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If $λ_{max}$ is 6563 Å, then wave length of second line for Balmer series will be |
$λ=\frac{16}{3R}$ $λ=\frac{36}{5R}$ $λ=\frac{4}{3R}$ None of the above |
$λ=\frac{16}{3R}$ |
$\text{Second line of Balmer series corresponds to transition from n= 2 to m=4}$ $ \frac{1}{\lambda} = R(\frac{1}{2^2} - \frac{1}{4^2}) = R(\frac{1}{4} - \frac{1}{16}) = \frac{3R}{16}$ $ \lambda = \frac{16}{3R}$ |
