Sept. 12, 2007 Chem. 1410 Problem Set 1, Solutions (1) (4 points) Note: since ν= c / λ , then, given a small change in wavelength of δλ about a central value of λ , the corresponding frequency change is δν= (d υ / d λ ) δλ = -(c/ λ2 ) δλ . [If frequency increases, wavelength decreases: hence the – sign.] Thus, appealing to the equation in point (ii), i.e., Dλ()λλ∆ = D ν () υν ∆ , we can write 8π λ 2 ⋅ ∆=υD ( υ ) ∆ ν [1] λ 4 c v Note, either side of this equation gives the number of modes in the selected interval, and thus ∆υ and ∆λ are the magnitudes of the (small) frequency and wavelength intervals, respectively. Finally, Eq. 1 above implies: 8π 8 πν 2 Dυ (υ ) = = , QED λ 2c c 3 (2) (a) (2 points) In general, the average of a property A over a discrete probability distribution < >= is given by A∑ pj A j , where p j is the normalized probability to be in state j, and Aj is j the value of the property A in associated with state j. In the case of interest here the states are labeled by j=0,1,2,… ∞ and the normalized probability to be in state j is the Boltzmann factor ∞ −ν − ν = jhkT/B jhkT / B pj e/ ∑ e . Furthermore, the energy of state j (corresponding to j photons in an j=0 υ = υ electromagnetic mode of frequency ) is Ej jh . Thus: ∞ ∞ −ν − ν <E(υ ) >= hje ν ∑jh/ kB T / ∑ e jh / k B T . j=0 j = 0 1 (b) (i) (1 point) The infinite geometric series 1++x x 2 + ... = for x < 1 (otherwise the series 1− x υ = −υ < diverges). In the case of our series expression for D( ) , we identify xexp( hkT /B ) 1 , and thus: − ν − D(υ )= [1 − e h/ kB T ] 1 , QED (ii) (2 points) Differentiating the series for D(υ ) term by term w.r.t. υ : 1 ∞ − jh − ν ∂D/ ∂υ = ∑ e jh/ kB T j=0 kB T −k T and thus: N()ν=B ∂ D ()/ υ ∂ υ , QED. h iii)(1 point) Given the expression in (ii) and the explicit form for D(υ ) in (i), we can evaluate: − υ h/ kB T υ = e N( ) − υ (1− e h/ kB T ) 2 and hence: hv <>=E()ν hN νυ ()/() D υ = , QED. [ehv/ kB T − 1] 2 P1.17) The observed lines in the emission spectrum of atomic hydrogen are given by 1 1 ~(cm −1 ) R (cm −1 ) cm −1 ,n n . In the notation favored by spectroscopists, ν = H 2 − 2 > 1 n1 n ~ 1 E -1 ν = = and RH = 109,677cm . The Lyman, Balmer, and Paschen series refers to n1 = λ hc 1, 2, and 3, respectively, for emission from atomic hydrogen. What is the highest value of ν~ and E in each of these series? 1 The highest value for ν~ corresponds to → 0. Therefore, n ~ 1 −1 −1 -18 ν = RH cm = 109,667cm or Emax = 2.18×10 J for the Lyman series. 12 ~ 1 −1 −1 -19 ν = RH cm = 27419cm or Emax = 5.45×10 J for the Balmer series, and 22 ~ 1 −1 −1 -19 ν = RH cm = 12186cm or Emax = 2.42×10 J for the Paschen series. 32 P1.19) If an electron passes through an electrical potential difference of 1 V, it has an energy of 1 electron-volt. What potential difference must it pass through in order to have a wavelength of 0.100 nm? 2 1 1 h h 2 E m v 2 m = e = e × = 2 2 2 meλ 2meλ (6.626×10−34 J s) 2 1 eV = = 2.41×10−17 J × = 150.4 eV 2× 9.109×10−31 kg × (10−10 m) 2 1.602×10−19 J The electron must pass through an electrical potential of 150.4 V. .