Question

wavelength of which balmer line for hydrogen atom is 4341A°
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Answer 1

Answer:

5

Explanation:

λ1 = R<lower H>[1/2² − 1/n²]

1/n² = 1/4 − 1/λ×R<lower H>

    = 1/4 − 1/4341 × 10^-18 × 109678

    = 0.04

n² = 1/0.04 = 25

n=5