PH300 Fall 2013 HW1 answer key 2. What is the series limit (that is, the smallest λ) for The series limit corresponds to a k value of ∞, which reduces the Rydberg equation to λ = n2/RH where RH is the Rydberg constant for hydrogen and has a value of 1.096776x107 m-1 a. The Lyman Series? n = 1 → λ = (1)2/ (1.096776 x107 m-1) = 91.18 nm b. The Balmer Series? n = 2 → λ = (2)2/ (1.096776 x107 m-1) = 364.7 nm c. The Brackett Series? n = 4 → λ = (4)2/ (1.096776 x107 m-1) = 1458.9 nm 3.Calculate the 4 largest wavelengths for the Brackett and Pfund series for Hydrogen. Using the Rydberg equation 1 1 1 λ= 2 − 2 RH n k −1 2a) four largest λ for Bracket series: n = 4 k = 5 → λ = {RH*(1/42 – 1/52)}-1 = 4.05 µm k = 6 → λ = {RH*(1/42 – 1/62)}-1 = 2.63 µm k = 7 → λ = {RH*(1/42 – 1/72)}-1 = 2.17 µm k = 8 → λ = {RH*(1/42 – 1/82)}-1 = 1.95 µm 2b) four largest λ for Pfund series: n = 5 k = 6 → λ = {RH*(1/52 – 1/62)}-1 = 7.46 µm k = 7 → λ = {RH*(1/52 – 1/72)}-1 = 4.65 µm k = 8 → λ = {RH*(1/52 – 1/82)}-1 = 3.74 µm k = 9 → λ = {RH*(1/52 – 1/92)}-1 = 3.29 µm 4. How much kinetic energy must an alpha particle have before its distance of closest approach to a gold nucleus is equal to the nuclear radius (7.0 x 10-15 m)? Assume initial kinetic energy ki, final kinetic energy kf=0, initial potential energy ui=0, and final potential energy uf. So we have uf-ui=ki-kf, thus ki=2e*79e/(4πЄr)=9*10^9*2*79*(1.6*10^-19)^2/(7*10^15) =5.2*10^-12J=32.5Mev