Calculate the wavelength of the first line in lyman series of the hydrogen spectrum (R = 109677 cm-1) how to do this?

2 Answers
Aug 18, 2017

#121.6 \text{nm}#

Explanation:

#1/lambda = \text{R}(1/(n_1)^2 - 1/(n_2)^2) * \text{Z}^2#

where,

R = Rydbergs constant (Also written is #\text{R}_\text{H}#)
Z = atomic number

Since the question is asking for #1^(st)# line of Lyman series therefore

#n_1 = 1#
#n_2 = 2#

since the electron is de-exited from #1(\text{st})# exited state (i.e #\text{n} = 2#) to ground state (i.e #text{n} = 1#) for first line of Lyman series.

enter image source here

Therefore plugging in the values

#1/lambda = \text{R}(1/(1)^2 - 1/(2)^2) * 1^2#

Since the atomic number of Hydrogen is 1.

By doing the math, we get the wavelength as

#lambda = 4/3*912 dot \text{A}#

since #1/\text{R} = 912 dot \text{A}#

therefore

#lambda = 1216 dot \text{A}#
or

#lambda = 121.6 \text{nm}#

#λ = "121.569 nm"#

Explanation:

The first emission line in the Lyman series corresponds to the electron dropping from #n = 2# to #n = 1#.

Lyman 1
(Adapted from Tes)

The wavelength is given by the Rydberg formula

#color(blue)(bar(ul(|color(white)(a/a) 1/λ = -R(1/n_f^2 -1/n_i^2)color(white)(a/a)|)))" "#

where

#R =# the Rydberg constant (#"109 677 cm"^"-1"#) and
#n_i# and #n_f# are the initial and final energy levels

For a positive wavelength, we set the initial as #n = 1# and final as #n = 2# for an absorption instead.

#1/λ = -"109 677 cm"^"-1" × (1/2^2 -1/1^2)#

#= "109 677" × 10^7color(white)(l) "m"^"-1" (1/4-1/1)#

#= "109 677 cm"^"-1" × (1-4)/(4×1)#

#= -"109 677 cm"^"-1" × (-3/4) = "82 257.8 cm"^"-1"#

#λ = 1/("82 257.8 cm"^"-1") = "1.215 69" × 10^"-5"color(white)(l) "cm"#

#= "1.215 69" × 10^"-7"color(white)(l) "m"#

#= ul"121.569 nm"#