JEE Main 2020 — Structure Of Atom Question with Solution
From: JEE Main 2020 (Online) 4th September Evening Slot
Question
The shortest wavelength of H atom in the
Lyman series is 1. The longest wavelength in
the Balmar series of He+ is :
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
For Shortest wavelength energy should be maximum.
For maximum energy transition must be form n = to n = 1.
For longest wavelength, E = minimum.
For Blamer series n = 3 to n = 2 will have E minimum for He+, Z = 2
So
For maximum energy transition must be form n = to n = 1.
For longest wavelength, E = minimum.
For Blamer series n = 3 to n = 2 will have E minimum for He+, Z = 2
So
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This is a previous-year question from JEE Main 2020, covering the Structure Of Atom chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.