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One of the hydrogen emission lines has a wavelength of 486.0 nm.
Identify the values for n-initial and n-final for the transition giving rise to this line. (Enter as two integers, separated by a comma: e.g. 2,1 or 3,2 etc.)

c:3.00E8 m/s
h:6.63E-34Js
R:2.18E-18J

Please could you explain how you got the nfinal and the n integer?

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Solution Summary

The solution describes how to obtain the n-final and n-initial values from a hydrogen emission spectrum with step-by-step calculations.

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because,
1/lambda = R*((1/n1^2) - (1/n2^2))
where, R = Rydberg constant = 1.09678 *10^7 /m
lambda = 486.0 nm = 486 *10^(-9) m
=> ...

Solution provided by:
Amrit Lal Ahuja, PhD (IP)
Education
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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