a b c d

a

$\text{Second line of Balmer series corresponds to transition from n= 2 to m=4}$ $ \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}$