### Nuclear & Particle Physics - Hydrogen Spectrum

In the spectrum of Hydrogen, what is the ratio of the longest wavelength in the Lyman series (n = 1) to the longest wavelength in the Balmer series (n = 2)?

A. 5/27
B. 1/3
C. 4/9
D. 3/2
E. 3
(GR9677 #09)

Solution:

Rydberg formula:

$\frac{1}{\lambda_{vac}}=R_H\bigg(\frac{1}{n^2_1}-\frac{1}{n^2_2}\bigg)$

λvac = the wavelength of the light emitted in vacuum
RH = Rydberg constant for Hydrogen
n1 and n2 are integers such that n$url=http://latex.codecogs.com/gif.latex?<&container=blogger&gadget=a&rewriteMime=image/*$ n2

For the longest wavelength take n2 = ∞

For Lyman-radiation (n2 = ∞ → n= 1):

1/λL = RH (1/1 − 0) = RH

For Balmer-radiation (n2 = ∞ → n1 = 2):

1/λB = RH (¼ − 0) = ¼RH

The Ratio:

λL/λB = (1/RH)(RH/4) = ¼ ≈ 5/27