 10-9 m) = 2745. m/s
and T = mv2/2 = (9.109 x 10-31 kg)(2745 m/s)2/2 = 3.43 x 10-24 J
2) For an electron with maximum kinetic energy, we may say
Ephoton = hc/ = + EK,max , where is the work function for the metal and EK,max is the maximum kinetic energy for the ejected electron.
So = Ephoton – EK,max = hc/ - mv2/2
= (6.626 x 10-34 J.s)(2.998 x 108 m/s) - (9.109 x 10-31 kg)(4.8 x 105 m/s)2
(254. x 10-9 m) 2
= 7.82 x 10-19 J – 1.05 x 10-19 J = 6.77 x 10-19 J
But = hc/0
So0 = hc/ = (6.626 x 10-34 J.s)(2.998 x 108 m/s)/(6.77 x 10-19 J)
= 2.93 x 10-7 m = 293. nm
3) a) d2/dx2 exp(ax) = d/dx aexp(ax) = a2 exp(ax)
An eigenfunction; eigenvalue is a2
b) [ (1/x) d/dx ] exp(ax2) = (1/x) (2ax) exp(ax2) = 2a exp(ax2)
An eigenfunction; eigenvalue is 2a
c) [ x2 – (d2/dx2) ] exp(-ax2)
= x2 exp(-ax2) – (d2/dx2) exp(-ax2)
= x2 exp(-ax2) – (d/dx) (-2ax) exp(-ax2)
= x2 exp(-ax2) – { -2a exp(-ax2) + 4a2x2 exp(-ax2) }
= 2a exp(-ax2) + x2 (1 – 4a2) exp(-ax2)
This will have the form Ô f(x) = (constant) f(x) if the second term drops out.
This will occur when (1 – 4a2) = 0, or a2 = ¼, and so a = ½.
So an eigenfunction for a = ½, with eigenvalue 2a = 1.
4) a) Boundary conditions are (0) = 0 and (L) = 0 (because the wavefunction outside the box is equal to zero, and the wavefunction must be continuous at x = 0 and x = L.
At x = 0A (0) cos(0) = A (0) (1) = 0, places no constraints on the value of b
At x = LA (L) cos(bL) = 0
Since A 0 and L 0, the only way this can happen is if cos(bL) = 0
So bL = /2, 3/2, 5/2, 7/2... = (2n+1)/2 , n = 0, 1, 2, 3, ...
and so b = /2L, 3/2L, 5/2L, 7/2L... = (2n+1)/2L , n = 0, 1, 2, 3, ...
b) To test if a function is an eigenfunction of Ĥ we can insert it into the eigenvalue equation
Ĥ A x cos(bx) = (-2/2m) d2/dx2 Ax cos(bx)
= (-2/2m) d/dx { A cos(bx) – Abx sin(bx) }
= (-2/2m) { - Ab sin(bx) – Ab sin(bx) – Ab2x cos(bx) }
= (2/2m) { 2Ab sin(bx) + Ab2x cos(bx) }
Since there is no positive values for A or b that will kill off the first term in the {} brackets, the function is not an eigenfunction of Ĥ.
5) a) The n = 4 wavefunction is (x) = (2/L)1/2 sin(4x/L). The maximum probability corresponds to the values of x within the box where sin(4x/L) = 1, or x = L/8, 3L/8, 5L/, and 7L/8.
b) P(0 < x < L/3) = 0L/3 { (2/L)1/2 sin(4x/L) }2 dx
= (2/L) 0L/3 { sin(4x/L) }2 dxUse integral # 296, with a = 4/L
= (2/L) { x/2 - (L/16) sin(8x/L) }0L/3
= (2/L) { (L/6) - (L/16) sin(8/3) }Since sin(8/3) = (3)1/2/2
= (1/3) - (31/2/16) 0.299
c) <x2> = 0L { (2/L)1/2 sin(4x/L) } x2 { (2/L)1/2 sin(4x/L) }dx
= (2/L) 0L x2 { sin(4x/L) }2 dxUse integral # 400, with a = 4/L
= (2/L) { (x3/6) – [ (x2L/16) - (L3/5123) ] sin(8x/L) – (xL2/642) cos(8x/L }0L
= (2/L) { (L3/6) – (L3/642) }
= L2 { (1/3) – (1/322) } 0.330 L2
6) The longest wavelength (lowest energy) transition will be n = 1 n = 2.
Since En = n2 E0, E0 = h2/8mL2
the energy for this transition will be E = Ephoton = hc/ = E2 – E1 = 4E0 – E0 = 3E0 = 3h2/8mL2
Solving for wavelength gives
= (8mL2c/3h)
= 8 (9.109 x 10-31 kg) (0.270 x 10-9 nm)2 (2.998 x 108 m/s)
3 (6.626 x 10-34 J.s)
= 8.0 x 10-8 m = 80. nm
Since the longest wavelength of light that ethene actually absorbs is 170 nm, this is not a terribly good model for the pi electrons.