Question

In hydrogen atom an electron jumps from n=3 to n=1 what is the wavelength associated with the transaction and also name the line and region

Answer 1

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Calculate the wavelength of the first line in lyman series of the hydrogen spectrum (R = 109677 cm-1) how to do this?
Chemistry Calculations with wavelength and frequency
2 Answers


Ernest Z.
and 1 other
Aug 18, 2017
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Answer:
#λ = "121.569 nm"#
Explanation:
The first emission line in the Lyman series corresponds to the electron dropping from #n = 2# to #n = 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"#
39

Jackie
Apr 6, 2018
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Answer:
#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.

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}#
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