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Group Problems #22 - Solutions

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Group Problems #22 - Solutions Group Problems #22 - Solutions Friday, October 21 Problem 1 Ritz Combination Principle Show that the longest wavelength of the Balmer series and the longest two wavelengths of the Lyman series satisfy the Ritz Combination Principle. The wavelengths corresponding to transitions between two energy levels in hydrogen is given by: n2 λ = λlimit 2 2 ; (1) n − n0 where n0 = 1 for the Lyman series, n0 = 2 for the Balmer series, and λlimit = 91:1 nm for the Lyman series and λlimit = 364:5 nm for the Balmer series. Inspection of this equation shows that the longest wavelength corresponds to the transition between n = n0 and n = n0 + 1. Thus, for the Lyman series, the longest two wavelengths are λ12 and λ13. For the Balmer series, the longest wavelength is λ23. The Ritz combination principle states that certain pairs of observed frequencies from the hydrogen spectrum add together to give other observed frequencies. For this particular problem, we then have: ν12 + ν23 = ν13 =) h(ν12 + ν23) = hν13 (2) 1 1 1 =) hc + = hc (3) λ12 λ23 λ13 1 1 1 =) + = : (4) λ12 λ23 λ13 Inverting Eqn. 1 above gives: 2 2 2 1 1 n − n0 1 n0 = 2 = 1 − 2 : (5) λ λlimit n λlimit n To calculate 1/λ12 and 1/λ13, we use the Lyman series numbers and find 1/λ12 = −1 −1 3=(4 ∗ 91:1) = 0:00823 nm and 1/λ13 = 8=(9 ∗ 91:1) = 0:00976 nm . To compute 1/λ23, we use the Balmer series numbers and find 1/λ23 = 5=(9 ∗ 364:5) = 0:00152 −1 nm . Computing the left-hand-side of Eqn. 4 gives 1/λ12 + 1/λ23 = 0:00823 + −1 0:00152 = 0:00975 nm , which is equal to 1/λ13 within rounding error. 1 2 Problem 2 The Rydberg Constant When an electron in an atom transitions between orbitals with principle quantum numbers n1 and n2, the emitted (or absorbed) photon has an energy hν = jEn1 − En2 j: (6) (a) Use Eq. (6) and the expression for the allowed electron energies, En, in the Bohr model to show that for hydrogen the emitted (or absorbed) photon has a wavelength 3 2 3 2 2 64π 0~ c n1n2 λ = 4 2 2 (7) mee n1 − n2 2 2 1 n1n2 = 2 2 ; (8) R1 n1 − n2 where R1 is called the Rydberg constant. The expression for the allowed energies in hydrogen is given by: 2 2 1 e 4π0~ En = − 2 ; with a0 = 2 (9) n 8π0a0 me where n is an integer 1 or greater, 0 is the permittivity of free space, and a0 is the so-called Bohr radius with m the mass and e the charge of an electron. The energy difference in Eqn. 6 is then given by: 2 2 2 2 hc e 1 1 e n1 − n2 hν = = jEn1 − En2 j = 2 − 2 = 2 2 : (10) λ 8π0a0 n2 n1 8π0a0 n1 n2 Inverting this equation and substituting for a0 gives: 2 2 hc n1 n2 λ = 8π0a0 2 2 2 (11) e n1 − n2 2 2 2 hc · 8π0 · 4π0~ n1 n2 = 2 2 2 2 (12) e · me n1 − n2 3 2 3 2 2 64π 0~ c n1n2 = 4 2 2 ; (13) me n1 − n2 where we have used the identity h = 2π~. (b) What is the numerical value of 1=R1 (don't forget the units!)? You may want to use the fine structure constant, α in your calculation: e2 1 α = = : (14) 4π0~c 137 course name PS # 3 First of all, we see from Eq. 8 above that 1=R1 must have units of length. Comparing Eq. 8 and Eq. 13 gives: 3 2 3 2 2 2 2 1 64π 0~ c 16π 0~ c 4π~c = 4 = 4 2 (15) R1 me e mc 1 2hc 2 · 1240 eV nm = = (137)2 (16) α2 mc2 511 × 103 eV = 91:09 nm: (17) (c) Find the wavelengths of the transitions from n1 = 3 to n2 = 2 and from n1 = 4 to n2 = 2. To which series do these two transitions belong? Right away we see that the transition n1 = 3 ! n2 = 2 belongs to the Balmer series with n2 = n0 = 2. The transition n1 = 4 ! n2 = 2 also belongs to the Balmer series for the same reason. Using Eq. 8 and our value for 1=R1 gives: 9 · 4 λ23 = 91:09 = 656 nm (18) 9 − 4 16 · 4 λ24 = 91:09 = 486 nm: (19) 16 − 4 As a final note, we compare the phenomenological expression for the transition wavelengths, Eq. 1, and the expression derived from the Bohr model, Eq. 7. When n2 = 1, then Eq. 7 becomes: 2 n1 λ1;n1 = 91:09 2 2 ; (20) n1 − 1 where n1 is an integer larger than 1. This expression is precisely that for the transition wavelengths in the Lyman series, where λlimit = 91:09 nm and n0 = 1. Similarly, when n2 = 2, we have: 2 2 n1 · 4 n1 λ2;n1 = 91:09 2 2 = 364:4 2 2 ; (21) n1 − 2 n1 − 2 where n1 is now an integer larger than 2. This expression is exactly that for the transition wavelengths in the Balmer series, where λlimit = 364:4 nm and n0 = 2. course name PS #.
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