Reason (R): In hydrogen spectrum Balmer series belongs to visible spectrum.
Solution:
Assertion (A) is true: Lyman series transitions end at \(n=1\) (higher energy, shorter \(lambda\)) while Balmer end at \(n=2\) (lower energy, longer \(\lambda\)).
Reason (R) is true: Balmer lines like \(H_\alpha\) and \(H_\beta\) are in the visible spectrum. However, (R) does not explain why Lyman has shorter wavelengths than Balmer, so it's not the correct explanation.
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