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
Concept. Use $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ for both lines and take the ratio, so $R$ cancels.
Step 1. Lyman first line ($2 \to 1$). $\dfrac{1}{\lambda_L} = R\cdot\dfrac{3}{4}$.
Step 2. Balmer first line ($3 \to 2$). $\dfrac{1}{\lambda_B} = R\left(\dfrac14 - \dfrac19\right) = R\cdot\dfrac{5}{36}$.
Step 3. Ratio.
$$\lambda_B = \lambda_L \times \frac{3/4}{5/36} = 1216 \times 5.4 \approx 6566\ \text{Å}$$
Answer. $6566$ Å (Option B).
Solution by Sreeraj P, M.Sc Physics