Q 12-12-017NEETJEE MainMedium
The wavelength of the first line of the Lyman series of hydrogen is $1216$ Å. The wavelength of the first line of the Balmer series is about
Answer: (B) $6566$ Å
Lyman first line ($2 \to 1$): $\dfrac{1}{\lambda_L} = R\cdot\dfrac{3}{4}$. Balmer first line ($3 \to 2$): $\dfrac{1}{\lambda_B} = R\cdot\dfrac{5}{36}$.
$$\lambda_B = \lambda_L \times \frac{3/4}{5/36} = 1216 \times 5.4 \approx 6566\ \text{Å}$$
Solution by Sreeraj P, M.Sc Physics