Physical Chemistry II – Exam 1 SOLUTIONS

Chemistry 362Spring 2013Dr. Jean M. StandardFebruary 8, 2013Name _____KEY_____________________Physical Chemistry II – Exam 1 SOLUTIONSConstantValue in SI Units€h6.62607×10 –34 J s = h /2π1.05457×10 –34 J sc2.99793×10 8 m/se1.60217×10 –19 Cm e9.10938×10 –31 kgε 0 8.85419×10 –12 C 2 J –1 m –1R (Rydberg const.) 109737.3 cm –11.) (17 points) When light with a wavelength of 3000 Å shines on a metal surface, thekinetic energy of ejected electrons is 2.960×10 –19 J. When light with a wavelength of5400 Å shines on the same metal surface, the kinetic energy of ejected electrons is1.362×10 –21 J. From this information, determine the value of Planck's constant (inunits of J s). [1 Å = 10 –10 m]For the photoelectric effect, KE = hν − φ. From this equation, we can see that a plot withkinetic energy on the y-axis and frequency on the x-axis yields a straight line with the slopeequal to Planck's constant h. Therefore, one way to determine Planck's constant from the datagiven is to calculate the slope,€h = slope = ΔKEΔνThe kinetic energy values are given in the problem, but the frequencies must be calculated fromthe wavelengths.€.ν 1 = c λ 1= 2.99793 ×108 m/s⎛⎛( 3000 Å) 10−10 m⎞⎞⎜⎜ ⎟⎟⎝⎝ 1 Å ⎠⎠ν 1 = 9.993 ×10 14 s –1 .ν 2 = c λ 2= 2.99793 ×108 m/s⎛⎛( 5400 Å) 10−10 m⎞⎞⎜⎜ ⎟⎟⎝⎝ 1 Å ⎠⎠ν 2 = 5.552 ×10 14 s –1 .€€

1.) continued2These frequency values may now be substituted, along with values of the kinetic energy, intothe equation for the slope,h = ΔKEΔν=1.362 ×10 −21 J – 2.960 ×10 −19 J5.552 ×10 14 s –1 – 9.993 ×10 14 s –1h = 6.634 ×10 −34 J s .€

2.) (17 points) Consider two operators,ˆ A = d /dx andˆ B = x. For the functionf (x) = x e −ax , where a is a constant, evaluate the following quantities.3a.)A ˆ B ˆ f ( x)€€A ˆ B ˆ f ( x) = d dx x ( x ) e−ax= d (dx x 2 e −ax)= 2x e −ax − a x 2 e −axA ˆ B ˆ f ( x) = ( 2 − ax) x e −ax .€b.) B ˆ A ˆ f ( x)( )( )B ˆ A ˆ f ( x) = x d dx x e−axB ˆ A ˆ f x= x e −ax − a x e −ax( ) = ( 1− ax) x e −ax .€c.) Do the operatorsˆ A andˆ B commute?(a yes or no answer is sufficient)NO. The operators€ €ˆ A andˆ B do not commute in this case.€€

3.) (17 points) An electron in a one-dimensional box of width 6.0 Å undergoes atransition from an initial state with n=2 to some final state. The wavelength of lightabsorbed in this transition was determined to be 56.5 nm. Calculate the quantumnumber of the final state. [1 Å = 10 –10 m; 1 nm = 10 –9 m]4The particle in a box energies are given byE n = n2 h 28mL 2 .For a transition from n=2 to some final state n, the energy difference€ΔE = E n − E 2€= n2 h 28mL 2 − 22 h 28mL 2( )ΔE = n 2 − 2 2h 28mL 2 .ΔE isSolving for the unknown quantum number n yields,€n 2 − 2 2n 2= 8mL2 ΔEh 2= 4 + 8mL2 ΔEh 2 .€A photon with an energy corresponding toE photon = ΔE = hν. Since, € for light, λν = c , we can substitute ν = λ , and calculate thecenergy difference€ΔE ,€ΔE would have a frequency given by€ΔE = hc€λ( 6.62607 ×10 −34 Js) 2.99793 ×10 8 m/s=56.5 ×10 −9 mΔE = 3.5158 ×10 −18 J .( )€

3.) continued5Substituting this energy differenceΔE into the expression above and solving,n 2= 4 + 8mL2 ΔE€ h 2( )( 6.0 ×10 −10 m) 2 ( 3.5158 ×10 −18 J)( 6.62607 ×10 −34 Js) 2= 4 + 8 9.10938 ×10−31 kg= 4 + 21n 2 = 25 ,or n = 5 .Therefore, the final state in the transition is to the quantum number n=5.€

€⎛⎛4.) (17 points) Consider the function ψ(x) = x 1− x ⎞⎞⎜⎜ ⎟⎟ on the interval 0 ≤ x ≤ L, where L⎝⎝ L⎠⎠is a constant. The function is equal to zero outside this region, ψ(x) = 0 for x < 0 andx > L.€€a.) Other than normalization, what are the conditions€for an acceptable €wavefunction? Does the wavefunction given above meet these conditions?Explain.An acceptable wavefunction must satisfy the following conditions:(1) It must be continuous.(2) It must be single-valued.(3) It must go to 0 as x →±∞.A sketch of the wavefunction shows that it is continuous and single-valued. (Note thatthe sketch given is € for L=1, but the same functional shape would be observed for anyvalue of L.)0.360.2ψ(x)0.10.00.0 0.2 0.4 0.6 0.8 1.0xSince the function equals 0 for forgoes to 0 as x →±∞.x < 0 andx > L, it also satisfies condition (3), that it€The main concern for condition € (1) € is that the function be continuous at the boundaries(x=0 € and x=L), so that it matches the wavefunction outside this region ψ(x) = 0 for x < 0and x > L. From the formula for the function given above in the region 0 ≤ x ≤ L, we seethat it equals 0 at the boundaries,€ €⎛⎛ψ(0) = 0⋅ 1− 0 ⎞⎞⎜⎜ ⎟⎟ = 0 €⎝⎝ L⎠⎠⎛⎛ψ(L) = L⋅ 1− L ⎞⎞⎜⎜ ⎟⎟ = 0 .⎝⎝ L⎠⎠Therefore, the function is continuous.€

4.) continued7b.) Verify whether or not the wavefunction given above on the interval 0 ≤ x ≤ L is aneigenfunction of the one-dimensional kinetic energy operator, T ˆ . If it is aneigenfunction, give the eigenvalue.€The one-dimensional kinetic energy operator isT ˆ = − 2 d 22m dx 2 .⎛⎛To verify whether or not ψ(x) € = x 1− x ⎞⎞⎜⎜ ⎟⎟ is an eigenfunction of the kinetic energy⎝⎝ L⎠⎠operator, we operate T ˆ on it,€T ˆ ψ( x) = − 22m= − 22m= − 22m= − 22md 2 ⎧⎧dx 2 x ⎛⎛1 − x ⎞⎞ ⎫⎫⎨⎨ ⎜⎜ ⎟⎟ ⎬⎬⎩⎩ ⎝⎝ L⎠⎠⎭⎭⎛⎛d 2dx 2 x − x 2⎜⎜L⎝⎝d ⎛⎛dx 1 − 2x ⎞⎞⎜⎜ ⎟⎟⎝⎝ L ⎠⎠d ⎛⎛dx 0 − 2 ⎞⎞⎜⎜ ⎟⎟⎝⎝ L⎠⎠⎞⎞⎟⎟⎠⎠T ˆ ψ( x) = 2mL .This result does not correspond to a constant times the original function; therefore, thisfunction is NOT an eigenfunction of the kinetic energy operator.€

5.) (16 points) The idea of quantization is one of the key elements of quantum theory.Explain the concept of quantization. Then, using one example system, describe atleast one property of the system that exhibits quantization. Make sure to include inyour discussion an equation that illustrates quantization of the property of interest.8Quantization is the idea that a physical property may take only certain allowed (discrete) values.Such properties would be predicted by classical theories to be continuous; that is, the propertycould take any value.A variety of systems could be used to demonstrate quantization of a physical property. Oneexample that we discussed in class was the Bohr theory of the hydrogen atom. The propertythat was quantized was the angular momentum, L = mvr = n, where L is the angularmomentum, m is the mass of the electron, v is the electron velocity, is a constant, and n is aninteger (the quantum number). This result also led to quantization of the radius of the orbit andthe electron energy.€€The quantization of the energy of the electron also was demonstrated by experiments involvingatomic line spectra. In that case, quantized energy levels in the atoms led to only specificwavelengths of light observed in the spectrum. For hydrogen, the observed wavelengths are1related to the formulaλ = R ⎛⎛ 1 2n − 1 ⎞⎞⎜⎜2 ⎟⎟ , where λ is the wavelength, R is a constant, and n 1⎝⎝ 1 n 2 ⎠⎠and n 2 are integers which label the quantized energy levels of the transition.€€€Also in the case€of the atomic line spectrum of hydrogen, the quantized property was the energy,which was found to be of the formE n = −me48ε 2 0 n 2 2, where m is the electron mass, e is thehelectron charge, ε 0 is the permittivity of free space, h is a constant, and n is an integer that labelsthe quantized energy levels.€Another€system we discussed that exhibits quantization is the particle in a box. In this case, theenergy of the particle is quantized and given by E n = n2 h 28mL 2 , where m is the mass, L is thebox width, h is a constant, and n is an integer that labels the quantized energy levels.€

6.) (16 points)9a.) For the particle in a one-dimensional box of width L, make a sketch of the secondexcited state (n=3) wavefunction and the corresponding probability density. Howmany nodes does this wavefunction possess? Where are they located?Graphs of the second excited state wavefunction, ψ 3 ( x), and probability density, ψ 2 3 ( x), areshown below. (Your sketches should capture the key features of these functions, and inaddition should be properly labeled.)2.0€2.0€1.01.5ψ3(x)0.00.0 0.2 0.4 0.6 0.8 1.0ψ3 2 (x)1.0-1.00.5-2.0x/L0.00.0 0.2 0.4 0.6 0.8 1.0x/LFrom the graphs we see that the second excited state wavefunction possesses two nodes, atx=L/3 and x =2L/3.b.) What is the most probable position for finding the particle in the second excitedstate?The most probable position (or positions) corresponds to the location of the maximum inthe probability density. From the graph of probability density for the second excited state,ψ 2 3 ( x), given above, there are three equal maxima and therefore three equivalent mostprobable locations: x = L/6, L/2, and 5L/6.€c.) Determine by explicit calculation the probability of finding the particle in the leftone third of the box if the particle is in the second excited state.ψ 2 ( x) dx, if we want the probability of finding the particleSince the probability is given byin the left one third of the box (between x=0 and L/3), we must evaluate the integral,€Probability =L / 3∫0ψ * x( ) ψ( x) dx .€

6 c.) continued10Substituting the particle in a box wavefunction for the second excited state (n=3),2 ⎛⎛ 3π x ⎞⎞ψ 3 (x) = sin⎜⎜⎟⎟ , the probability integral becomesL ⎝⎝ L ⎠⎠€Probability = 2 LL / 3sin 2 ⎛⎛ 3π x ⎞⎞∫ ⎜⎜ ⎟⎟ dx .⎝⎝ L ⎠⎠0This integral can be evaluated using the list provided. From the list of integrals given, wefind the indefinite € integral:∫ sin 2 b x dx = x 2 − sin2b x4b.Replacing b with€3πL yields€Probability = 2 L⎡⎡ ⎛⎛sin 6π x ⎞⎞ ⎤⎤⎢⎢ ⎜⎜ ⎟⎟x⎥⎥⎢⎢2 − ⎝⎝ L ⎠⎠⎛⎛4⎜⎜3π ⎥⎥⎢⎢⎞⎞⎟⎟⎥⎥⎣⎣ ⎢⎢ ⎝⎝ L ⎠⎠ ⎦⎦x=L / 3⎥⎥ x=0.Finally, evaluating the expression at the limits leads to€Probability = 2 L= 2 L⎡⎡ ⎛⎛ L6 − L⎜⎜ sin 2π⎝⎝ 12π ( ) ⎞⎞ ⎛⎛⎢⎢⎟⎟ − ⎜⎜ 0 −⎣⎣⎠⎠ ⎝⎝⎡⎡ L⎣⎣⎢⎢6 − 0 − 0 + 0 ⎤⎤⎦⎦⎥⎥L12π sin( 0)⎞⎞ ⎤⎤⎟⎟ ⎥⎥⎠⎠ ⎦⎦Probability = 1 3 .[Note:€This result for the probability makes sense because the probability density for thesecond excited state is divided into three equal humps; therefore, the probability in the leftthird of the box has to be 1/3.]