Taking Rydberg's constant $R_H=1.097 \times 10^7~m$ first and second wavelength of Balmer series in hydrogen spectrum is
Answer & explanation
Correct answer: option 4
$\frac{1}{\lambda}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$.
For first wavelength, $n_1=2$, $n_2=3 \Rightarrow \lambda_1=6563$ Å.
For second wavelength, $n_1=2, n_2=4 \Rightarrow \lambda_2=4861 ~Å$