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